• Author / Uploaded
• amrdif_20066729

Citation preview

William A. Adkins, Mark G. Davidson

ORDINARY DIFFERENTIAL EQUATIONS Solution Manual August 15, 2009

Springer Berlin Heidelberg New York Hong Kong London Milan Paris Tokyo

1 Solutions

Section 1.1 1. The rate of change in the population 𝑃 (𝑡) is the derivative 𝑃 ′ (𝑡). The Malthusian Growth Law states that the rate of change in the population is proportional to 𝑃 (𝑡). Thus 𝑃 ′ (𝑡) = 𝑘𝑃 (𝑡), where 𝑘 is the proportionality constant. Without reference to the 𝑡 variable, the differential equation becomes 𝑃 ′ = 𝑘𝑃 2. a. This statement mathematically is 𝑏(𝑡) = 𝑏0 𝑃 (𝑡) where we have used 𝑏0 to represent the proportionality constant. b. This statement translates as 𝑑(𝑡) = 𝑑0 𝑃 2 (𝑡) where we have used 𝑑0 to represent the proportionality constant. c. The overall growth rate is 𝑃 ′ (𝑡). Thus the Logistic Growth Law is 𝑃 ′ (𝑡) = 𝑏(𝑡) − 𝑑(𝑡) = 𝑏0 𝑃 (𝑡) − 𝑑0 𝑃 2 (𝑡)

= (𝑏0 − 𝑑0 𝑃 (𝑡))𝑃 (𝑡).

3. Torricelli’s law states that the√change in height, ℎ′√ (𝑡) is proportional to the square root of the height, ℎ(𝑡). Thus ℎ′ (𝑡) = 𝜆 ℎ(𝑡), where 𝜆 is the proportionality constant. 4. The highest order derivative is 𝑦 ′ so the order is 1 and the standard form is 𝑦 ′ = 𝑡3 /𝑦 2 . 5. The highest order derivative is 𝑦 ′′ so the order is 2. The standard form is 𝑦 ′′ = 𝑡3 /𝑦 ′ . 6. The highest order derivative is 𝑦 ′ so the order is 1 and the standard form is 𝑦 ′ = (𝑒𝑡 − 𝑡𝑦)/𝑡2 .

4

1 Solutions

7. The highest order derivative is 𝑦 ′′ so the order is 2. The standard form is 𝑦 ′′ = (3𝑦 + 𝑡𝑦 ′ )/𝑡2 . 8. The highest order derivative is 𝑦 ′′ so the order is 2 and the standard form is 𝑦 ′′ = 𝑡2 − 3𝑦 ′ − 2𝑦. 9. The highest order derivative is 𝑦 (4) so the order is 4. Solving for 𝑦 (4) gives √ 3 (4) the standard form: 𝑦 = (1 − (𝑦 ′′′ )4 )/𝑡.

10. The highest order derivative is 𝑦 ′ so the order is 1 and the standard form is 𝑦 ′ = 𝑡𝑦 4 − 𝑡2 𝑦.

11. The highest order derivative is 𝑦 ′′′ so the order is 3. Solving for 𝑦 ′′′ gives the standard form: 𝑦 ′′′ = 2𝑦 ′′ − 3𝑦 ′ + 𝑦. 12. The following table summarizes the needed calculations: 𝑦 ′ (𝑡)

Function

2𝑦(𝑡)

𝑦1 (𝑡) = 0

𝑦1′ (𝑡) = 0

2𝑦1 (𝑡) = 0

𝑦2 (𝑡) = 𝑡2

𝑦2′ (𝑡) = 2𝑡

2𝑦2 (𝑡) = 2𝑡2

𝑦3 (𝑡) = 3𝑒2𝑡

𝑦3′ (𝑡) = 6𝑒2𝑡

2𝑦3 (𝑡) = 6𝑒2𝑡

𝑦4 (𝑡) = 2𝑒3𝑡

𝑦4′ (𝑡) = 6𝑒3𝑡

2𝑦4 (𝑡) = 4𝑒3𝑡

Thus 𝑦1 and 𝑦3 are the only solutions. 13. The following table summarizes the needed calculations: 𝑡𝑦 ′ (𝑡)

Function 𝑦1 (𝑡) = 0 𝑦2 (𝑡) = 3𝑡 𝑦3 (𝑡) = −5𝑡 𝑦4 (𝑡) = 𝑡

3

𝑡𝑦1′ (𝑡) 𝑡𝑦2′ (𝑡) 𝑡𝑦3′ (𝑡) 𝑡𝑦4′ (𝑡)

𝑦(𝑡)

=0

𝑦1 (𝑡) = 0

= 3𝑡

𝑦2 (𝑡) = 3𝑡

= −5𝑡 = 3𝑡

𝑦3 (𝑡) = −5𝑡

3

𝑦4 (𝑡) = 𝑡3

Thus 𝑦1 , 𝑦2 , and 𝑦3 are solutions. 14. We first write the differential equation in standard form: 𝑦 ′′ = −4𝑦. The following table summarizes the needed calculations: 𝑦 ′′ (𝑡)

Function 𝑦1 (𝑡) = 𝑒

2𝑡

𝑦2 (𝑡) = sin 2𝑡 𝑦3 (𝑡) = cos(2𝑡 − 1) 𝑦4 (𝑡) = 𝑡

2

𝑦1′′ (𝑡) 𝑦2′′ (𝑡) 𝑦3′′ (𝑡) 𝑦4′′ (𝑡)

= 4𝑒

2𝑡

−4𝑦(𝑡)

−4𝑦1 (𝑡) = −4𝑒2𝑡

= −4 sin 2𝑡

−4𝑦2 (𝑡) = −4 sin 2𝑡

=2

−4𝑦4 (𝑡) = −4𝑡2

− 4 cos(2𝑡 − 1) −4𝑦3 (𝑡) = −4 cos(2𝑡 − 1)

1 Solutions

5

Thus 𝑦2 and 𝑦3 are solutions. 15. The following table summarizes the needed calculations: 𝑦 ′ (𝑡)

Function 𝑦1 (𝑡) = 0 𝑦2 (𝑡) = 1 𝑦3 (𝑡) = 2 𝑦4 (𝑡) =

1 1−𝑒2𝑡

𝑦1′ (𝑡) 𝑦2′ (𝑡) 𝑦3′ (𝑡)

2𝑦(𝑡)(𝑦(𝑡) − 1)

=0

2𝑦1 (𝑡)(𝑦1 (𝑡) − 1) = 2 ⋅ 0 ⋅ (−1) = 0

=0

2𝑦2 (𝑡)(𝑦2 (𝑡) − 1) = 2 ⋅ 1 ⋅ 0 = 0

=0

𝑦4′ (𝑡) =

2𝑡

2𝑒 (1−𝑒2𝑡 )2

2𝑦3 (𝑡)(𝑦3 (𝑡) − 1) = 2 ⋅ 2 ⋅ 1 = 4 ( ) 1 1 2𝑦4 (𝑡)(𝑦4 (𝑡) − 1) = 2 1−𝑒 − 1 2𝑡 2𝑡 1−𝑒 1 = 2 1−𝑒 2𝑡

𝑒2𝑡 1−𝑒2𝑡

=

2𝑒2𝑡 (1−𝑒2𝑡 )2

Thus 𝑦1 , 𝑦2 , and 𝑦4 are solutions. 16. The following table summarizes the needed calculations: Function 𝑦1 (𝑡) = 1 𝑦2 (𝑡) = 𝑡 𝑦3 (𝑡) = ln 𝑡 √ 𝑦4 (𝑡) = 𝑡 − 4

2𝑦1 (𝑡)𝑦1′ (𝑡) 2𝑦2 (𝑡)𝑦2′ (𝑡) 2𝑦3 (𝑡)𝑦3′ (𝑡) 2𝑦4 (𝑡)𝑦4′ (𝑡)

2𝑦(𝑡)𝑦 ′ (𝑡)

1

=0

1

= 2𝑡 =

2 1𝑡

ln 𝑡 √ =2 𝑡−

1 𝑡 = 2 ln 𝑡 4 2√1𝑡−4

1 =1

1

Thus 𝑦4 is the only solution. 17. The following table summarizes the needed calculations: Function √ 𝑦1 (𝑡) = −𝑡 √ 𝑦2 (𝑡) = − 𝑒𝑡 − 𝑡 √ 𝑦3 (𝑡) = 𝑡 √ 𝑦4 (𝑡) = − −𝑡

2𝑦(𝑡)𝑦 ′ (𝑡) √ −1 = −1 2 −𝑡 √ 2 −𝑡 √ −(𝑒𝑡 − 1) −2 𝑒𝑡 − 𝑡 √ 𝑡 = 𝑒𝑡 − 1 2 𝑒 −𝑡 √ 1 2 𝑡 √ =1 2 𝑡 √ 1 = −1 2(− −𝑡) √ 2 −𝑡

𝑦2 + 𝑡 − 1 √ ( −𝑡)2 + 𝑡 − 1 = −1 √ (− 𝑒𝑡 − 𝑡)2 + 𝑡 − 1 = 𝑒𝑡 − 1 √ ( 𝑡)2 + 𝑡 − 1 = 2𝑡 − 1 √ (− −𝑡))2 + 𝑦 − 1 = −1

Thus 𝑦1 , 𝑦2 , and 𝑦4 are solutions. 18. The following table summarizes the needed calculations for the first three functions:

6

1 Solutions

Function

𝑦 ′ (𝑡)

𝑦1 (𝑡) = 𝑡

1

𝑦2 (𝑡) = 2𝑡

2

𝑦3 (𝑡) = 3𝑡

3

𝑦 2 (𝑡) − 4𝑦(𝑡)𝑡 + 6𝑡2 𝑡2 2 2 𝑡 − 4𝑡 + 6𝑡2 =3 𝑡2 2 2 2 4𝑡 − 8𝑡 + 6𝑡 =2 𝑡2 9𝑡2 − 12𝑡2 + 6𝑡2 =3 𝑡2

3𝑡 + 2𝑡2 𝑡(3 + 2𝑡) = the quotient rule and simplifying gives 1+𝑡 1+𝑡 2𝑡2 + 4𝑡 + 3 𝑦4′ (𝑡) = . On the other hand, (1 + 𝑡)2

For 𝑦4 (𝑡) =

𝑦42 (𝑡)

𝑡2 (3 + 2𝑡)2 4𝑡2 (3 + 2𝑡) − + 6𝑡2 2 − 4𝑦4 (𝑡)𝑡 + 6𝑡 (1 + 𝑡) (1 + 𝑡) = 𝑡2 𝑡2 (3 + 2𝑡)2 − 4(3 + 2𝑡)(1 + 𝑡) + 6(1 + 𝑡)2 = (1 + 𝑡)2 2 2𝑡 + 4𝑡 + 3 = . (1 + 𝑡)2 2

It follows that 𝑦2 , 𝑦3 , and 𝑦4 are solutions. 19. 𝑦 ′ (𝑡) = 3𝑐𝑒3𝑡 3𝑦 + 12 = 3(𝑐𝑒3𝑡 − 4) + 12 = 3𝑐𝑒3𝑡 − 12 + 12 = 3𝑐𝑒3𝑡 . Note that 𝑦(𝑡) is defined for all 𝑡 ∈ ℝ. 20. 𝑦 ′ (𝑡) = −𝑐𝑒−𝑡 + 3

−𝑦(𝑡) + 3𝑡 = −𝑐𝑒−𝑡 − 3𝑡 + 3 + 3𝑡 = −𝑐𝑒−𝑡 + 3. Note that 𝑦(𝑡) is defined for all 𝑡 ∈ ℝ. 21. 𝑐𝑒𝑡 (1 − 𝑐𝑒𝑡 )2 1 1 1 − (1 − 𝑐𝑒𝑡 ) 𝑐𝑒𝑡 𝑦 2 (𝑡) − 𝑦(𝑡) = − = = . (1 − 𝑐𝑒𝑡 )2 1 − 𝑐𝑒𝑡 (1 − 𝑐𝑒𝑡 )2 (1 − 𝑐𝑒𝑡 )2 𝑦 ′ (𝑡) =

If 𝑐 ≤ 0 then the denominator 1 − 𝑐𝑒𝑡 > 0 and 𝑦(𝑡) has domain ℝ. If 𝑐 > 0 then 1 − 𝑐𝑒𝑡 = 0 if 𝑡 = ln 1𝑐 = − ln 𝑐. Thus 𝑦(𝑡) is defined either on the interval (−∞, − ln 𝑐) or (− ln 𝑐, ∞).

1 Solutions

7

22. 2

𝑦 ′ (𝑡) = 𝑐𝑒𝑡 2𝑡 = 2𝑐𝑡𝑒𝑡

2

2

2𝑡𝑦(𝑡) = 2𝑡𝑐𝑒𝑡 . 23. 𝑦 ′ (𝑡) =

−𝑐𝑒𝑡 𝑐𝑒𝑡 − 1

−𝑒𝑦 − 1 = −𝑒− ln(𝑐𝑒

𝑡

−1)

−1=

−1 −𝑐𝑒𝑡 − 1 = . 𝑐𝑒𝑡 − 1 𝑐𝑒𝑡 − 1

24. We first calculate 𝑦 ′ (𝑡) = −𝑐(𝑡 + 1)−2 so (𝑡 + 1)𝑦 ′ (𝑡) + 𝑦(𝑡) = (𝑡 + 1)

−𝑐 𝑐 −𝑐 𝑐 + = + = 0. (𝑡 + 1)2 𝑡+1 𝑡+1 𝑡+1

Observe that 𝑦(𝑡) is not defined at 𝑡 = −1 so the two intervals where 𝑦 is defined are (−∞, −1) and (−1, ∞). 25. 𝑦 ′ (𝑡) = −(𝑐 − 𝑡)−2 (−1) = 𝑦 2 (𝑡) =

1 . (𝑐 − 𝑡)2

1 (𝑐 − 𝑡)2

The denominator of 𝑦(𝑡) is 0 when 𝑡 = 𝑐. Thus the two intervals where 𝑦(𝑡) is defined are (−∞, 𝑐) and (𝑐, ∞). 26. This is a differential equation we can solve by simple integration: We get 2 𝑦(𝑡) = 𝑡2 + 3𝑡 + 𝑐. 27. Integration gives 𝑦(𝑡) =

𝑒2𝑡 2

− 𝑡 + 𝑐.

28. Integration (by parts) gives 𝑦(𝑡) = −𝑡𝑒−𝑡 − 𝑒−𝑡 + 𝑐. 29. Observe that

𝑡+1 𝑡

= 1 + 1𝑡 . Integration gives 𝑦(𝑡) = 𝑡 + ln ∣𝑡∣ + 𝑐.

30. We integrate two times. First, 𝑦 ′ (𝑡) = 𝑡2 + 𝑡 + 𝑐1. Second, 𝑦(𝑡) = 𝑐1 𝑡 + 𝑐 2 .

𝑡3 3

2

+ 𝑡2 +

31. We integrate two times. First, 𝑦 ′ (𝑡) = −2 cos 3𝑡 + 𝑐1 . Second, 𝑦(𝑡) = −2 3 sin 3𝑡 + 𝑐1 𝑡 + 𝑐2 . 32. From Problem 19 the general solution is 𝑦(𝑡) = 𝑐𝑒3𝑡 − 4. At 𝑡 = 0 we get −2 = 𝑦(0) = 𝑐𝑒0 − 4 = 𝑐 − 4. It follows that 𝑐 = 2 and 𝑦(𝑡) = 2𝑒3𝑡 − 4.

8

1 Solutions

33. From Problem 20 the general solution is 𝑦(𝑡) = 𝑐𝑒−𝑡 + 3𝑡 − 3. At 𝑡 = 0 we get 0 = 𝑦(0) = 𝑐𝑒0 + 3(0) − 3 = 𝑐 − 3. It follows that 𝑐 = 3 and 𝑦(𝑡) = 3𝑒−𝑡 + 3𝑡 − 3. 34. From Problem 21 the general solution is 𝑦(𝑡) = 1/(1 − 𝑐𝑒𝑡 ). At 𝑡 = 0 we 1 . It follows that 𝑐 = −1 and 𝑦(𝑡) = 1/(1 + 𝑒𝑡 ). get 1/2 = 𝑦(0) = 1−𝑐 35. From Problem 24 the general solution is 𝑦(𝑡) = 𝑐(𝑡 + 1)−1 . At 𝑡 = 1 we get −9 = 𝑦(1) = 𝑐(1 + 1)−1 = 𝑐/2. It follows that 𝑐 = −18 and 𝑦(𝑡) = −18(𝑡 + 1)−1 . 36. From Problem 27 the general solution is 𝑦(𝑡) = 𝑒2𝑡 /2 − 𝑡 + 𝑐. Evaluation at 𝑡 = 0 gives 4 = 𝑒0 /2 − 0 + 𝑐 = 1/2 + 𝑐. Hence 𝑐 = 7/2 and 37. From Problem 28 the general solution is 𝑦(𝑡) = −𝑡𝑒−𝑡 −𝑒−𝑡 +𝑐. Evaluation at 𝑡 = 0 gives −1 = 𝑦(0) = −1 + 𝑐 so 𝑐 = 0. Hence 𝑦(𝑡) = −𝑡𝑒−𝑡 − 𝑒−𝑡 . 38. From Problem 31 the general solution is 𝑦(𝑡) = −2 3 sin 3𝑡 + 𝑐1 𝑡 + 𝑐2 and a 𝑦 ′ (𝑡) = −2 cos 3𝑡 + 𝑐1 . Evaluation at 𝑡 = 0 gives 1 = 𝑦(0) = 𝑐2 and 2 = 𝑦 ′ (0) = −2 + 𝑐1 . If follows that 𝑐1 = 4 and 𝑐2 = 1. Thus 𝑦(𝑡) = −2 3 sin 3𝑡 + 4𝑡. 39. Implicit differentiation with respect to 𝑡 gives 6𝑡 + 8𝑦𝑦 ′ = 0. 40. Implicit differentiation with respect to 𝑡 gives 2𝑦𝑦 ′ − 2𝑡 − 3𝑡2 = 0. 41. Differentiation gives 𝑦 ′ = 2𝑐𝑒2𝑡 + 1. However, from the given function we have 𝑐𝑒2𝑡 = 𝑦 − 𝑡. Substitution gives 𝑦 ′ = 2(𝑦 − 𝑡) + 1 = 2𝑦 − 2𝑡 + 1. 42. Differentiation gives 𝑦 ′ = 3𝑐𝑡2 + 2𝑡. However, from the given function 2 ′ we have 𝑐𝑡3 = 𝑦 − 𝑡2 and hence 𝑐𝑡2 = 𝑦−𝑡 𝑡 . Substitution gives 𝑦 = 2

3 𝑦−𝑡 + 2𝑡 = 𝑡

3𝑦 𝑡

− 𝑡.

1 Solutions

Section 1.2 𝑦′ = 𝑡

1. 5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

−4

−3

−2

−1

0 t

1

2

3

4

5

2

3

4

5

3

4

5

𝑦′ = 𝑦2

2. 5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

−4

−3

−2

−1

0 t

1

𝑦 ′ = 𝑦(𝑡 + 𝑡)

3. 5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

−4

−3

−2

−1

0 t

1

2

9

10

1 Solutions

4.

4 3 2 1 0 −1 −2 −3 −4

5.

−4

−3

−2

−1

0

1

2

3

4

−4

−3

−2

−1

0 t

1

2

3

4

5

−4

−3

−2

−1

0 t

1

2

3

4

5

5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

6.

5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

1 Solutions

7.

11

5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

8.

−4

−3

−2

−1

0 t

1

2

3

4

5

−4

−3

−2

−1

0 t

1

2

3

4

5

−4

−3

−2

−1

0 t

1

2

3

4

5

5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

9.

5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

10. We set 𝑦 2 = 0 and see that 𝑦 = 0 is the only constant (= equilibrium) solution. 11. We set 𝑦(𝑦 + 𝑡) = 0. We look for constant solutions to 𝑦(𝑦 + 𝑡) = 0, and we see that 𝑦 = 0 is the only constant (= equilibrium) solution.

12

1 Solutions

12. The equation 𝑦 − 𝑡 = 0 has no constant solution. Thus, there are no equilibrium solutions. 13. The equation 1 − 𝑦 2 = 0 has two constant solutions: 𝑦 = 1 and 𝑦 = −1 14. We substitute 𝑦 = 𝑎𝑡 + 𝑏 into 𝑦 ′ = 𝑦 − 𝑡 to get 𝑎 = (𝑎 − 1)𝑡 + 𝑏. Equality for all 𝑡 forces 𝑎 − 1 = 0 and 𝑎 = 𝑏. Thus 𝑎 = 1 and 𝑏 = 1 and the only linear solution is 𝑦 = 𝑡 + 1. 15. We substitute 𝑦 = 𝑎𝑡 + 𝑏 into 𝑦 ′ = cos(𝑡 + 𝑦) to get 𝑎 = cos((𝑎 + 1)𝑡 + 𝑏). Equality for all 𝑡 means that cos((𝑎 + 1)𝑡 + 𝑏) must be a constant function, which can occur only if the coefficient of 𝑡 is 0. This forces 𝑎 = −1 leaving us with the equation −1 = cos 𝑏. This implies 𝑏 = (2𝑛 + 1)𝜋, where 𝑛 is an integer. Hence 𝑦 = −𝑡 + (2𝑛 + 1)𝜋, 𝑛 ∈ ℤ is a family of linear solutions.

Section 1.3 1. separable; ℎ(𝑡) = 1 and 𝑔(𝑦) = 2𝑦(5 − 𝑦) 2. In standard form we get 𝑦 ′ = (1 − 𝑦)/𝑦. This is separable; ℎ(𝑡) = 1 and 𝑔(𝑦) = (1 − 𝑦)/𝑦. 1−2𝑡𝑦 as a 3. First write in standard form: 𝑦 ′ = 1−2𝑡𝑦 𝑡2 . We cannot write 𝑡2 product of a function of 𝑡 and a function of 𝑦. It is not separable.

4. In standard form we get 𝑦 ′ = 𝑦(𝑦 − 𝑡). We cannot write 𝑦(𝑦 − 𝑡) as a product of a function of 𝑡 and a function of 𝑦. It is not separable. 5. Write in standard form to get: 𝑦 ′ = (𝑦 − 2𝑦𝑡)/𝑦. Here we can write (𝑦 − 2𝑡𝑦)/𝑦 = 1 − 2𝑡. It is separable; ℎ(𝑡) = 1 − 2𝑡 and 𝑔(𝑦) = 1. 6. We can factor to get 𝑦 ′ = 𝑦 2 (𝑡 − 1) + 𝑡 − 1 = (𝑦 2 + 1)(𝑡 − 1). It is separable; ℎ(𝑡) = 𝑡 − 1 and 𝑔(𝑦) = 𝑦 2 + 1. −2𝑡𝑦 ′ 7. In standard form we get 𝑦 ′ = 𝑡2−2𝑡𝑦 +3𝑦 2 . We cannot write 𝑦 = 𝑡2 +3𝑦 2 as a product of a function of 𝑡 and a function of 𝑦. It is not separable

8. It is not separable as 𝑡2 + 𝑦 2 cannot be written as a product of a function of 𝑡 and a function of 𝑦. 9. In standard form we get: 𝑦 ′ = 𝑒−𝑡 (𝑦 3 − 𝑦) It is separable; ℎ(𝑡) = 𝑒−𝑡 and 𝑔(𝑦) = 𝑦 3 − 𝑦 10. The variables are already separated, so integrate both sides of the equation to get 𝑦 2 /2 = 𝑡2 /2 + 𝑐, which we can rewrite as 𝑦 2 − 𝑡2 = 𝑘 where 𝑘 = 2𝑐 ∈ ℝ is a constant. Since 𝑦(2) = −1, substitute 𝑡 = 2 and 𝑦 = −1 to get that 𝑘 = (−1)2 − 22 = −3. Thus the solution is given implicitly by

1 Solutions

13

√ the equation 𝑦 2 − 𝑡2 = −3 or we can solve explicitly to get 𝑦 = − 𝑡2 − 3, where the negative square root is used since 𝑦(2) = −1 < 0. 2

11. In standard form we get 𝑦 ′ = 1−𝑦 𝑡𝑦 . Clearly, 𝑦 = ±1 are equilibrium solutions. Separating the variables gives 𝑦 1 𝑑𝑦 = 𝑑𝑡. 2 1−𝑦 𝑡 Integrating both sides of this equation (using the substitution 𝑢 = 1 − 𝑦 2 , 𝑑𝑢 = −2𝑦 𝑑𝑦 for the integral on the left) gives 1 − ln ∣1 − 𝑦 2 ∣ = ln ∣𝑡∣ + 𝑐. 2 Multiplying by −2, taking the exponential of both sides, and removing the absolute values gives 1 − 𝑦 2 = 𝑘𝑡−2 where 𝑘 is a nonzero constant. However, when 𝑘 = 0 the equation becomes 1 − 𝑦 2 = 0 and hence 𝑦 = ±1. By considering an arbitrary constant (which we will call 𝑐), the implicit equation 𝑡2 (1 − 𝑦 2 ) = 𝑐 includes the two equilibrium solutions for 𝑐 = 0. 12. The variables are already separated, so integrate both sides to get 𝑦 4 /4 = 𝑡2 /2 + 𝑐, 𝑐 a real constant. This can be simplified to 𝑦 4 = 2𝑡2 + 𝑐. (where we replace 4𝑐 by 𝑐) We leave the answer in implicit form. 13. The variables are already separated, so integrate both sides to get 𝑦 5 /5 = 𝑡2 /2 + 2𝑡 + 𝑐, 𝑐 a real constant. Simplifying gives 𝑦 5 = 25 𝑡2 + 10𝑡 + 𝑐. We leave the answer in implicit form 14. There is an equilibrium solution 𝑦 = 0. Separating variables give 𝑦 −2 𝑦 ′ = 𝑡 and integrating gives −𝑦 −1 = 𝑡2 /2 + 𝑐. Thus 𝑦 = −2/(𝑡2 + 2𝑐), 𝑐 a real constant. This is equivalent to writing 𝑦 = −2/(𝑡2 + 𝑐), 𝑐 a real constant, since twice an arbitrary constant is still an arbitrary constant. 15. In standard form we get 𝑦 ′ = (1 − 𝑦) tan 𝑡 so 𝑦 = 1 is a solution. Sepa𝑑𝑦 rating variables gives 1−𝑦 = tan 𝑡 𝑑𝑡. The function tan 𝑡 is continuous on the interval (−𝜋/2, 𝜋/2) and so has an antiderivative. Integration gives − ln ∣1 − 𝑦∣ = − ln ∣cos 𝑡∣+𝑘1 . Multiplying by −1 and exponentiating gives ∣1 − 𝑦∣ = 𝑘2 ∣cos 𝑡∣ where 𝑘2 is a positive constant. Removing the absolute value signs gives 1−𝑦 = 𝑘3 cos 𝑡, with 𝑘3 ∕= 0. If we allow 𝑘3 = 0 we get the equilibrium solution 𝑦 = 1. Thus the solution can be written 𝑦 = 1−𝑐 cos 𝑡, 𝑐 any real constant. 16. An equilibrium solution is 𝑦 = 0. Separating variables gives 𝑦 −𝑛 𝑑𝑦 = 1−𝑛 𝑚+1 𝑡𝑚 𝑑𝑡 and integrating gives 𝑦1−𝑛 = 𝑡𝑚+1 + 𝑐, 𝑐 a real constant. Simplifying 1−𝑛 𝑚+1 gives 𝑦 1−𝑛 = 𝑚+1 𝑡 + 𝑐, and the equilibrium solution 𝑦 = 0.

14

1 Solutions

17. There are two equilibrium solutions; 𝑦 = ( 0 and 𝑦)= 4. Separating vari1 1 1 𝑑𝑦 = 𝑑𝑡. Integrating ables and using partial fractions gives 4 𝑦 + 4−𝑦 𝑦 𝑦 and simplifying gives ln 4−𝑦 = 4𝑡 + 𝑘1 which is equivalent to 4−𝑦 = 𝑐𝑒4𝑡 , 4𝑡

4𝑐𝑒 𝑐 a nonzero constant. Solving for 𝑦 gives 𝑦 = 1+𝑐𝑒 4𝑡 . When 𝑐 = 0 we get the equilibrium solution 𝑦 = 0. However, there is no 𝑐 which gives the other equilibrium solution 𝑦 = 4.

18. There are no equilibrium solutions. Separating variables gives 𝑦2𝑦+1 𝑑𝑦 = 𝑑𝑡 and integrating gives 21 ln(𝑦 2 + 1) = 𝑡 + 𝑘. Solving for 𝑦 2 gives 𝑦 2 = 𝑐𝑒2𝑡 − 1, where 𝑐 > 0. −1 19. Separating variables gives 𝑦2𝑑𝑦 𝑦 = 𝑡+𝑐. +1 = 𝑑𝑡 and integrating gives tan Thus 𝑦 = tan(𝑡 + 𝑐), 𝑐 a real constant. ( ) 𝑦2 20. Separating variables gives 𝑦 𝑑𝑦 = −1 𝑡 − 𝑡 𝑑𝑡 and integrating gives 2 = 2 − ln ∣𝑡∣ − 𝑡2 + 𝑐. Simplifying gives 𝑦 2 + 𝑡2 + ln 𝑡2 = 𝑐, 𝑐 a real constant. 1 21. In standard form we get 𝑦 ′ = −(𝑦+1) 𝑦−1 1+𝑡2 from which we see that 𝑦 = −1 equilibrium solution. Separating variables and simplifying gives ( is an ) 𝑑𝑡 2 2 𝑦+1 − 1 𝑑𝑦 = 𝑡2 +1 . Integrating and simplifying gives ln(𝑦 + 1) − 𝑦 =

tan−1 𝑡 + 𝑐.

22. Separating variables gives 2𝑦 𝑑𝑦 = 𝑒𝑡 𝑑𝑡 and integrating gives 𝑦 2 = 𝑒𝑡 + 𝑐, 𝑐 a constant. 23. The equilibrium solution is 𝑦 = 0. Separating variables gives 𝑦 −2 𝑑𝑦 = 𝑑𝑡 1 1−𝑡 . Integrating and simplifying gives 𝑦 = ln∣1−𝑡∣+𝑐 , 𝑐 real constant. 24. In standard form we get 𝑦 ′ = 𝑦(𝑦+1) from which we see 𝑦 = 0 and 𝑦 = −1 are equilibrium solutions. The equilibrium solution 𝑦(𝑡) = 0 satisfies the initial condition 𝑦(0) = 0 so 𝑦(𝑡) = 0 is the required solution. 25. 𝑦 = 0 is the only equilibrium solution. The equilibrium solution 𝑦(𝑡) = 0 satisfies the initial condition 𝑦(1) = 0 so 𝑦(𝑡) = 0 is the required solution. 𝑑𝑦 26. Rewriting we get 𝑦 ′ = 𝑑𝑥 = 𝑥+2 = 0 is 𝑥 𝑦 from which we see (that 𝑦 ) 𝑑𝑦 2 an equilibrium solution. Separating variables gives 𝑦 = 1 + 𝑥 𝑑𝑥 and integrating gives ln ∣𝑦∣ = 𝑥 + ln 𝑥2 + 𝑘, 𝑘 a constant. Solving for 𝑦 by taking the exponential of both sides gives 𝑦 = 𝑐𝑥2 𝑒𝑥 , and allowing 𝑐 = 0 gives the equilibrium solution. The initial condition gives 𝑒 = 𝑦(1) = 𝑐𝑒 so 𝑐 = 1. Thus 𝑦 = 𝑥2 𝑒𝑥 .

27. In standard form we get 𝑦 ′ = −2𝑡𝑦 so 𝑦 = 0 is a solution. Separating vari2 ables and integrating gives ln ∣𝑦∣ = −𝑡2 + 𝑘. Solving for 𝑦 gives 𝑦 = 𝑐𝑒−𝑡

1 Solutions

15

and allowing 𝑐 = 0 gives the equilibrium solution. The initial condition 2 implies 4 = 𝑦(0) = 𝑐𝑒0 = 𝑐. Thus 𝑦 = 4𝑒−𝑡 . 28. Since cot 𝑦 = 0 at 𝑦 = 𝜋2 + 𝑚𝜋 for all integers 𝑚 we have equilibrium lines at 𝑦 = 𝜋2 + 𝑚𝜋, none of which satisfy the initial condition 𝑦(1) = 𝜋4 . Separating variables gives tan 𝑦 𝑑𝑦 = 𝑑𝑡 𝑡 and integrating gives − ln ∣cos 𝑦∣ = ln 𝑡 + 𝑐. We can ( solve for 𝑐 here using the initial condition: we get 𝑐 = √ ) 𝜋 2 − ln cos 4 = − ln 2 . Solving for 𝑦 gives 𝑦 = cos−1 √12𝑡 𝑢 29. Separating variables gives 𝑑𝑦 𝑦 = 𝑢2 +1 𝑑𝑢√and integrating gives ln ∣𝑦∣ = √ ln 𝑢2 + 1 + 𝑘. Solving for 𝑦 gives 𝑦 √ = 𝑐 𝑢2 + 1, for 𝑐 ∕= 0. The initial condition gives 2 = 𝑦(0) = 𝑐. So 𝑦 = 2 𝑢2 + 1.

30. In standard form we get 𝑦 ′ = Separating variables gives

𝑑𝑦 𝑦

𝑡 𝑦 𝑡+2 (

so 𝑦 = 0 is an equilibrium solution. ) 2 = 1 − 𝑡+2 𝑑𝑡. Integrating we get ln ∣𝑦∣ = 𝑡

𝑒 𝑡 − 2 ln ∣𝑡 + 2∣ + 𝑘. Solving for 𝑦 we get 𝑦 = 𝑐 (𝑡+2) 2 , for 𝑐 ∕= 0. However, allowing 𝑐 = 0 gives the equilibrium solution.

31. We assume the decay model 𝑁 (𝑡) = 𝑁 (0)𝑒−𝜆𝑡 . If 𝑡 is the age of the bone then 𝑁 (𝑡) = 31 𝑁 (0). Thus 13 = 𝑒−𝜆𝑡 . Solving for 𝑡 gives 𝑡 = ln𝜆3 = 5730 ln 3 ≈ 9082 years ln 2 32. Let 𝑚 denote the number of Argon-40 atoms in the sample. Then 8𝑚 is the number of Potassium-40 atoms. Let 𝑡 be the age of the rock. Then 𝑡 years ago there were 𝑚 + 8𝑚 = 9𝑚 atoms of Potassium-40. Hence 𝑁 (0) = 9𝑚. On the other hand, 8𝑚 = 𝑁 (𝑡) = 𝑁 (0)𝑒−𝜆𝑡 = 9𝑚𝑒−𝜆𝑡 . This − ln 8 −𝜏 8 implies that 98 = 𝑒−𝜆𝑡 and hence 𝑡 = 𝜆 9 = ln 2 ln 9 ≈ 212 million years old. 33. We need only solve .3𝑁 (0) = 𝑁 (0)𝑒−𝜆𝑡 for 𝑡. We get 𝑡 = − ln𝜆.3 = ln .3 − 5.27 = 9.15 years. ln 2 34. The ambient temperature is 32∘ F, the temperature of the ice water. From Equation (12) we get 𝑇 (𝑡) = 32 + 𝑘𝑒𝑟𝑡 . At 𝑡 = 0 we get 70 = 32 + 𝑘, so 𝑘 = 38 and 𝑇 (𝑡) = 32 + 38𝑒𝑟𝑡 . After 30 minutes we have 1 55 = 𝑇 (30) = 32 + 38𝑒30𝑟 and solving for 𝑟 gives 𝑟 = 30 ln 23 38 . To find the 𝑟𝑡 time 𝑡 when 𝑇 (𝑡) = 45 we solve 45 = 32 + 38𝑒 , with 𝑟 as above. We get ln 13−ln 38 𝑡 = 30 ln 23−ln 38 ≈ 64 minutes. 35. The ambient temperature is 𝑇𝑎 = 70∘ . Equation (12) gives 𝑇 (𝑡) = 70 + 𝑘𝑒𝑟𝑡 for the temperature of the coffee at time 𝑡. Since the initial temperature of the coffee is 𝑇 (0) = 180 we get 180 = 𝑇 (0) = 70 + 𝑘. Thus 𝑘 = 110. The constant 𝑟 is determined from the temperature at a 7 second time: 140 = 𝑇 (3) = 70 + 110𝑒3𝑟 so 𝑟 = 13 ln 11 ≈ −.1507. Thus

16

1 Solutions

𝑇 (𝑡) = 70 + 110𝑒𝑟𝑡, with 𝑟 as calculated. The temperature requested is ( 7 ) 53 𝑇 (5) = 70 + 110 11 ≈ 121.8∘.

36. The ambient temperature is 𝑇𝑎 = 65∘ . Equation (12) gives 𝑇 (𝑡) = 65 + 𝑘𝑒𝑟𝑡 for the temperature at time 𝑡. Since the initial temperature of the thermometer is 𝑇 (0) = 90 we get 90 = 𝑇 (0) = 65 + 𝑘. Thus 𝑘 = 25. The constant 𝑟 is determined from the temperature at a second time: 85 = 𝑇 (2) = 65 + 25𝑒2𝑟 so 𝑟 = 12 ln 45 . Thus 𝑇 (𝑡) = 65 + 25𝑒𝑟𝑡 , with 𝑟 = 12 ln 54 . To answer the first question we solve the equation 75 = 𝑇 (𝑡) = 65 + 25𝑒 𝑟𝑡 ln 2−ln 5 for 𝑡. We get 𝑡 = 2 ln ≈ 8.2 minutes. The temperature at 𝑡 = 20 is ( 4 )104−ln 5 ∘ 𝑇 (20) = 65 + 25 5 ≈ 67.7 .

37. The ambient temperature is 𝑇𝑎 = 70∘ . Equation (12) gives 𝑇 (𝑡) = 70 + 𝑘𝑒𝑟𝑡 for the temperature of the soda at time 𝑡. Since the initial temperature of the soda is 𝑇 (0) = 40 we get 40 = 𝑇 (0) = 70 + 𝑘. Thus 𝑘 = −30. The constant 𝑟 is determined from the temperature at a second time: 60 = 𝑇 (2) = 70 − 30𝑒2𝑟 so 𝑟 = 12 ln 13 . Thus 𝑇 (𝑡) = 70 − 30𝑒𝑟𝑡 , with 𝑟 = 12 ln 31 . 1 1 30 The temperature at 𝑡 = 1 is 𝑇 (1) = 70 − 30𝑒 2 ln 3 = 70 − √ ≈ 52.7∘ . 3 38. The ambient temperature is 𝑇𝑎 = 70∘ . Equation (12) gives 𝑇 (𝑡) = 70 + 𝑘𝑒𝑟𝑡 for the temperature of the coffee at time 𝑡. We are asked to determine the initial temperature of the coffee so 𝑇 (0) is unknown. However, we have the equations 150 = 𝑇 (5) = 70 + 𝑘𝑒5𝑟 142 = 𝑇 (6) = 70 + 𝑘𝑒6𝑟 or 80 = 𝑘𝑒5𝑟 72 = 𝑘𝑒4𝑟 . 𝑟 Dividing the second equation by the first gives 72 80 = 𝑒 so 𝑟 = ln 0.9. −5𝑟 From the first equation we get 𝑘 = 80𝑒 ≈ 135.5. We now calculate 𝑇 (0) = 70 + 𝑘 ≈ 205.5∘

39. The ambient temperature is 𝑇𝑎 = 40∘ . Equation (12) gives 𝑇 (𝑡) = 40 + 𝑘𝑒𝑟𝑡 for the temperature of the beer at time 𝑡. Since the initial temperature of the beer is 𝑇 (0) = 80 we get 80 = 𝑇 (0) = 40 + 𝑘. Thus 𝑘 = 40. The constant 𝑟 is determined from the temperature at a second time: 60 = 𝑇 (1) = 40 + 40𝑒𝑟 so 𝑟 = − ln 2. Thus 𝑇 (𝑡) = 40 + 40𝑒𝑟𝑡 , with 𝑟 = − ln 2. We now solve the equation 50 = 𝑇 (𝑡) = 40 + 40𝑒𝑟𝑡 for 𝑡 and ln 4 get 𝑡 = − − ln 2 = 2. She should therefore put the beer in the refrigerator at 2 p.m. 40. Let us start time 𝑡 = 0 at 1980. Then 𝑃 (0) = 290. The Malthusian growth model gives 𝑃 (𝑡) = 290𝑒𝑟𝑡. At 𝑡 = 10 (1990) we have 370 = 290𝑒10𝑟

1 Solutions

17

1 30𝑟 ln 37 = and hence 𝑟 = 10 29 . At 𝑡 = 30 (2010) we have 𝑃 (30) = 290𝑒 ( 37 )3 290 29 ≈ 602.

41. The initial population is 40 = 𝑃 (0). Since the population doubles in 3 hours we have 𝑃 (3) = 80 or 80 = 40𝑒3𝑟 . Hence 𝑟 = ln32 . Now we can compute the population after 30 hours: 𝑃 (30) = 40𝑒30𝑟 = 40(210 ) = 40, 960. 42. We have 3𝑃 (0) = 𝑃 (5) = 𝑃 (0)𝑒3𝑟 . So 𝑟 = ln53 . Now we solve the equation 2𝑃 (0) = 𝑃 (𝑡) = 𝑃 (0)𝑒𝑟𝑡 for 𝑡. We get 𝑡 = ln𝑟2 = 5lnln32 ≈ 3.15 years.

43. In the logistic growth equation 𝑚 = 800 and 𝑃 (0) = 290. Thus 𝑃 (𝑡) = 800⋅290 800⋅290 290+510𝑒−𝑟𝑡 . To determine 𝑟 we use 𝑃 (10) = 370 to get 370 = 290+510𝑒−10𝑟 . 1 ln 1887 A simple calculation give 𝑟 = 10 1247 . Now the population in 2010 is 800⋅290 𝑃 (30) = ≈ 530 1247 3 290+510( 1887 ) 44. In the logistics equation 𝑚 = 5000 and 𝑃0 = 2000. Thus 𝑃 (𝑡) = 10,000 10,000 10,000,000 2,000+3,000𝑒−𝑟𝑡 = 2+3𝑒−𝑟𝑡 . Since 𝑃 (2) = 3000 we get 3000 = 2+3𝑒−𝑟𝑡 . Solv10,000 10,000 ing this equation for 𝑟 gives 𝑟 = ln 32 . Now 𝑃 (4) = 2+3𝑒 4 ≈ −4𝑟 = 2+3( 23 ) 3857 45. Let 𝑥 = 𝑒−𝑟𝑡0 . Then 𝑥2 = 𝑒−2𝑟𝑡 . The equation 𝑃 (𝑡0 ) = 𝑃1 implies that 𝑃0 (𝑚−𝑃1 ) 0 (𝑚−𝑃2 ) . The equation 𝑃 (2𝑡0 ) = 𝑃2 implies 𝑥2 = 𝑃 𝑥= 𝑃 𝑃2 (𝑚−𝑃0 ) . These 1 (𝑚−𝑃0 ) 𝑃02 (𝑚−𝑃1 )2 0 (𝑚−𝑃2 ) = 𝑃 𝑃2 (𝑚−𝑃0 ) . Cross multiplying and 𝑃12 (𝑚−𝑃0 )2 (𝑃0 𝑃2 −𝑃12 )𝑚+(𝑃12 𝑃0 +𝑃12 𝑃2 −2𝑃0 𝑃1 𝑃2 ) = 0. Solving −𝑟𝑡0

equation together imply

simplifying leads to for 𝑚 gives the result. Now replace the formula for 𝑚 into 𝑒 =𝑥= 𝑃0 (𝑚−𝑃1 ) 𝑃0 𝑃2 −𝑃1 −𝑟𝑡0 . Simplifying gives 𝑒 = . The formula for 𝑟 follows 𝑃1 (𝑚−𝑃0 ) 𝑃2 𝑃1 −𝑃0 after taking the natural log of both sides. 46. We have 𝑃 (0) = 𝑃0 = 400, 𝑃 (3) = 𝑃1 = 700, and 𝑃 (6) = 𝑃2 = 1000. Us= ing the result of the previous problem we get 𝑚 = 700(700(400+1000)−2⋅400⋅1000) (700)2 −400⋅1000 1, 400

Section 1.4 1. This equation is already in standard form with 𝑝(𝑡) = 3. An antiderivative ∫ of 𝑝(𝑡) is 𝑃 (𝑡) = 3 𝑑𝑡 = 3𝑡 so the integrating factor is 𝜇(𝑡) = 𝑒3𝑡 . If we multiply the differential equation 𝑦 ′ + 3𝑦 = 𝑒𝑡 by 𝜇(𝑡), we get the equation 𝑒3𝑡 𝑦 ′ + 3𝑒3𝑡 𝑦 = 𝑒4𝑡 , and the left hand side of this equation is a perfect derivative, namely, (𝑒3𝑡 𝑦)′ . Thus, (𝑒3𝑡 𝑦)′ = 𝑒4𝑡 . Now take antiderivatives of both sides and multiply by 𝑒−3𝑡 . This gives

18

1 Solutions

𝑦=

1 𝑡 𝑒 + 𝑐𝑒−3𝑡 4

for the general solution of the equation. To find the constant 𝑐 to satisfy the initial condition 𝑦(0) = −2, substitute 𝑡 = 0 into the general solution to get −2 = 𝑦(0) = 41 + 𝑐. Hence 𝑐 = − 49 , and the solution of the initial value problem is 1 9 𝑦 = 𝑒𝑡 − 𝑒−3𝑡 . 4 4 2. Divide by cos 𝑡 to put the equation in the standard form 𝑦 ′ + (tan 𝑡)𝑦 = sec 𝑡. In this case 𝑝(𝑡) = tan 𝑡, an antiderivative is 𝑃 (𝑡) = ln(sec 𝑡), and the integrating factor is 𝜇(𝑡) = sec 𝑡. (We do not need ∣ sec 𝑡∣ since we are working near 𝑡 = 0 where sec 𝑡 > 0.) Now multiply by the integrating factor to get (sec 𝑡)𝑦 ′ + (sec 𝑡 tan 𝑡)𝑦 = sec2 𝑡, the left hand side of which is a perfect derivative. Thus ((sec 𝑡)𝑦)′ = sec2 𝑡 and taking antiderivatives of both sides gives (sec 𝑡)𝑦 = tan 𝑡 + 𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by 1/ sec 𝑡 = cos 𝑡 to get 𝑦 = sin 𝑡+𝑐 cos 𝑡 for the general solution. Letting 𝑡 = 0 gives 5 = 𝑦(0) = sin 0 + 𝑐 cos 0 = 𝑐 so 𝑐 = 5 and 𝑦 = sin 𝑡 + 5 cos 𝑡. 3. This equation is already in standard form. In this case 𝑝(𝑡) = −2, an antiderivative is 𝑃 (𝑡) = −2𝑡, and the integrating factor is 𝜇(𝑡) = 𝑒−2𝑡 . Now multiply by the integrating factor to get 𝑒−2𝑡 𝑦 ′ − 2𝑒−2𝑡 𝑦 = 1, the left hand side of which is a perfect derivative ((𝑒−2𝑡 )𝑦)′ . Thus ((𝑒−2𝑡 )𝑦)′ = 1 and taking antiderivatives of both sides gives (𝑒−2𝑡 )𝑦 = 𝑡 + 𝑐, where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑒2𝑡 to get 𝑦 = 𝑡𝑒2𝑡 + 𝑐𝑒2𝑡 for the general solution. Letting 𝑡 = 0 gives 4 = 𝑦(0) = 𝑐 so 𝑦 = 𝑡𝑒2𝑡 + 4𝑒2𝑡 . 4. Divide by 𝑡 to put the equation in the standard form 𝑒𝑡 1 𝑦′ + 𝑦 = 𝑡 𝑡 In this case 𝑝(𝑡) = 1/𝑡, an antiderivative is 𝑃 (𝑡) = ln 𝑡, and the integrating factor is 𝜇(𝑡) = 𝑡. Now multiply the standard form equation by the integrating factor to get 𝑡𝑦 ′ + 𝑦 = 𝑒𝑡 , the left hand side of which is a perfect

1 Solutions

19

derivative (𝑡𝑦)′ . (Note that this is just the original left hand side of the equation. Thus if we had recognized that the left hand side was already a perfect derivative, the preliminary steps could have been skipped for this problem, and we could have proceeded directly to the next step.) Thus the equation can be written as (𝑡𝑦)′ = 𝑒𝑡 and taking antiderivatives of both sides gives 𝑡𝑦 = 𝑒𝑡 + 𝑐 where 𝑐 ∈ ℝ is a constant. Now divide by 𝑡 to get 𝑒𝑡 𝑐 𝑦= + 𝑡 𝑡 for the general solution. 5. The general solution from Problem 4 is 𝑦 = 𝑡 0 = 𝑒 + 𝑐. So 𝑐 = −𝑒 and 𝑦 = 𝑒𝑡 − 𝑒𝑡 .

𝑒𝑡 𝑡

+ 𝑐𝑡 . Now let 𝑡 = 1 to get

6. Divide by 𝑡 to put the equation in the standard form 𝑦′ +

𝑚 𝑦 = ln 𝑡. 𝑡

𝑚 In this case 𝑝(𝑡) = 𝑚 𝑡 , an antiderivative is 𝑃 (𝑡) = 𝑚 ln 𝑡 = ln 𝑡 , and the 𝑚 integrating factor is 𝜇(𝑡) = 𝑡 . Now multiply the standard form equation by the integrating factor to get 𝑡𝑚 𝑦 ′ + 𝑚𝑡𝑚−1 𝑦 = 𝑡𝑚 ln 𝑡, the left hand side of which is a perfect derivative ((𝑡𝑚 )𝑦)′ . Thus ((𝑡𝑚 )𝑦)′ = 𝑡𝑚 ln 𝑡. To integrate 𝑡𝑚 ln 𝑡 we consider the cases 𝑚 = −1 and 𝑚 ∕= −1 separately. ∫ 2 Case 𝑚 = −1: A simple substitution gives 𝑡−1 ln 𝑡 𝑑𝑡 = (ln2𝑡) + 𝑐.

Hence, 𝑡−1 𝑦 =

Case 𝑚 ∕= −1: 𝑡𝑚+1 (𝑚+1)2

(ln 𝑡)2 2

+ 𝑐 and so 𝑦 =

𝑡(ln 𝑡)2 2

+ 𝑐𝑡

Use integration by parts to get

+ 𝑐. Then 𝑦 =

𝑡 ln 𝑡 𝑚+1

−

𝑡 (𝑚+1)2

+

𝑐

𝑡𝑚 .

∫

𝑡𝑚 ln 𝑡 𝑑𝑡 =

𝑡𝑚+1 ln 𝑡 𝑚+1

−

7. We first put the equation in standard form and get 1 𝑦 ′ + 𝑦 = cos(𝑡2 ). 𝑡 In this case 𝑝(𝑡) = 1𝑡 , an antiderivative is 𝑃 (𝑡) = ln 𝑡, and the integrating factor is 𝜇(𝑡) = 𝑡. Now multiply by the integrating factor to get 𝑡𝑦 ′ + 𝑦 = 𝑡 cos(𝑡2 ), the left hand side of which is a perfect derivative (𝑡𝑦)′ . Thus (𝑡𝑦)′ = 𝑡 cos(𝑡2 ) and taking antiderivatives of both sides gives 𝑡𝑦 = 21 sin(𝑡2 ) + 𝑐 where 𝑐 ∈ ℝ is a constant. Now divide by 𝑡 to get 𝑦 = general solution.

sin(𝑡2 ) 2𝑡

∫

+ 𝑐𝑡 . for the

8. In this case 𝑝(𝑡) = 2 and the integrating factor is 𝑒 2 𝑑𝑡 = 𝑒2𝑡 . Now multiply to get 𝑒2𝑡 𝑦 ′ + 2𝑒2𝑡 𝑦 = 𝑒2𝑡 sin 𝑡, which simplifies to (𝑒2𝑡 𝑦)′ =

new: please check

20

1 Solutions

1 𝑒2𝑡 sin 𝑡. Now integrate both sides to get 𝑒2𝑡 𝑦 = (− cos 𝑡 + 2 sin 𝑡)𝑒2𝑡 + 𝑐, 5 ∫ where we computed 𝑒2𝑡 sin 𝑡 by parts two times. Dividing by 𝑒2𝑡 gives 1 𝑦 = (2 sin 𝑡 − cos 𝑡) + 𝑐𝑒−2𝑡 . 5 ∫

9. In this case 𝑝(𝑡) = −3 and the integrating factor is 𝑒 −3 𝑑𝑡 = 𝑒−3𝑡 . Now multiply to get 𝑒−3𝑡 𝑦 ′ + 2𝑒−3𝑡 𝑦 = 25𝑒−3𝑡 cos 4𝑡, which simplifies to (𝑒−3𝑡 𝑦)′ = 25𝑒−3𝑡 cos 4𝑡. Now integrate both∫ sides to get 𝑒−3𝑡 𝑦 = (4 sin 4𝑡 − 3 cos 4𝑡)𝑒−3𝑡 + 𝑐, where we computed 25𝑒−3𝑡 cos 4𝑡 by parts twice. Dividing by 𝑒−3𝑡 gives 𝑦 = 4 sin 4𝑡 − 3 cos 4𝑡 + 𝑐𝑒3𝑡 . 10. In standard form this equation becomes 𝑦′ −

2 1 𝑦= . 𝑡(𝑡 + 1) 𝑡(𝑡 + 1)

−1 1 Using partial fractions we get 𝑝(𝑡) = 𝑡(𝑡+1) = 𝑡+1 − 1𝑡 , an antiderivative is ( 𝑡+1 ) 𝑃 (𝑡) = ln(𝑡 + 1) − ln 𝑡 = ln 𝑡 , and the integrating factor is 𝜇(𝑡) = 𝑡+1 𝑡 . Now multiply by the integrating factor to get

𝑡+1 ′ 1 2 𝑦 − 2𝑦 = 2, 𝑡 𝑡 𝑡

′ the left hand side of which is a perfect derivative ( 𝑡+1 𝑡 𝑦) . Thus

(

2 𝑡+1 ′ 𝑦) = 2 𝑡 𝑡

−2 and taking antiderivatives of both sides gives 𝑡+1 𝑡 𝑦 = 𝑡 + 𝑐 where 𝑐 ∈ ℝ 𝑡 −2 𝑐𝑡 is a constant. Now multiply by 𝑡+1 to get 𝑦 = 𝑡+1 + 𝑡+1 = 𝑐𝑡−2 𝑡+1 for the general solution.

11. In we get 𝑧 ′ − 2𝑡𝑧 = −2𝑡3 . An integrating factor is ∫ standard form 2 2 −2𝑡 𝑑𝑡 −𝑡2 𝑒 = 𝑒 . Thus (𝑒−𝑡 𝑧)′ = −2𝑡3 𝑒−𝑡 . Integrating both sides gives 2 2 𝑒−𝑡 𝑧 = (𝑡2 + 1)𝑒−𝑡 + 𝑐, where the integral of the right hand side is done 2 2 by parts. Now divide by the integrating factor 𝑒−𝑡 to get 𝑧 = 𝑡2 +1+𝑐𝑒𝑡 . 12. The given differential equation is in standard form, 𝑝(𝑡) = 𝑎, an antiderivative is 𝑃 (𝑡) = 𝑎𝑡, and the integrating factor is 𝜇(𝑡) = 𝑒𝑎𝑡 . Now multiply by the integrating factor to get 𝑒𝑎𝑡 𝑦 ′ + 𝑎𝑒𝑎𝑡 𝑦 = 𝑏𝑒𝑎𝑡 , the left hand side of which is a perfect derivative ((𝑒𝑎𝑡 )𝑦)′ . Thus ((𝑒𝑎𝑡 )𝑦)′ = 𝑏𝑒𝑎𝑡 . If 𝑎 ∕= 0 then taking antiderivatives of both sides gives 𝑒𝑎𝑡 𝑦 = 𝑎𝑏 𝑒𝑎𝑡 + 𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑒−𝑎𝑡 to get 𝑦 = 𝑎𝑏 + 𝑐𝑒−𝑎𝑡 for the general solution. In the case 𝑎 = 0 then 𝑦 ′ = 𝑏 and 𝑦 = 𝑏𝑡 + 𝑐.

1 Solutions

21

13. The given equation is in standard form, 𝑝(𝑡) = cos 𝑡, an antiderivative is 𝑃 (𝑡) = − sin 𝑡, and the integrating factor is 𝜇(𝑡) = 𝑒− sin 𝑡 . Now multiply by the integrating factor to get 𝑒− sin 𝑡 𝑦 ′ + (cos 𝑡)𝑒− sin 𝑡 𝑦 = (cos 𝑡)𝑒− sin 𝑡 , the left hand side of which is a perfect derivative ((𝑒− sin 𝑡 )𝑦)′ . Thus ((𝑒− sin 𝑡 )𝑦)′ = (cos 𝑡)𝑒− sin 𝑡 and taking antiderivatives of both sides gives (𝑒− sin 𝑡 )𝑦 = 𝑒− sin 𝑡 +𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑒sin 𝑡 to get 𝑦 = 1 + 𝑐𝑒sin 𝑡 for the general solution. To satisfy the initial condition, 0 = 𝑦(0) = 1 + 𝑐𝑒sin 0 = 1 + 𝑐, so 𝑐 = −1. Thus, the solution of the initial value problem is 𝑦 = 1 − 𝑒sin 𝑡 −2 14. The given equation is in standard form, 𝑝(𝑡) = 𝑡+1 , an antiderivative is −2 𝑃 (𝑡) = −2 ln(𝑡 + 1) = ln((𝑡 + 1) ), and the integrating factor is 𝜇(𝑡) = (𝑡 + 1)−2 . Now multiply by the integrating factor to get

(𝑡 + 1)−2 𝑦 ′ −

2 𝑦 = 1, (𝑡 + 1)3

the left hand side of which is a perfect derivative (((𝑡 + 1)−2 )𝑦)′ . Thus (((𝑡 + 1)−2 )𝑦)′ = 1 and taking antiderivatives of both sides gives ((𝑡 + 1)−2 )𝑦 = 𝑡 + 𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by (𝑡 + 1)2 to get 𝑦 = (𝑡 + 𝑐)(𝑡 + 1)2 for the general solution. 15. The given linear differential equation is in standard form, 𝑝(𝑡) = −2 𝑡 , an antiderivative is 𝑃 (𝑡) = −2 ln 𝑡 = ln 𝑡−2 , and the integrating factor is 𝜇(𝑡) = 𝑡−2 . Now multiply by the integrating factor to get 𝑡−2 𝑦 ′ −

2 𝑡+1 𝑦 = 3 = 𝑡−2 + 𝑡−3 , 𝑡3 𝑡

the left hand side of which is a perfect derivative (𝑡−2 𝑦)′ . Thus (𝑡−2 𝑦)′ = 𝑡−2 + 𝑡−3 −2

and taking antiderivatives of both sides gives (𝑡−2 )𝑦 = −𝑡−1 − 𝑡 2 +𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑡2 to and we get 𝑦 = −𝑡 − 12 + 𝑐𝑡−2 −3 for the general solution. Letting 𝑡 = 1 gives −3 = 𝑦(1) = −3 2 + 𝑐 so 𝑐 = 2 and 1 3 𝑦(𝑡) = −𝑡 − − 𝑡−2 . 2 2

22

1 Solutions

16. The given equation is in standard form, 𝑝(𝑡) = 𝑎, an antiderivative is 𝑃 (𝑡) = 𝑎𝑡, and the integrating factor is 𝜇(𝑡) = 𝑒𝑎𝑡 . Now multiply by the integrating factor to get 𝑒𝑎𝑡 𝑦 ′ + 𝑎𝑒𝑎𝑡 𝑦 = 1, the left hand side of which is a perfect derivative (𝑒𝑎𝑡 𝑦)′ . Thus (𝑒𝑎𝑡 𝑦)′ = 1 and taking antiderivatives of both sides gives 𝑒𝑎𝑡 𝑦 = 𝑡 + 𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑒−𝑎𝑡 to get 𝑦 = 𝑡𝑒−𝑎𝑡 + 𝑐𝑒−𝑎𝑡 for the general solution. 17. The given equation is in standard form, 𝑝(𝑡) = 𝑎, 𝑝(𝑡) = 𝑎, an antiderivative is 𝑃 (𝑡) = 𝑎𝑡, and the integrating factor is 𝜇(𝑡) = 𝑒𝑎𝑡 . Now multiply by the integrating factor to get 𝑒𝑎𝑡 𝑦 ′ + 𝑎𝑒𝑎𝑡 𝑦 = 𝑒(𝑎+𝑏)𝑡 , the left hand side of which is a perfect derivative (𝑒𝑎𝑡 𝑦)′ . Thus (𝑒𝑎𝑡 𝑦)′ = 𝑒(𝑎+𝑏)𝑡 and taking antiderivatives of both sides gives (𝑒𝑎𝑡 )𝑦 =

1 (𝑎+𝑏)𝑡 𝑒 +𝑐 𝑎+𝑏

where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑒−𝑎𝑡 to get 𝑦=

1 𝑏𝑡 𝑒 + 𝑐𝑒−𝑎𝑡 𝑎+𝑏

for the general solution. 18. The given differential equation is in standard form, 𝑝(𝑡) = 𝑎, an antiderivative is 𝑃 (𝑡) = 𝑎𝑡, and the integrating factor is 𝜇(𝑡) = 𝑒𝑎𝑡 . Now multiply by the integrating factor to get 𝑒𝑎𝑡 𝑦 ′ + 𝑎𝑒𝑎𝑡 𝑦 = 𝑡𝑛 , the left hand side of which is a perfect derivative (𝑒𝑎𝑡 𝑦)′ . Thus (𝑒𝑎𝑡 𝑦)′ = 𝑡𝑛 . Now assume 𝑛 ∕= −1. Then taking antiderivatives of both sides gives 𝑛+1 (𝑒𝑎𝑡 )𝑦 = 𝑡𝑛+1 + 𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑒−𝑎𝑡 to 𝑛+1

get 𝑦 = 𝑡𝑛+1 𝑒−𝑎𝑡 + 𝑐𝑒−𝑎𝑡 for the general solution. If 𝑛 = 1 then taking antiderivatives leads to 𝑒𝑎𝑡 𝑦 = ln 𝑡 + 𝑐 and hence 𝑦 = (ln 𝑡)𝑒−𝑎𝑡 + 𝑐𝑒−𝑎𝑡 is the general solution in this case.

19. In standard form we get 𝑦 ′ − (tan 𝑡)𝑦 = sec 𝑡. In this case 𝑝(𝑡) = − tan 𝑡, an antiderivative is 𝑃 (𝑡) = ln cos 𝑡, and the integrating factor is 𝜇(𝑡) = 𝑒𝑃 (𝑡) = cos 𝑡. Now multiply by the integrating factor to get (cos 𝑡)𝑦 ′ − (sin 𝑡)𝑦 = 1, the left hand side of which is a perfect derivative ((cos 𝑡)𝑦)′ . Thus ((cos 𝑡)𝑦)′ = 1 and taking antiderivatives of both sides gives (cos 𝑡)𝑦 = 𝑡 + 𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by 1/ cos 𝑡 = sec 𝑡 and we get 𝑦 = (𝑡 + 𝑐) sec 𝑡 for the general solution. 20. Divide by 𝑡 to put the equation in the standard form 𝑦′ +

2 ln 𝑡 4 ln 𝑡 𝑦= . 𝑡 𝑡

𝑡 2 In this case 𝑝(𝑡) = 2 ln 𝑡 , an antiderivative is 𝑃 (𝑡) = (ln 𝑡) , and the inte(ln 𝑡)2 grating factor is 𝜇(𝑡) = 𝑒 . Now multiply by the integrating factor to get

1 Solutions 2

𝑒(ln 𝑡) 𝑦 ′ +

23

2 ln 𝑡 (ln 𝑡)2 4 ln 𝑡 (ln 𝑡)2 𝑒 𝑦= 𝑒 , 𝑡 𝑡 2

the left hand side of which is a perfect derivative (𝑒(ln 𝑡) 𝑦)′ . Thus 2 𝑡 (ln 𝑡)2 (𝑒(ln 𝑡) 𝑦)′ = 4 ln and taking antiderivatives of both sides gives 𝑡 𝑒 2

2

𝑒(ln 𝑡) 𝑦 = 2𝑒(ln 𝑡) + 𝑐 2

where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑒−(ln 𝑡) and we get 𝑦 = 2 2 + 𝑐𝑒−(ln 𝑡) for the general solution. 21. The given differential equation is in standard form, 𝑝(𝑡) = −𝑛/𝑡, an antiderivative is 𝑃 (𝑡) = −𝑛 ln 𝑡 = ln(𝑡−𝑛 ), and the integrating factor is 𝜇(𝑡) = 𝑡−𝑛 . Now multiply by the integrating factor to get 𝑡−𝑛 𝑦 ′ − 𝑛𝑡−𝑛−1 𝑦 = 𝑒𝑡 , the left hand side of which is a perfect derivative (𝑡−𝑛 𝑦)′ . Thus (𝑡−𝑛 𝑦)′ = 𝑒𝑡 and taking antiderivatives of both sides gives (𝑡−𝑛 )𝑦 = 𝑒𝑡 + 𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑡𝑛 to and we get 𝑦 = 𝑡𝑛 𝑒𝑡 + 𝑐𝑡𝑛 for the general solution. 22. The given differential equation is in standard form, 𝑝(𝑡) = −1, an antiderivative is 𝑃 (𝑡) = −𝑡, and the integrating factor is 𝜇(𝑡) = 𝑒−𝑡 . Now multiply by the integrating factor to get 𝑒−𝑡 𝑦 ′ − 𝑒−𝑡 𝑦 = 𝑡𝑒𝑡 , the left hand side of which is a perfect derivative (𝑒−𝑡 𝑦)′ . Thus (𝑒−𝑡 𝑦)′ = 𝑡𝑒𝑡 . Taking antiderivatives of both sides and using integration by parts gives 𝑒−𝑡 𝑦 = 𝑡𝑒𝑡 − 𝑒𝑡 + 𝑐 = (𝑡 − 1)𝑒𝑡 + 𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑒𝑡 to get 𝑦 = (𝑡 − 1)𝑒2𝑡 + 𝑐𝑒𝑡 for the general solution. Letting 𝑡 = 0 gives 𝑎 = 𝑦(0) = −1 + 𝑐 so 𝑐 = 𝑎 + 1 and 𝑦 = (𝑡 − 1)𝑒2𝑡 + (𝑎 + 1)𝑒𝑡 . 23. Divide by 𝑡 to put the equation in the standard form 3 𝑦 ′ + 𝑦 = 𝑡. 𝑡 In this case 𝑝(𝑡) = 3/𝑡, an antiderivative is 𝑃 (𝑡) = 3 ln 𝑡 = ln(𝑡3 ), and the integrating factor is 𝜇(𝑡) = 𝑡3 . Now multiply the standard form equation by the integrating factor to get 𝑡3 𝑦 ′ +3𝑡2 𝑦 = 𝑡4 , the left hand side of which is a perfect derivative (𝑡3 𝑦)′ . Thus (𝑡3 𝑦)′ = 𝑡4 and taking antiderivatives of both sides gives 𝑡3 𝑦 = 51 𝑡5 + 𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑡−3 and we get 𝑦 = 51 𝑡2 + 𝑐𝑡−3 for the general solution. Letting 𝑡 = −1 gives 2 = 𝑦(−1) = 51 − 𝑐 so 𝑐 = −9 5 and 𝑦=

1 2 9 −3 𝑡 − 𝑡 . 5 5

24

1 Solutions

24. In this case the given differential equation is in standard form. We have 𝑝(𝑡) = 2𝑡, an antiderivative is 𝑃 (𝑡) = 𝑡2 , and the integrating factor is 2 2 2 𝜇(𝑡) = 𝑒𝑡 . Now multiply by the integrating factor to get 𝑒𝑡 𝑦 ′ + 2𝑡𝑒𝑡 𝑦 = 2 2 2 𝑒𝑡 , the left hand side of which is a perfect derivative (𝑒𝑡 𝑦)′ . Thus (𝑒𝑡 𝑦)′ = ∫ 2 2 2 𝑒𝑡 and taking antiderivatives of both sides gives 𝑒𝑡 𝑦 = 𝑒𝑡 𝑑𝑡 + 𝑐 where 𝑐 ∈ ℝ is a constant. However, the right hand side has no closed form antiderivative. Using Corollary 8 to can write ∫ 𝑡 ∫ 𝑡 2 2 2 2 2 2 𝑦 = 𝑒−𝑡 𝑒𝑢 𝑑𝑢 + 𝑦(0)𝑒−𝑡 = 𝑒−𝑡 𝑒𝑢 𝑑𝑢 + 𝑒−𝑡 . 0

0

25. Divide by 𝑡2 to put the equation in the standard form 2 𝑦 ′ + 𝑦 = 𝑡−2 . 𝑡 In this case 𝑝(𝑡) = 2/𝑡, an antiderivative is 𝑃 (𝑡) = 2 ln 𝑡 = ln 𝑡2 , and the integrating factor is 𝜇(𝑡) = 𝑡2 . Now multiply by the integrating factor to get 𝑡2 𝑦 ′ + 2𝑡𝑦 = 1, the left hand side of which is a perfect derivative (𝑡2 𝑦)′ . Thus (𝑡2 𝑦)′ = 1 and taking antiderivatives of both sides gives 𝑡2 𝑦 = 𝑡 + 𝑐 where 𝑐 ∈ ℝ is a constant. Now multiply by 𝑡−2 to get 𝑦 = 1𝑡 + 𝑐𝑡−2 for the general solution. Letting 𝑡 = 2 gives 𝑎 = 𝑦(2) = 21 + 4𝑐 so 𝑐 = 4𝑎 − 2 and 1 𝑦 = + (4𝑎 − 2)𝑡−2 . 𝑡 gal lbs lbs ×0 =0 . min gal min gal 𝑦(𝑡) lbs 4 lbs output rate = 4 × = 𝑦(𝑡) . min 100 gal 100 min

26. input rate: input rate = 4 output rate: Since

𝑦 ′ = input rate − output rate we get the initial value problem 𝑦′ = 0 −

4 𝑦, 100

𝑦(0) = 80.

1 Simplifying and putting in standard form gives 𝑦 ′ + 𝑦 = 0. The coeffi25 ∫ cient function is 𝑝(𝑡) = 1/25, 𝑃 (𝑡) = 𝑝(𝑡) 𝑑𝑡 = 𝑡/25, and the integrating factor is 𝜇(𝑡) = 𝑒𝑡/25 . Thus (𝑒𝑡/25 𝑦)′ = 0. Integrating and simplifying gives 𝑦 = 𝑐𝑒−𝑡/25 . The initial condition implies 𝑐 = 80 so 𝑦 = 80𝑒−𝑡/25 . The concentration of the brine solution is now obtained by dividing by the 80 −𝑡/25 volume which is 100 gallons: 100 𝑒 = 0.8𝑒−𝑡/25 .

1 Solutions

25

27. Let 𝑉 (𝑡) denote the volume of fluid in the tank at time 𝑡. Initially, there are 10 gal of brine. For each minute that passes there is a net decrease of 4 − 3 = 1 gal of brine. Thus 𝑉 (𝑡) = 10 − 𝑡 gal. gal lbs lbs ×1 =3 . input rate: input rate = 3 min gal min gal 𝑦(𝑡) lbs 4𝑦(𝑡) lbs output rate: output rate = 4 × = . min 𝑉 (𝑡) gal 10 − 𝑡 min Since 𝑦 ′ = input rate − output rate, it follows that 𝑦(𝑡) satisfies the initial value problem 4 𝑦′ = 3 − 𝑦(𝑡) , 𝑦(0) = 2. 10 − 𝑡 Put in standard form, this equation becomes 𝑦′ +

4 𝑦 = 3. 10 − 𝑡

∫ 4 The coefficient function is 𝑝(𝑡) = 10−𝑡 , 𝑃 (𝑡) = 𝑝(𝑡) 𝑑𝑡 = −4 ln(10 − 𝑡) = ln(10 − 𝑡)−4 , and the integrating factor is 𝜇(𝑡) = (10 − 𝑡)−4 . Multiplying the standard form equation by the integrating factor gives ((10 − 𝑡)−4 𝑦)′ = 3(10 − 𝑡)−4 . Integrating and simplifying gives 𝑦 = (10 − 𝑡) + 𝑐(10 − 𝑡)4 . The initial condition 𝑦(0) = 2 implies 2 = 𝑦(0) = 10 + 𝑐104 and hence 𝑐 = −8/104 so 𝑦 = (10 − 𝑡) −

8 (10 − 𝑡)4 . 104

Of course, this formula is valid for 0 ≤ 𝑡 ≤ 10. After 10 minutes there is no fluid and hence no salt in the tank. L g g × 10 = 10 . min L min L 𝑦(𝑡) g 𝑦(𝑡) g output rate: output rate = 1 × = . min 10 L 10 min ′ Since 𝑦 = input rate − output rate, it follows that 𝑦(𝑡) satisfies the initial value problem 1 𝑦 ′ = 10 − 𝑦 , 𝑦(0) = 0. 10

28. input rate: input rate = 1

1 Put in standard form, this equation becomes 𝑦 ′ + 10 𝑦 = 10. The coefficient ∫ 1 function is 𝑝(𝑡) = 10 , 𝑃 (𝑡) = 𝑝(𝑡) 𝑑𝑡 = 𝑡/10, and the integrating factor is 𝜇(𝑡) = 𝑒𝑡/10 . Multiplying the standard form equation by the integrating factor gives (𝑒𝑡/10 𝑦)′ = 10𝑒𝑡/10 . Integrating and simplifying gives 𝑦 = 100 + 𝑐𝑒−𝑡/10 . The initial condition 𝑦(0) = 0 implies 𝑐 = −100 so 𝑦 = 100 − 100𝑒−𝑡/10. After 10 minutes we have 𝑦(10) = 100 − 100𝑒−1 g of salt. The concentration is thus (100 − 100𝑒−1)/10 = 10 − 10𝑒−1 g/L

26

1 Solutions

29. Let 𝑉 (𝑡) denote the volume of fluid in the container at time 𝑡. Initially, there are 10 L. For each minute that passes there is a net gain of 4−2 = 2 L of fluid. So 𝑉 (𝑡) = 10 + 2𝑡. The container overflows when 𝑉 (𝑡) = 10 + 2𝑡 = 30 or 𝑡 = 10 minutes. L g g × 20 = 80 . input rate: input rate = 4 min L min L 𝑦(𝑡) g 2𝑦(𝑡) g output rate: output rate = 2 × = . min 10 + 𝑡 L 10 + 𝑡 min Since 𝑦 ′ = input rate − output rate, it follows that 𝑦(𝑡) satisfies the initial value problem 2𝑦 𝑦 ′ = 80 − , 𝑦(0) = 0. 10 + 𝑡 Simplifying and putting in standard form gives the equation 𝑦′ +

1 𝑦 = 80. 5+𝑡

∫ 1 , 𝑃 (𝑡) = 𝑝(𝑡) 𝑑𝑡 = ln(5 + 𝑡), and the The coefficient function is 𝑝(𝑡) = 5+𝑡 integrating factor is 𝜇(𝑡) = 5 + 𝑡. Multiplying the standard form equation by the integrating factor gives ((5 + 𝑡)𝑦)′ = 80(5 + 𝑡). Integrating and simplifying gives 𝑦 = 40(5+𝑡)+𝑐(5+𝑡)−1, where 𝑐 is a constant. The initial condition 𝑦(0) = 0 implies 𝑐 = −1000 so 𝑦 = 40(5 + 𝑡) − 1000(5 + 𝑡) −1 . At the time the container overflows 𝑡 = 10 we have 𝑦(10) = 600 − 1000 15 ≈ 533.33 g of salt. 30. Let 𝑉 (𝑡) denote the volume of fluid in the tank at time 𝑡. Initially, there are 10 gallons of fluid. For each minute that goes by there is a net increase of 4 − 2 = 2 gallons. It follows that 𝑉 (𝑡) = 10 + 2𝑡. The tank will overflow when 100 = 𝑉 (𝑡). Solving 100 = 10 + 2𝑡 gives 𝑡 = 45. Thus 𝑇 = 45 minutes. Next we find 𝑦(𝑡): gal lbs lbs input rate: input rate = 4 × 0.5 =2 . min gal min gal 𝑦(𝑡) lbs 2𝑦(𝑡) lbs output rate: output rate = 2 × = . min 10 + 2𝑡 gal 10 + 2𝑡 min Since 𝑦 ′ = input rate − output rate, it follows that 𝑦(𝑡) satisfies the initial value problem 𝑦 𝑦′ = 2 − , 𝑦(0) = 0. 5+𝑡 Putting this equation in standard form gives 𝑦′ +

1 𝑦 = 2. 5+𝑡

∫ 1 The coefficient function is 𝑝(𝑡) = 5+𝑡 , 𝑃 (𝑡) = 𝑝(𝑡) 𝑑𝑡 = ln(5 + 𝑡), and the integrating factor is 𝜇(𝑡) = 5 + 𝑡. Thus ((5 + 𝑡)𝑦)′ = 2(5 + 𝑡). Integrating

1 Solutions

27

and simplifying gives 𝑦 = (5+𝑡)+𝑐(5+𝑡)−1 . The initial condition 𝑦(0) = 0 implies 𝑐 = −25 so 𝑦 = (5 + 𝑡) − 25(5 + 𝑡)−1, for 0 ≤ 𝑡 ≤ 45. At 𝑡 = 𝑇 = 45 25 we get 𝑦(45) = 50 − 50 = 49.5 lbs salt. Once the tank is full, the inflow and outflow rates will be equal and the brine in the tank will (in the limit as 𝑡 → ∞) stabilize to the concentration of the incoming brine, i.e., 0.5 lb/gal. Since the tank holds 100 gal, the total amount present will approach 0.5 × 100 = 50 lbs. Thus lim𝑡→∞ 𝑦(𝑡) = 50. 31. input rate: input rate = 𝑟𝑐 output rate:

output rate = 𝑟 𝑃𝑉(𝑡)

Let 𝑃0 denote the amount of pollutant at time 𝑡 = 0. Since 𝑃 ′ = input rate − output rate it follows that 𝑃 (𝑡) is a solution of the initial value problem 𝑟𝑃 (𝑡) 𝑃 ′ = 𝑟𝑐 − , 𝑃 (0) = 𝑃0 . 𝑉 Rewriting this equation in standard form gives the differential equation 𝑃 ′ + 𝑉𝑟 𝑃 = 𝑟𝑐. The coefficient function is 𝑝(𝑡) = 𝑟/𝑉 and the integrating 𝑟𝑡 𝑟𝑡 factor is 𝜇(𝑡) = 𝑒𝑟𝑡/𝑉 . Thus (𝑒 𝑉 𝑃 )′ = 𝑟𝑐𝑒 𝑉 . Integrating and simplifying −𝑟𝑡 gives 𝑃 (𝑡) = 𝑐𝑉 +𝑘𝑒 𝑉 , where 𝑘 is the constant of integration. The initial −𝑟𝑡 condition 𝑃 (0) = 𝑃0 implies 𝑐 = 𝑃0 − 𝑐𝑉 so 𝑃 (𝑡) = 𝑐𝑉 + (𝑃0 − 𝑐𝑉 )𝑒 𝑉 . (a) lim𝑡→∞ 𝑃 (𝑡) = 𝑐𝑉.

(b) When the river is cleaned up at 𝑡 = 0 we assume the input concentration is 𝑐 = 0. The amount of pollutant is therefore given by −𝑟𝑡 𝑃 (𝑡) = 𝑃0 𝑒 𝑉 . This will reduce by 1/2 when 𝑃 (𝑡) = 12 𝑃0 . We solve the −𝑟𝑡 equation 21 𝑃0 = 𝑃0 𝑒 𝑉 for 𝑡 and get 𝑡1/2 = 𝑉 ln𝑟2 . Similarly, the pollutant will reduce by 1/10 when 𝑡1/10 = 𝑉 ln𝑟10 . (c) Letting 𝑉 and 𝑟 be given as stated for each lake gives: Lake Erie: 𝑡1/2 = 1.82 years, 𝑡1/10 = 6.05 years. Lake Ontario: 𝑡1/2 = 5.43 years, 𝑡1/10 = 18.06 years 32. Let 𝑦1 (𝑡) and 𝑦2 (𝑡) denote the amount of salt in Tank 1 and Tank 2, respectively, at time 𝑡. L g g input rate for Tank 1: input rate = 4 × 100 = 400 . min L min L 𝑦1 (𝑡) g 4𝑦1 (𝑡) g output rate for Tank 1: output rate = 4 × = . min 10 L 10 m The initial value problem for Tank 1 is thus: 𝑦1′ = 400 −

4 𝑦1 , 10

𝑦1 (0) = 0.

4 Simplifying and putting this equation in standard form gives 𝑦1′ + 10 𝑦1 = 4𝑡/10 4𝑡/10 ′ 400. The integrating factor is 𝜇(𝑡) = 𝑒 . Thus (𝑒 𝑦1 ) = 400𝑒4𝑡/10 .

28

1 Solutions

Integrating and simplifying gives 𝑦1 = 1000 + 𝑐𝑒−4𝑡/10 . The initial condition 𝑦1 (0) = 0 implies 𝑐 = −1000 so 𝑦1 = 1000 − 1000𝑒−4𝑡/10. Now the brine solution in Tank 1 has concentration 𝑦1 (𝑡)/10 = 100 − 100𝑒−4𝑡/10 and flows into Tank 2 at a rate of 4 liters per minute. Thus L g input rate for Tank 2: input rate = 4 × (100 − 100𝑒−4𝑡/10 ) = min L g . 400 − 400𝑒−4𝑡/10 min L 𝑦2 (𝑡) g 4𝑦2 (𝑡) g output rate for Tank 2: output rate = 4 × = . min 10 L 10 min The initial value problem for Tank 2 is thus: 𝑦2′ = 400 − 400𝑒−4𝑡/10 −

4 𝑦2 , 10

𝑦2 (0) = 0.

Simplifying and putting this equation in standard form gives 𝑦2′ +

4 𝑦2 = 400 − 400𝑒−4𝑡/10. 10

The integrating factor is again (as for the Tank 1 equation) 𝜇(𝑡) = 𝑒4𝑡/10 . Thus multiplying by the integrating factor gives (𝑒4𝑡/10 𝑦2 )′ = 400𝑒4𝑡/10 − 400. Integrating and simplifying gives 𝑦2 (𝑡) = 1000 − 400𝑡𝑒−4𝑡/10 + 𝑐𝑒−4𝑡/10 . The initial condition 𝑦2 (0) = 0 implies 𝑐 = −1000 so 𝑦2 (𝑡) = 1000 − 400𝑡𝑒−4𝑡/10 − 1000𝑒−4𝑡/10.

33. Let 𝑦1 (𝑡) and 𝑦2 (𝑡) denote the amount of salt in Tank 1 and Tank 2, respectively, at time 𝑡. The volume of fluid at time 𝑡 in Tank 1 is 𝑉1 (𝑡) = 10 + 2𝑡 and Tank 2 is 𝑉2 (𝑡) = 5 + 𝑡. L g g input rate for Tank 1: input rate = 4 × 10 = 40 . min L min L 𝑦1 (𝑡) g output rate for Tank 1: output rate = 2 × = min 10 + 2𝑡 L 2𝑦(𝑡) g . The initial value problem for Tank 1 is thus 10 + 2𝑡 min 𝑦1′ = 40 −

2 𝑦1 , 10 + 2𝑡

𝑦1 (0) = 0.

Simplifying this equation and putting it in standard form gives

1 Solutions

𝑦1′ +

29

1 𝑦1 = 40. 5+𝑡

The integrating factor is 𝜇(𝑡) = 5 + 𝑡. Thus ((5 + 𝑡)𝑦1 )′ = 40(5 + 𝑡). Integrating and simplifying gives 𝑦1 (𝑡) = 20(5 + 𝑡) + 𝑐/(5 + 𝑡). The initial condition 𝑦(0) = 0 implies 𝑐 = −500 so 𝑦1 = 20(5 + 𝑡) − 500/(5 + 𝑡). L 𝑦1 (𝑡) g input rate for Tank 2: input rate = 2 × = 20 − min 10 + 2𝑡 L 500 g . (5 + 𝑡)2 min output rate for Tank 2:

output rate = 1

The initial value problem for Tank 2 is thus 𝑦2′ = 20 − 500/(5 + 𝑡)2 −

L 𝑦2 (𝑡) g 𝑦2 (𝑡) g × = . min 5 + 𝑡 L 5 + 𝑡 min

1 𝑦2 , (5 + 𝑡)

𝑦2 (0) = 0.

When this equation is put in standard form we get 𝑦2′ +

1 500 𝑦2 = 20 − . (5 + 𝑡) (5 + 𝑡)2

The integrating factor is 𝜇(𝑡) = 5 + 𝑡. Thus ((5 + 𝑡)𝑦2 )′ = 20(5 + 𝑡) −

500 . 5+𝑡

Integrating and simplifying gives 𝑦2 (𝑡) = 10(5 + 𝑡) −

500 ln(5 + 𝑡) 𝑐 + . 5+𝑡 5+𝑡

The initial condition 𝑦2 (0) = 0 implies 𝑐 = 500 ln 5 − 250 so 𝑦2 (𝑡) = 10(5 + 𝑡) −

500 ln(5 + 𝑡) 500 ln 5 − 250 + . 5+𝑡 5+𝑡

Section 1.5 𝑦 2 + 𝑦𝑡 + 𝑡2 1. In standard form we get 𝑦 ′ = which is homogeneous since 𝑡2 the degrees of the numerator and denominator are each two. Let 𝑦 = 𝑡𝑣. Then 𝑣 + 𝑡𝑣 ′ = 𝑣 2 + 𝑣 + 1 and so 𝑡𝑣 ′ = 𝑣 2 + 1. Separating variables gives 𝑑𝑣 𝑑𝑡 = . Integrating gives tan−1 𝑣 = ln ∣𝑡∣ + 𝑐. So 𝑣 = tan(ln ∣𝑡∣ + 𝑐). 𝑣2 + 1 𝑡 Substituting 𝑣 = 𝑦/𝑡 gives 𝑦 = 𝑡 tan(ln ∣𝑡∣ + 𝑐).

30

1 Solutions

2. Since the numerator and denominator are homogeneous of degree 1 the 4 − 3𝑣 quotient is homogeneous. Let 𝑦 = 𝑡𝑣. Then 𝑣 + 𝑡𝑣 ′ = and so 1−𝑣 2 (𝑣 − 2) 𝑡𝑣 ′ = . Clearly, 𝑣 = 2 is an equilibrium solution. Separating the 1−𝑣 −𝑑𝑣 𝑑𝑣 𝑑𝑡 variables and using partial fractions gives − = . Integrat(𝑣 − 2)2 𝑣 − 2 𝑡 ing gives (𝑣 − 2)−1 − ln ∣𝑣 − 2∣ = ln ∣𝑡∣ + 𝑐. Simplifying and exponentiating 𝑡 −1 gives 𝑒(𝑣−2) = 𝑘𝑡(𝑣 − 2), 𝑘 ∕= 0. Now let 𝑣 = 𝑦/𝑡 then 𝑒 𝑦−2𝑡 = 𝑘(𝑦 − 2𝑡) for 𝑘 ∕= 0. The equilibrium solution 𝑣 = 2 gives 𝑦 = 2𝑡 as another solution. 3. Since the numerator and denominator are homogeneous of degree 2 the quotient is homogeneous. Let 𝑦 = 𝑡𝑣. Then 𝑣 + 𝑡𝑣 ′ = 𝑣 2 − 4𝑣 + 6. So 𝑡𝑣 ′ = 𝑣 2 − 5𝑣 + 6 = (𝑣 − 2)(𝑣 − 3). There are two equilibrium solutions 𝑣 = 2, 3. Separating the variables and using partial fractions gives ( ) 𝑣 − 3 1 1 𝑑𝑡 = − 𝑑𝑣 = . Integrating and simplifying gives ln 𝑣−3 𝑣−2 𝑡 𝑣 − 2 3 − 2𝑘𝑡 3𝑡 − 2𝑘𝑡2 ln ∣𝑡∣ + 𝑐. Solving for 𝑣 gives 𝑣 = , for 𝑘 ∕= 0, and so 𝑦 = , 1 − 𝑘𝑡 1 − 𝑘𝑡 for 𝑘 ∕= 0. When 𝑘 = 0 we get 𝑣 = 3 or 𝑦 = 3𝑡, which is the same as the equilibrium solution 𝑣 = 3. The equilibrium solution 𝑣 = 2 gives 𝑦 = 2𝑡. 3𝑡 − 2𝑘𝑡2 Thus we can write the solutions as 𝑦 = , 𝑘 ∈ ℝ and 𝑦 = 2𝑡. 1 − 𝑘𝑡 4. Since the numerator and denominator are homogeneous of degree 2 the 𝑡2 𝑣 2 + 2𝑡2 𝑣 quotient is homogeneous. Let 𝑦 = 𝑡𝑣. Then 𝑣 + 𝑡𝑣 ′ = = 𝑡2 + 𝑡 2 𝑣 2 𝑣 + 2𝑣 𝑣 . Subtract 𝑣 from both sides to get 𝑡𝑣 ′ = . Separate the 1+𝑣 𝑣+1 variables to get ) ( 1 1 𝑑𝑣 = 𝑑𝑡. 1+ 𝑣 𝑡 Integrating gives 𝑣 + ln ∣𝑣∣ = ln ∣𝑡∣ + 𝑐. Now exponentiate, substitute 𝑣 = 𝑦/𝑡, and simplify to get 𝑦𝑒𝑦/𝑡 = 𝑘𝑡2 , 𝑘 ∈ ℝ. 5. Since the numerator and denominator are homogeneous of degree 2 the 3𝑣 2 − 1 quotient is homogeneous. Let 𝑦 = 𝑡𝑣. Then 𝑣 + 𝑡𝑣 ′ = . Subtract 2𝑣 2 𝑣 −1 𝑣 from both sides to get 𝑡𝑣 ′ = . The equilibrium solutions are 2𝑣 2𝑣 𝑑𝑣 𝑑𝑡 𝑣 = ±1. Separating variables gives 2 = and integrating gives 𝑣 − 1 𝑡 2 2 ln 𝑣 − 1 = ln ∣𝑡∣+𝑐. Exponentiating gives 𝑣 −1 = 𝑘𝑡 and by simplifying √ √ we get 𝑣 = ± 1 + 𝑘𝑡. Now 𝑣 = 𝑦/𝑡 so 𝑦 = ±𝑡 1 + 𝑘𝑡. The equilibrium solutions 𝑣 = ±1 become 𝑦 = ±𝑡. These occur when 𝑘 = 0, so are already included in the general formula.

1 Solutions

31

6. Since 𝑡2 +𝑦 2 and 𝑡𝑦 are homogeneous of degree 2 their quotient (𝑡2 +𝑦 2 )/𝑡𝑦 is a homogeneous function. Let 𝑦 = 𝑡𝑣. Then 𝑦 ′ = 𝑣 + 𝑡𝑣 ′ and the given differential equation becomes 𝑣 + 𝑡𝑣 ′ =

1 + 𝑣2 1 = + 𝑣. 𝑣 𝑣

Simplifying and separating variables√gives 𝑣 𝑑𝑣 = 𝑑𝑡/𝑡. Integrating we get 𝑣 2 /2 √ = ln 𝑡 + 𝑐 and so 𝑣 = ± 2 ln 𝑡 + 2𝑐. Since 𝑣 = 𝑦/𝑡 we get 𝑦 = ±𝑡 2 ln 𝑡 + 2𝑐. √ √ 𝑦 + 𝑡2 − 𝑦 2 ′ . Since (𝛼𝑡)2 − (𝛼𝑦)2 = 7. In standard form we get 𝑦 = 𝑡 √ √ 𝛼2√ (𝑡2 − 𝑦 2 ) = 𝛼 𝑡2 − 𝑦 2 for 𝛼 > 0 it is easy to see that 𝑦 ′ = √ 𝑦 + 𝑡2 − 𝑦 2 is homogeneous. Let 𝑦 = 𝑡𝑣. Then 𝑣 + 𝑡𝑣 ′ = 𝑣 + 1 − 𝑣 2 . 𝑡 √ Simplifying gives 𝑡𝑣 ′ = 1 − 𝑣 2 . Clearly 𝑣 = ±1 are equilibrium solution. 𝑑𝑣 𝑑𝑡 = . Integrating gives sin−1 𝑣 = Separating variables gives √ 2 𝑡 1−𝑣 ln ∣𝑡∣ + 𝑐 and so 𝑣 = sin(ln ∣𝑡∣ + 𝑐). Now substitute 𝑣 = 𝑦/𝑡 to get 𝑦 = 𝑡 sin(ln ∣𝑡∣ + 𝑐). The equilibrium solutions imply 𝑦 = ±𝑡 are also solutions. √ 𝑦 𝑦 𝑡2 + 𝑦 2 ′ 8. In standard form we get 𝑦 = + . It is straightforward to see 𝑡 𝑡2 √ √ 𝑦 𝑦 𝑡2 + 𝑦 2 ′ 2 2 + that 𝑡 + 𝑦 is homogeneous of degree one. So 𝑦 = 𝑡 𝑡2 ′ is a homogeneous differential equation. Let 𝑦 = 𝑡𝑣 then 𝑣 + 𝑡𝑣 = √ √ 𝑣 + 𝑣 1 + 𝑣 2 or 𝑡𝑣 ′ = 𝑣 1 + 𝑣 2 . It follows that 𝑣 = 0 is an equilib𝑑𝑣 𝑑𝑡 = . Integrating rium solution. Separating variables gives √ 2 𝑡 𝑣 1+𝑣 𝑣 = ln ∣𝑡∣ + 𝑐. (To integrate the left hand side use √ gives ln 2 1+ 1+𝑣 𝑣 √ = 𝑘𝑡, the trig substitution 𝑣 = tan 𝜃.) Exponentiating gives 1 + 1 + 𝑣2 𝑦 √ 𝑘 ∕= 0. Now let 𝑣 = 𝑦/𝑡. Then = 𝑘𝑡. The case where 𝑘 = 0 𝑡 + 𝑡2 + 𝑦 2 gives the equilibrium solution. 9. Note that although 𝑦 = 0 is part of the general solution it does not satisfy the initial value. Divide both sides by 𝑦 2 to get 𝑦 −2 𝑦 ′ − 𝑦 −1 = 𝑡. Let 𝑧 = 𝑦 −1 . Then 𝑧 ′ = −𝑦 −2 𝑦 ′ . Substituting gives −𝑧 ′ − 𝑧 = 𝑡 or 𝑧 ′ + 𝑧 = −𝑡. An integrating factor is 𝑒𝑡 . So (𝑒𝑡 𝑧) = −𝑡𝑒𝑡 . Integrating both sides gives 𝑡 𝑡 𝑡 by parts to compute ∫𝑒 𝑧 = 𝑡−𝑡𝑒 + 𝑒 + 𝑐, where we have used integration −𝑡𝑒 𝑑𝑡. Solving for 𝑧 gives 𝑧 = −𝑡 + 1 + 𝑐𝑒−𝑡 . Now substitute 𝑧 = 𝑦 −1

32

1 Solutions

1 . The initial condition implies −𝑡 + 1 + 𝑐𝑒−𝑡 1 1 1= and so 𝑐 = 0. The solution is thus 𝑦 = . 1+𝑐 1−𝑡

and solve for 𝑦 to get 𝑦 =

10. Note that 𝑦 = 0 is a solution. Divide by 𝑦 2 to get 𝑦 −2 𝑦 ′ + 𝑦 −1 = 1. Let 𝑧 = 𝑦 −1 . Then 𝑧 ′ = −𝑦 −2 𝑦 ′ and substituting gives −𝑧 ′ + 𝑧 = 1. In the standard form for linear equations this becomes 𝑧 ′ − 𝑧 = −1. Multiplying by the integrating factor 𝑒−𝑡 gives (𝑒−𝑡 𝑧)′ = −𝑒−𝑡 so that 𝑒−𝑡 𝑧 = 𝑒−𝑡 + 𝑐. Hence 𝑧 = 1 + 𝑐𝑒𝑡 . Now go back to the original function 𝑦 by solving 1 𝑧 = 𝑦 −1 for 𝑦. Thus 𝑦 = 𝑧 −1 = (1 + 𝑐𝑒𝑡 )−1 = 1+𝑐𝑒 𝑡 . The general solution 1 is 𝑦 = 1+𝑐𝑒𝑡 and 𝑦 = 0. 11. Note that 𝑦 = 0 is a solution. First divide both sides by 𝑦 3 to get 𝑦 −3 𝑦 ′ + 𝑧′ 𝑡𝑦 −2 = 𝑡. Let 𝑧 = 𝑦 −2 . Then 𝑧 ′ = −2𝑦 −3 𝑦 ′ , so = 𝑦 −3 𝑦 ′ . Substituting −2 𝑧′ +𝑡𝑧 = 𝑡, which in standard form is 𝑧 ′ −2𝑡𝑧 = −2𝑡. An integrating gives −2 ∫ 2 2 2 factor is 𝑒 −2𝑡 𝑑𝑡 = 𝑒−𝑡 , so that (𝑒−𝑡 𝑧)′ = −2𝑡𝑒−𝑡 . Integrating both 2 2 sides gives 𝑒−𝑡 𝑧 = 𝑒−𝑡 + 𝑐, where the integral of the right hand side is 2 done by the substitution 𝑢 = −𝑡2 . Solving for 𝑧 gives 𝑧 = 1 + 𝑐𝑒𝑡 . Since 1 𝑧 = 𝑦 −2 we find 𝑦 = ± √ . 1 + 𝑐𝑒𝑡2 12. Note that 𝑦 = 0 is a solution. First divide both sides by 𝑦 3 to get 𝑧′ 𝑦 −3 𝑦 ′ + 𝑡𝑦 −2 = 𝑡3 . Let 𝑧 = 𝑦 −2 . Then 𝑧 ′ = −2𝑦 −3 𝑦 ′ , so = 𝑦 −3 𝑦 ′ . Sub−2 𝑧′ stituting gives + 𝑡𝑧 = 𝑡3 , which in standard form is 𝑧 ′ − 2𝑡𝑧 = −2𝑡3 . −2 ∫ 2 2 2 An integrating factor is 𝑒 −2𝑡 𝑑𝑡 = 𝑒−𝑡 . Thus (𝑒−𝑡 𝑧)′ = −2𝑡3 𝑒−𝑡 . Inte2 2 grating both sides gives 𝑒−𝑡 𝑧 = (𝑡2 + 1)𝑒−𝑡 + 𝑐, where the integral of the right hand side is computed using integration by parts with 𝑢 = 𝑡2 and 2 2 𝑑𝑣 = −2𝑡𝑒−𝑡 𝑑𝑡. Solving for 𝑧 gives 𝑧 = 𝑡2 + 1 + 𝑐𝑒𝑡 . Since 𝑧 = 𝑦 −2 we 1 find 𝑦 = ± √ . 2 𝑡 + 1 + 𝑐𝑒𝑡2 13. Note that 𝑦 = 0 is a solution. Divide by 𝑦 2 and (1 − 𝑡2 ) to get 𝑦 −2 𝑦 ′ − 𝑡 5𝑡 𝑦 −1 = . Let 𝑧 = 𝑦 −1 . Then 𝑧 ′ = −𝑦 −2 𝑦 ′ and substituting 2 1−𝑡 1 − 𝑡2 𝑡 5𝑡 𝑡 gives −𝑧 ′ − 𝑦′ = . In standard form we get 𝑧 ′ + 𝑧= 1 − 𝑡2 (1 − 𝑡2 ) 1 − 𝑡2 −5𝑡 . Multiplying by the integrating factor 1 − 𝑡2 𝜇(𝑡) = 𝑒

∫

𝑡 1−𝑡2

𝑑𝑡

1

= 𝑒− 2 ln(1−𝑡

2

)

= (1 − 𝑡2 )−1/2

1 Solutions

33

gives (𝑧(1 − 𝑡2 )−1/2 )′ = −5𝑡(1 − 𝑡2)−3/2 . Integrating gives 𝑧(1 − 𝑡2)−1/2 = √ 2 −1/2 2 −5(1 − 𝑡 ) + 𝑐 and hence 𝑧 = −5 + 𝑐 1 − 𝑡 . Since 𝑧 = 𝑦 −1 we have 1 √ 𝑦= . −5 + 𝑐 1 − 𝑡2 14. Note that 𝑦 = 0 is a solution. Divide both sides by 𝑦 2/3 to get 𝑦 −2/3 𝑦 ′ + 𝑦 1/3 1 𝑧 = 1. Let 𝑧 = 𝑦 1/3 . Then 𝑧 ′ = 𝑦 −2/3 𝑦 ′ and hence 3𝑧 ′ + = 1. In 𝑡 3 𝑡∫ 𝑧 1 1 ′ standard form we get 𝑧 + = . The integrating factor is 𝑒 3𝑡 𝑑𝑡 = 3𝑡 3 1 ln 𝑡 𝑒 3 = 𝑡1/3 . Multiplying by 𝑡1/3 gives (𝑡1/3 𝑧)′ = 𝑡1/3 and integrating 3 1 𝑡 gives 𝑡1/3 𝑧 = 𝑡4/3 + 𝑐. Solving for 𝑧 we get 𝑧 = + 𝑐𝑡−1/3 . Since 𝑧 = 𝑦 1/3 4 4 we can solve for 𝑦 to get 𝑦 = (𝑡/4 + 𝑐𝑡−1/3 )3 . 15. If we divide by 𝑦 we get 𝑦 ′ + 𝑡𝑦 = 𝑡𝑦 −1 which is a Bernoulli equation with 𝑛 = −1. Note that since 𝑛 < 0, 𝑦 = 0 is not a solution. Dividing by 𝑦 −1 𝑧′ gets us back to 𝑦𝑦 ′ + 𝑡𝑦 2 = 𝑡. Let 𝑧 = 𝑦 2 . Then 𝑧 ′ = 2𝑦𝑦 ′ so + 𝑡𝑧 = 𝑡 2 2 and in standard form we get 𝑧 ′ + 2𝑡𝑧 = 2𝑡. An integrating factor is 𝑒𝑡 so 2 2 2 2 2 (𝑒𝑡 𝑧)′ = 2𝑡𝑒𝑡 . Integration gives 𝑒𝑡 𝑧 = 𝑒𝑡 + 𝑐 so 𝑧 = 1 + 𝑐𝑒−𝑡 . Since √ 𝑧 = 𝑦 2 we get 𝑦 = ± 1 + 𝑐𝑒−𝑡2 . 1 𝑡 − 1 −1 16. First divide by 2𝑦 to get 𝑦 ′ − 𝑦 = 𝑦 , which is a Bernoulli equation 2 2 with 𝑛 = −1. Since 𝑛 < 0, 𝑦 = 0 is not a solution. Now divide by 𝑦 −1 to 1 𝑡−1 get 𝑦𝑦 ′ − 𝑦 2 = . Let 𝑧 = 𝑦 2 . Then 𝑧 ′ = 2𝑦𝑦 ′ and substituting gives 2 2 1 ′ 1 𝑡−1 𝑧 − 𝑧= . In standard form we get 𝑧 ′ − 𝑧 = (𝑡 − 1). An integrating 2 2 ∫ 2 factor is 𝑒 −1 𝑑𝑡 = 𝑒−𝑡 . Multiplying by 𝑒−𝑡 gives (𝑒−𝑡 𝑧)′ = (𝑡 − 1)𝑒−𝑡 . −𝑡 −𝑡 𝑡 Integration by parts gives √ 𝑒 𝑧 = −𝑡𝑒 + 𝑐 and thus 𝑧 = −𝑡 + 𝑐𝑒 . Since 2 𝑡 𝑧 = 𝑦 we have 𝑦 = ± 𝑐𝑒 − 𝑡, 𝑐 ∈ ℝ. 17. Note that 𝑦 = 0 is a solution. First divide both sides by 𝑦 3 to get 𝑦 −3 𝑦 ′ + 𝑧′ 𝑦 −2 = 𝑡. Let 𝑧 = 𝑦 −2 . Then 𝑧 ′ = −2𝑦 −3 𝑦 ′ . So +𝑧 = 𝑡. In standard form −2 ∫ we get 𝑧 ′ − 2𝑧 = −2𝑡. An integrating factor is 𝑒 −2 𝑑𝑡 = 𝑒−2𝑡 and hence 1 (𝑒−2𝑡 𝑧)′ = −2𝑡𝑒−2𝑡 . Integration by parts gives 𝑒−2𝑡 𝑧 = (𝑡 + )𝑒−2𝑡 + 𝑐 and 2 1 1 hence 𝑧 = 𝑡 + + 𝑐𝑒2𝑡 . Since 𝑧 = 𝑦 −2 we get 𝑦 = ± √ . 2 𝑡 + 12 + 𝑐𝑒2𝑡

𝑃 18. The logistic differential equation is 𝑃 ′ = 𝑟(1 − )𝑃 which can be written 𝑚 𝑟 𝑃 ′ − 𝑟𝑃 = − 𝑃 2 . Note that 𝑃 = 0 is a solution. Divide by 𝑃 2 to get 𝑚

34

1 Solutions

−𝑟 −𝑟 . Let 𝑧 = 𝑃 −1 . Then 𝑧 ′ = −𝑃 −2 𝑃 ′ and −𝑧 ′ −𝑟𝑧 = 𝑃 −2 𝑃 ′ −𝑟𝑃 −1 = 𝑚 𝑚 𝑟 𝑟 or 𝑧 ′ + 𝑟𝑧 = . An integrating factor is 𝑒𝑟𝑡 so (𝑒𝑟𝑡 𝑧)′ = 𝑒𝑟𝑡 . Integrating 𝑚 𝑚 𝑒𝑟𝑡 1 1 + 𝑚𝑐𝑒−𝑟𝑡 −𝑟𝑡 𝑟𝑡 + 𝑐. Solving for 𝑧 we get 𝑧 = + 𝑐𝑒 = . gives 𝑒 𝑧 = 𝑚 𝑚 𝑚 𝑚 𝑚 Since 𝑧 = 𝑃 −1 we get 𝑃 = . Now 𝑃0 = 𝑃 (0) = 1 + 𝑚𝑐𝑒−𝑟𝑡 1 + 𝑚𝑐 𝑚 − 𝑃0 . Substituting and simplifying gives and solving for 𝑐 we get 𝑐 = 𝑚𝑃0 𝑚𝑃0 𝑃 (𝑡) = . 𝑃0 + (𝑚 − 𝑃0 )𝑒−𝑟𝑡 19. Let 𝑧 = 2𝑡 − 2𝑦 + 1. Then 𝑧 ′ = 2 − 2𝑦 ′ and so 𝑦 ′ =

2 − 𝑧′ . Substituting 2

2 − 𝑧′ = 𝑧 −1 and in standard form we get 𝑧 ′ = 2 − 2𝑧 −1 , a we get 2 separable differential equation. Clearly, 𝑧 = 1 is an equilibrium solution. Assume for now that 𝑧 ∕= 1. Then separating variables and simplifying ( ) 𝑧 1 1 −1 using 1/(1−𝑧 ) = 𝑧−1 = 1+ 𝑧−1 gives 1 + 𝑧−1 𝑑𝑧 = 2 𝑑𝑡. Integrating we get 𝑧 + ln ∣𝑧 − 1∣ = 2𝑡 + 𝑐. Now substitute 𝑧 = 2𝑡 − 2𝑦 + 1 and simplify to get −2𝑦 + ln ∣2𝑡 − 2𝑦∣ = 𝑐, 𝑐 ∈ ℝ. (We absorb the constant 1 in 𝑐.) The equilibrium solution 𝑧 = 1 becomes 𝑦 = 𝑡.

20. Let 𝑧 = 𝑡 − 𝑦. Then 𝑧 ′ = 1 − 𝑦 ′ and so 𝑦 ′ = 1 − 𝑧 ′ . Substituting we get 1 − 𝑧 ′ = 𝑧 2 and in standard form we get 𝑧 ′ = 1 − 𝑧 2 . We see that 𝑧 = ±1 𝑑𝑧 are equilibrium solutions. Separating variables we get = 𝑑𝑡. Partial 1 − 𝑧2 ( ) 1 1 fractions gives + 𝑑𝑧 = 2 𝑑𝑡. Integrating and simplifying 1 − 𝑧 1 + 𝑧 2𝑡 1 + 𝑧 = 2𝑡 + 𝑐, 𝑐 ∈ ℝ. Solving for 𝑧 we get 𝑧 = 𝑘𝑒 − 1 , 𝑘 ∕= 0. gives ln 1 − 𝑧 𝑘𝑒2𝑡 + 1 However, the case 𝑘 = 0 gives the equilibrium solution 𝑧 = −1. Now 𝑘𝑒2𝑡 − 1 substitute 𝑧 = 𝑡 − 𝑦 and simplify to get 𝑦 = 𝑡 − 2𝑡 , 𝑘 ∈ ℝ. The 𝑘𝑒 + 1 equilibrium solution 𝑧 = 1 becomes 𝑦 = 𝑡 − 1. 21. Let 𝑧 = 𝑡 + 𝑦. Then 𝑧 ′ = 1 + 𝑦 ′ and substituting we get 𝑧 ′ − 1 = 𝑧 −2 . In 1 + 𝑧2 . Separating variables and simplifying standard form we get 𝑧 ′ = 𝑧2 ( ) 1 we get 1 − 𝑑𝑧 = 𝑑𝑡. Integrating we get 𝑧 − tan−1 𝑧 = 𝑡 + 𝑐. Now 1 + 𝑧2 let 𝑧 = 𝑡 + 𝑦 and simplify to get 𝑦 − tan−1 (𝑡 + 𝑦) = 𝑐, 𝑐 ∈ ℝ. 22. Let 𝑧 = 𝑡 − 𝑦. Then 𝑧 ′ = 1 − 𝑦 ′ and substituting we get 1 − 𝑧 ′ = sin 𝑧. In 𝜋 standard form we get 𝑧 ′ = 1 − sin 𝑧. Notice that 𝑧 = + 2𝜋𝑛, 𝑛 ∈ ℤ, are 2

1 Solutions

equilibrium solutions. Separating variable gives

35

𝑑𝑧 = 𝑑𝑡. Now 1 − sin 𝑧

1 1 + sin 𝑧 1 + sin 𝑧 = = sec2 𝑧 + sec 𝑧 tan 𝑧. = 2 1 − sin 𝑧 cos2 𝑧 1 − sin 𝑧 So integrating gives tan 𝑧 + sec 𝑧 = 𝑡 + 𝑐. Substituting, we get the implicit solution tan(𝑡 − 𝑦) + sec(𝑡 − 𝑦) = 𝑡 + 𝑐. For the equilibrium solution 𝜋 𝜋 𝑧 = + 2𝜋𝑛 we get 𝑦 = 𝑡 − − 2𝜋𝑛. 2 2 23. This is the same as Exercise 16 where the Bernoulli equation technique there used the substitution 𝑧 = 𝑦 2 . Here use the given substitution to get 𝑧 ′ = 2𝑦𝑦 ′ + 1. Substituting we get 𝑧 ′ − 1 = 𝑧 and in standard form 𝑧 ′ = 1+𝑧. Clearly, 𝑧 = −1 is an equilibrium solution. Separating variables 𝑑𝑧 gives = 𝑑𝑡 and integrating gives ln ∣1 + 𝑧∣ = 𝑡 + 𝑐, 𝑐 ∈ ℝ. Solving 1+𝑧 for 𝑧 we get 𝑧 = 𝑘𝑒𝑡 − 1, where 𝑘 ∕= 0. Since√𝑧 = 𝑦 2 + 𝑡 − 1 we get 𝑡 𝑦 2 + 𝑡 − 1 = 𝑘𝑒𝑡 − 1 and solving for 𝑦 gives √ 𝑦 = ± 𝑘𝑒 − 𝑡. The case 𝑘 = 0 gives the equilibrium solutions 𝑦 = ± −𝑡. 24. If 𝑧 = sin 𝑦 then 𝑧 ′ = (cos 𝑦)𝑦 ′ . Multiply the given differential equation by cos 𝑦 to get (cos 𝑦)𝑦 ′ = sin 𝑦 + 2 cos 𝑡. Substituting we get 𝑧 ′ = 𝑧 + 2 cos 𝑡 and in standard form 𝑧 ′ − 𝑧 = 2 cos 𝑡. An integrating factor is 𝑒−𝑡 so (𝑧𝑒−𝑡 )′ = 2(cos 𝑡)𝑒−𝑡 . Integrating by parts twice leads to 𝑧𝑒−𝑡 = (sin 𝑡 − cos 𝑡)𝑒−𝑡 + 𝑐 and hence 𝑧 = sin 𝑡 − cos 𝑡 + 𝑐𝑒𝑡 . Solving for 𝑦 gives 𝑦 = sin−1 (sin 𝑡 − cos 𝑡 + 𝑐𝑒𝑡 ), 𝑐 ∈ ℝ. 25. If 𝑧 = ln 𝑦 then 𝑧 ′ =

𝑦′ . Divide the given differential equation by 𝑦. Then 𝑦

𝑦′ + ln 𝑦 = 𝑡 and substitution gives 𝑧 ′ + 𝑧 = 𝑡. An integrating factor is 𝑦 𝑒𝑡 so (𝑒𝑡 𝑧)′ = 𝑡𝑒𝑡 . Integration (by parts) gives 𝑒𝑡 𝑧 = (𝑡 − 1)𝑒𝑡 + 𝑐 and so −𝑡 𝑧 = 𝑡 − 1 + 𝑐𝑒−𝑡 . Finally, solving for 𝑦 we get 𝑦 = 𝑒𝑡−1+𝑐𝑒 , 𝑐 ∈ ℝ. −𝑧 ′ 26. Let 𝑧 = 𝑒−𝑦 . Then 𝑧 ′ = −𝑒−𝑦 𝑦 ′ so −𝑧 ′ /𝑧 = 𝑦 ′ . Substituting gives = 𝑧 −1 −1. Multiply both sides by 𝑧 and put in standard form to get 𝑧 ′ −𝑧 = 1. 𝑧 An integrating factor is 𝑒−𝑡 so (𝑒−𝑡 𝑧)′ = 𝑒−𝑡 . Integrating we get 𝑒−𝑡 𝑧 = −𝑒−𝑡 + 𝑐 and so 𝑧 = 𝑐𝑒𝑡 − 1. Since 𝑧 = 𝑒−𝑦 , 𝑧 > 0 and this requires 𝑐 > 0. We thus get 𝑦 = − ln(𝑐𝑒𝑡 − 1), 𝑐 > 0.

Section 1.6 1. This can be written in the form 𝑀 (𝑡, 𝑦) + 𝑁 (𝑡, 𝑦)𝑦 ′ = 0 where 𝑀 (𝑡, 𝑦) = 𝑦 2 + 2𝑡 and 𝑁 (𝑡, 𝑦) = 2𝑡𝑦. Since 𝑀𝑦 (𝑡, 𝑦) = 2𝑦 = 𝑁𝑡 (𝑡, 𝑦), the equation is exact (see Equation (3.2.2)), and the general solution is given implicitly

36

1 Solutions

by 𝐹 (𝑡, 𝑦) = 𝑐 where the function 𝐹 (𝑡, 𝑦) is determined by 𝐹𝑡 (𝑡, 𝑦) = 𝑀 (𝑡, 𝑦) = 𝑦 2 + 2𝑡 and 𝐹𝑦 (𝑡, 𝑦) = 𝑁 (𝑡, 𝑦) = 2𝑡𝑦. These equations imply that 𝐹 (𝑡, 𝑦) = 𝑡2 + 𝑡𝑦 2 will work so the solutions are given implicitly by 𝑡2 + 𝑡𝑦 2 = 𝑐. 1 2. 𝑡𝑦 + 𝑦 2 − 𝑡2 = 𝑐 2 3. Not Exact 4. 𝑡𝑦 2 + 𝑡3 = 𝑐 5. Not Exact 6. 𝑡2 𝑦 + 𝑦 3 = 2 7. (𝑦 − 𝑡2 )2 − 2𝑡4 = 𝑐 8. 𝑦 =

1 2 𝑐 𝑡 − 3 𝑡

9. 𝑦 4 = 4𝑡𝑦 + 𝑐 10. 𝑏 + 𝑐 = 0

Section 1.7 1. We first change the variable 𝑡 to 𝑢 and write 𝑦 ′ (𝑢) = 𝑢𝑦(𝑢). Now integrate ∫𝑡 ∫𝑡 both sides from 1 to 𝑡 to get 1 𝑦 ′ (𝑢) 𝑑𝑢 = 1 𝑢𝑦(𝑢) 𝑑𝑢. Now the left side ∫𝑡 ′ ∫𝑡 is 1 𝑦 (𝑢) 𝑑𝑢 = 𝑦(𝑡) − 𝑦(1) = 𝑦(𝑡) − 1. Thus 𝑦(𝑡) = 1 + 1 𝑢𝑦(𝑢) 𝑑𝑢.

2. Change the variable 𝑡 to 𝑢 and write 𝑦 ′ (𝑢) = 𝑦 2 (𝑢). Now integrate both ∫𝑡 ∫𝑡 sides from 0 to 𝑡 to get 0 𝑦 ′ (𝑢) 𝑑𝑢 = 0 𝑦 2 (𝑢) 𝑑𝑢. The left side is 𝑦(𝑡) + 1 ∫𝑡 2 so 𝑦(𝑡) = −1 + 0 𝑦 (𝑢) 𝑑𝑢.

𝑢 − 𝑦(𝑢) . Now integrate 𝑢 + 𝑦(𝑢) 𝑢 − 𝑦(𝑢) 𝑑𝑢. The left side is 𝑢 + 𝑦(𝑢)

3. Change the variable 𝑡 to 𝑢 and write 𝑦 ′ (𝑢) = both sides from 0 to 𝑡 to get 𝑦(𝑡) − 1 so 𝑦(𝑡) = 1 +

∫𝑡 0

𝑦 ′ (𝑢) 𝑑𝑢 =

∫ 𝑡 𝑢 − 𝑦(𝑢) 𝑑𝑢. 0 𝑢 + 𝑦(𝑢)

∫𝑡 0

4. Change the variable 𝑡 to 𝑢 and write 𝑦 ′ (𝑢) = 1 + 𝑢2 . Now integrate both ∫𝑡 ∫𝑡 sides from 0 to 𝑡 to get 0 𝑦 ′ (𝑢) 𝑑𝑢 = 0 (1 + 𝑢2 ) 𝑑𝑢. The left side is 𝑦(𝑡) ∫𝑡 so 𝑦(𝑡) = 0 (1 + 𝑢2 ) 𝑑𝑢.

5. The corresponding integral equation is 𝑦(𝑡) = 1 + have

∫𝑡 1

𝑢𝑦(𝑢) 𝑑𝑢. We then

1 Solutions

37

𝑦0 (𝑡) = 1 ) 𝑡 𝑢2 𝑡2 1 1 + 𝑡2 𝑦1 (𝑡) = 1 + 𝑢 ⋅ 1 𝑑𝑢 = 1 + = 1 + − = 2 1 2 2 2 1 ) ( ) ∫ 𝑡 ( 𝑡 1 + 𝑢2 𝑢2 𝑢4 5 𝑡2 𝑡4 𝑦2 (𝑡) = 1 + 𝑢 𝑑𝑢 = 1 + + = + + 2 4 8 1 8 4 8 1 ) ( 2 ) ∫ 𝑡( 3 5 4 6 𝑡 5𝑢 𝑢 𝑢 5𝑢 𝑢 𝑢 𝑦3 (𝑡) = 1 + + + 𝑑𝑢 = 1 + + + 8 4 8 16 16 48 1 1 ∫

𝑡

(

29 5𝑡2 𝑡4 𝑡6 + + + . 48 16 16 48

=

6. The corresponding integral equation is ∫ 𝑡 𝑦(𝑡) = 1 + (𝑢 − 𝑦(𝑢)) 𝑑𝑢. 0

We then have 𝑦0 (𝑡) = 1

) 𝑡 𝑡2 𝑢2 − 𝑢 = 1 − 𝑡 + 2 2 0 0 ( )) ( ) 𝑡 ∫ 𝑡( 𝑢3 𝑢2 𝑦2 (𝑡) = 1 + 𝑢− 1−𝑢+ 𝑑𝑢 = 1 + −𝑢 + 𝑢2 − 2 6 0 0

𝑦1 (𝑡) = 1 +

∫

𝑡

(𝑢 − 1) 𝑑𝑢 = 1 +

(

𝑡3 = 1 − 𝑡 + 𝑡2 − 6 ( )) ∫ 𝑡( 𝑡3 𝑡4 𝑢3 2 𝑦3 (𝑡) = 1 + 𝑢− 1−𝑢+𝑢 − 𝑑𝑢 = 1 − 𝑡 + 𝑡2 − + 6 3 4! 0 ( )) ∫ 𝑡( 3 4 𝑢 𝑢 𝑡3 𝑡4 𝑡5 𝑦4 (𝑡) = 1 + 𝑢 − 1 − 𝑢 + 𝑢2 − + 𝑑𝑢 = 1 − 𝑡 + 𝑡2 − + − . 3 4! 3 12 5! 0

7. The corresponding integral equation is 𝑦(𝑡) = have

∫𝑡 0

(𝑢 + 𝑦 2 (𝑢)) 𝑑𝑢. We then

𝑦0 (𝑡) = 0 ∫ 𝑡 𝑡2 𝑦1 (𝑡) = (𝑢 + 0) 𝑑𝑢 = 2 0 ( 2 )2 ) ) ∫ 𝑡( ∫ 𝑡( 𝑢 𝑢4 𝑡2 𝑡5 𝑦2 (𝑡) = 𝑢+ 𝑑𝑢 = 𝑢+ 𝑑𝑢 = + 2 4 2 20 0 0 ) ( 2 ) ( ) ∫ 𝑡( ∫ 2 𝑡 𝑢 𝑢5 𝑢4 𝑢7 𝑢10 𝑦3 (𝑡) = 𝑢+ + 𝑑𝑢 = 𝑢+ + + 𝑑𝑢 2 20 4 20 400 0 0 =

𝑡2 𝑡5 𝑡8 𝑡11 + + + . 2 20 160 4400

38

1 Solutions

8. The corresponding integral equation is ∫ 𝑡 𝑦(𝑡) = 1 + ((𝑦(𝑢))3 − 𝑦(𝑢)) 𝑑𝑢. 0

We now have 𝑦0 (𝑡) = 1 𝑦1 (𝑡) = 1 + 𝑦2 (𝑡) = 1

∫

𝑡 0

(13 − 1) 𝑑𝑢 = 1 +

∫

𝑡

0 𝑑𝑢 = 1 0

𝑦3 (𝑡) = 1 9. The corresponding integral equation is ∫ 𝑡 𝑦(𝑡) = (1 + (𝑢 − 𝑦(𝑢))2 ) 𝑑𝑢. 0

We then have 𝑦0 (𝑡) = 0 ) 𝑡 ( ∫ 𝑡 ( ) 𝑡3 𝑢3 2 𝑦1 (𝑡) = 1 + (𝑢 − 0) 𝑑𝑢 = 𝑢 + =𝑡+ 3 0 3 0 ) ( ( )) ( ) ∫ 𝑡( ∫ 2 𝑡 𝑢3 𝑢6 𝑦2 (𝑡) = 1+ 𝑢− 𝑢+ 𝑑𝑢 = 1+ 𝑑𝑢 3 9 0 0 ) 𝑡 ( 𝑡7 𝑢7 = 𝑡 + = 𝑢+ 63 0 7 ⋅ 32 ) ∫ 𝑡( 14 𝑢 𝑡15 𝑦3 (𝑡) = 1 + 2 4 𝑑𝑢 = 𝑡 + 7 ⋅3 15 ⋅ 72 ⋅ 34 0 ) ∫ 𝑡( 𝑢30 𝑡31 𝑦4 (𝑡) = 1 + 2 4 8 𝑑𝑢 = 𝑡 + 15 ⋅ 7 ⋅ 3 31 ⋅ 152 ⋅ 74 ⋅ 38 0 ) ∫ 𝑡( 𝑡63 𝑢62 𝑦5 (𝑡) = 1+ 2 𝑑𝑢 = 𝑡 + 4 8 1 2 31 ⋅ 15 ⋅ 7 ⋅ 3 6 63 ⋅ 31 ⋅ 154 ⋅ 78 ⋅ 316 0 10. The right hand side is 𝐹 (𝑡, 𝑦) = 1 + 𝑦 2 . Then 𝐹𝑦 (𝑡, 𝑦) = 2𝑦. Both 𝐹 and 𝐹𝑦 are continuous in the whole (𝑡, 𝑦)-plane and thus are continuous on any rectangle containing the origin (0, 0). Picard’s theorem applies and we can conclude there is a unique solution on an interval about 0. √ 11. The right hand side is 𝐹 (𝑡, 𝑦) = 𝑦. If ℛ is any rectangle about (1, 0) then there are 𝑦-coordinates that are negative. Hence 𝐹 is not defined on ℛ and Picards’ theorem does not apply.

1 Solutions

39

1 √ . Choose a 2 𝑦 rectangle ℛ about (0, 1) that lies above the 𝑡-axis. Then both 𝐹 and 𝐹𝑦 are continuous on ℛ. Picard’s theorem applies and we can conclude there is a unique solution on an interval about 0.

12. The right hand side is 𝐹 (𝑡, 𝑦) =

√

𝑦. Then 𝐹𝑦 (𝑡, 𝑦) =

𝑡−𝑦 −2𝑡 . Then 𝐹𝑦 (𝑡, 𝑦) = . Choose 𝑡+𝑦 (𝑡 + 𝑦)2 a rectangle ℛ about (0, −1) that contains no points on the line 𝑡 + 𝑦 = 0. Then both 𝐹 and 𝐹𝑦 are continuous on ℛ. Picard’s theorem applies and we can conclude there is a unique solution on an interval about 0.

13. The right hand side is 𝐹 (𝑡, 𝑦) =

𝑡−𝑦 , which is not defined at the initial 𝑡+𝑦 condition (𝑡0 , 𝑦0 ) = (1, −1). Thus Picard’s theorem does not apply. ∫𝑡 15. The corresponding integral equation is 𝑦(𝑡) = 1 + 0 𝑎𝑦(𝑢) 𝑑𝑢. We thus have 14. The right hand side is 𝐹 (𝑡, 𝑦) =

𝑦0 (𝑡) = 1 𝑦1 (𝑡) = 1 +

∫

𝑡

𝑎 𝑑𝑢 = 1 + 𝑎𝑡 0

∫ 𝑡 𝑎 2 𝑡2 𝑎(1 + 𝑎𝑢) 𝑑𝑢 = 1 + (𝑎 + 𝑎2 𝑢) 𝑑𝑢 = 1 + 𝑎𝑡 + 2 0 0 ) ∫ 𝑡 ( 𝑎 2 𝑢2 𝑎 2 𝑡2 𝑎 3 𝑡3 𝑦3 (𝑡) = 1 + 𝑎 1 + 𝑎𝑢 + 𝑑𝑢 = 1 + 𝑎𝑡 + + 2 2 3! 0 .. . 𝑎 2 𝑡2 𝑎 𝑛 𝑡𝑛 𝑦𝑛 (𝑡) = 1 + 𝑎𝑡 + +⋅⋅⋅ + . 2 𝑛! 𝑦2 (𝑡) = 1 +

∫

𝑡

𝑎 𝑘 𝑡𝑘 . We recognize this sum as the first 𝑛 𝑘! terms of the Taylor series expansion for 𝑒𝑎𝑡 . Thus the limiting function is 𝑦(𝑡) = lim𝑛→∞ 𝑦𝑛 (𝑡) = 𝑒𝑎𝑡 . It is straightforward to verify that it is a solution. If 𝐹 (𝑡, 𝑦) = 𝑎𝑦 then 𝐹𝑦 (𝑡, 𝑦) = 𝑎. Both 𝐹 and 𝐹𝑦 are continuous on the whole (𝑡, 𝑦)-plane. By Picard’s theorem, Theorem 5, 𝑦(𝑡) = 𝑒𝑎𝑡 is the only solution to the given initial value problem. We can write 𝑦𝑛 (𝑡) =

∑𝑛

𝑘=0

16. 1. The equation is separable so separate the variables to get 𝑦 −2 𝑑𝑦 = 𝑑𝑡. Integrating gives −𝑦 −1 = 𝑡 + 𝑐 and the initial condition 𝑦(0) = 1 implies that the integration constant 𝑐 = −1, so that the exact solution is 1 𝑦(𝑡) = = 1 + 𝑡 + 𝑡2 + 𝑡3 + 𝑡4 + ⋅ ⋅ ⋅ ; ∣𝑡∣ < 1. 1−𝑡 2. To apply Picard’s method, let 𝑦0 = 1 and define

40

1 Solutions

𝑦1 (𝑡) = 1 + 𝑦2 (𝑡) = 1 +

∫

∫

∫

𝑡

(𝑦0 (𝑢))2 𝑑𝑢 = 1 +

0 𝑡

(𝑦1 (𝑢))2 𝑑𝑢 = 1 +

0 𝑡

2

∫

∫

𝑡

1 𝑑𝑢 = 1 + 𝑡; 0 𝑡

(1 + 𝑢)2 𝑑𝑢 = 1 + 𝑡 + 𝑡2 +

0

∫ 𝑡(

𝑢3 1+𝑢+𝑢 + 3 2

𝑡3 ; 3

)2

𝑑𝑢 (𝑦2 (𝑢)) 𝑑𝑢 = 1 + 0 ) ∫ 𝑡( 8 5 2 1 = 1+ 1 + 2𝑢 + 3𝑢2 + 𝑢3 + 𝑢4 + 𝑢5 + 𝑢6 𝑑𝑢 3 3 3 9 0 2 1 1 1 = 1 + 𝑡 + 𝑡 2 + 𝑡 3 + 𝑡4 + 𝑡5 + 𝑡6 + 𝑡7 . 3 3 9 63

𝑦3 (𝑡) = 1 +

0

Comparing 𝑦3 (𝑡) to the exact solution, we see that the series agree up to order 3. 17. Let 𝐹 (𝑡, 𝑦) = cos(𝑡 + 𝑦). Then 𝐹𝑦 (𝑡, 𝑦) = − sin(𝑡 + 𝑦). Let 𝑦1 and 𝑦2 be arbitrary real numbers. Then by the mean value theorem there is a number 𝑦0 in between 𝑦1 and 𝑦2 such that ∣𝐹 (𝑡, 𝑦1 ) − 𝐹 (𝑡, 𝑦2 )∣ = ∣sin(𝑡 + 𝑦0 )∣ ∣𝑦1 − 𝑦2 ∣ ≤ ∣𝑦1 − 𝑦2 ∣ . It follows that 𝐹 (𝑡, 𝑦) is Lipschitz on any strip. Theorem 10 implies there is a unique solution on all of ℝ. 18. Let 𝐹 (𝑡, 𝑦) = −𝑝(𝑡)𝑦 + 𝑓 (𝑡). Since 𝑝(𝑡) and 𝑓 (𝑡) are continuous on [𝑎, 𝑏] it follows that 𝐹 (𝑡, 𝑦) is continuous on the strip {(𝑡, 𝑦) : 𝑡 ∈ [𝑎, 𝑏], 𝑦 ∈ ℝ}. Further more 𝑝(𝑡) is bounded: i.e. there is a number 𝐴 such that ∣𝑝(𝑡)∣ ≤ 𝐴, for all 𝑡 ∈ [𝑎, 𝑏]. Let 𝑡 ∈ [𝑎, 𝑏] and 𝑦1 and 𝑦2 be arbitrary real numbers. Then ∣𝐹 (𝑡, 𝑦1 ) − 𝐹 (𝑡, 𝑦2 )∣ ≤ ∣−𝑝(𝑡)𝑦1 + 𝑓 (𝑡) − (−𝑝(𝑡)𝑦2 + 𝑓 (𝑡))∣ = ∣𝑝(𝑡)∣ ∣𝑦1 − 𝑦2 ∣ ≤ 𝐴 ∣𝑦1 − 𝑦2 ∣ .

It follows that 𝐹 (𝑡, 𝑦) is Lipschitz with Lipschitz constant 𝐴. By Theorem 10, 𝑦 ′ + 𝑝(𝑡)𝑦 = 𝑓 (𝑡), 𝑦(𝑡0 ) = 𝑦0 has a unique solution on the entire interval [𝑎, 𝑏]. 19. 1. First assume that 𝑡 ∕= 0. Then 𝑡𝑦 ′ = 2𝑦 − 𝑡 is linear and in standard ∫form becomes 𝑦 ′ − 2𝑦/𝑡 = −1. An integrating factor is 𝜇(𝑡) = 𝑒 (−2/𝑡) 𝑑𝑡 = 𝑡−2 and multiplying both sides by 𝜇 gives 𝑡−2 𝑦 ′ − 2𝑡−3 𝑦 = −𝑡−2 . This simplifies to (𝑡−2 𝑦)′ = −𝑡−2 . Now integrate to get 𝑡−2 𝑦 = 𝑡−1 + 𝑐 or 𝑦(𝑡) = 𝑡 + 𝑐𝑡2 . We observe that this solution is also valid for 𝑡 = 0. Graphs are given below for various values of 𝑐.

1 Solutions

41

4 3 2 1 0 −1 −2 −3 −4 −6

−4

−2

0 t

2

4

6

Graph of 𝑦(𝑡) = 𝑡 + 𝑐𝑡3 for various 𝑐 2. Every solution satisfies 𝑦(0) = 0. There is no contradiction to Theorem 2 5 since, in standard form, the equation is 𝑦 ′ = 𝑦 − 1 = 𝐹 (𝑡, 𝑦) and 𝑡 𝐹 (𝑡, 𝑦) is not continuous for 𝑡 = 0. 20. 1. If 𝐹 (𝑡, 𝑦) = 𝑦 2 then 𝐹𝑦 (𝑡, 𝑦) = 2𝑦. Both are continuous on any rectangle that contains (𝑡0 , 𝑦0 ). Hence Theorem 5 applies and implies there is a unique solution on an interval that contains 𝑡0 . 1 2. 𝑦(𝑡) = 0 is a solution defined for all 𝑡; 𝑦(𝑡) = is a solution defined 1−𝑡 on (−∞, 1) . 21. No. Both 𝑦1 (𝑡) and 𝑦2 (𝑡) would be solutions to the initial value problem 𝑦 ′ = 𝐹 (𝑡, 𝑦), 𝑦(0) = 0. If 𝐹 (𝑡, 𝑦) and 𝐹𝑦 (𝑡, 𝑦) are both continuous near (0, 0), then the initial value problem would have a unique solution by Theorem 5. 22. No, Both 𝑦1 (𝑡) and 𝑦2 (𝑡) would be solutions to the initial value problem 𝑦 ′ = 𝐹 (𝑡, 𝑦), 𝑦(0) = 1. If 𝐹 (𝑡, 𝑦) and 𝐹𝑦 (𝑡, 𝑦) are both continuous near (0, 1), then the initial value problem would have a unique solution by Theorem 5. 23. For 𝑡 < 0 we have 𝑦1′ (𝑡) = 0 and for 𝑡 > 0 we have 𝑦1′ (𝑡) = 3𝑡2 . For 𝑡 = 0 𝑦1 (ℎ) − 𝑦1 (0) 𝑦1 (ℎ) we calculate 𝑦1′ (0) = limℎ→0 = limℎ→0 . To compute ℎ−0 ℎ this limit we show the left hand and right hand limits agree. We get 𝑦1 (ℎ) ℎ ℎ→0 𝑦1 (ℎ) lim ℎ ℎ→0− { 0, It follows that 𝑦1′ (𝑡) = 3𝑡2 lim+

ℎ3 = lim+ ℎ2 = 0 ℎ ℎ→0 ℎ→0 0 = lim =0 ℎ→0+ ℎ = lim+

for 𝑡 < 0 and so for 𝑡 ≥ 0

42

1 Solutions

=

{

3𝑦1 (𝑡) =

{

𝑡𝑦1′ (𝑡)

0, 3𝑡3

for 𝑡 < 0 for 𝑡 ≥ 0

On the other hand, 0, 3𝑡3

for 𝑡 < 0 for 𝑡 ≥ 0

It follows that 𝑦1 is a solution. It is trivial to see that 𝑦2 (𝑡) is a solution. 3 There is no contraction to Theorem 5 since, in standard form 𝑦 ′ = 𝑦 = 𝑡 𝐹 (𝑡, 𝑦) has a discontinuous 𝐹 (𝑡, 𝑦) near (0, 0). So Picard’s theorem does not even apply.

Section 2.1 1. ℒ {3𝑡 + 1} (𝑠) ∫ ∞ = (3𝑡 + 1)𝑒−𝑠𝑡 𝑑𝑡 0 ∫ ∞ ∫ ∞ =3 𝑡𝑒−𝑠𝑡 𝑑𝑡 + 𝑒−𝑠𝑡 𝑑𝑡 0 0 ∞ ∞ ( ) ∫ 1 ∞ −𝑠𝑡 −1 −𝑠𝑡 𝑡 −𝑠𝑡 =3 𝑒 + 𝑒 𝑑𝑡 + 𝑒 −𝑠 𝑠 0 𝑠 0 ∞ ) (( ) ( 0 ) 1 −1 −𝑠𝑡 1 =3 + 𝑒 𝑠 𝑠 𝑠 0 3 1 = 2+ . 𝑠 𝑠 2. { } ℒ 5𝑡 − 9𝑒𝑡 (𝑠) ∫ ∞ = 𝑒−𝑠𝑡 (5𝑡 − 9𝑒𝑡 ) 𝑑𝑡 0 ∫ ∞ ∫ ∞ −𝑠𝑡 =5 𝑡𝑒 𝑑𝑡 − 9 𝑒−𝑠𝑡 𝑒𝑡 𝑑𝑡 0 0 ∞ ( ) ∫ ∫ ∞ −𝑡 −𝑠𝑡 1 ∞ −𝑠𝑡 =5 𝑒 + 𝑒 𝑑𝑡 − 9 𝑒−(𝑠−1)𝑡 𝑑𝑡 𝑠 𝑠 0 0 0 ( ) 1 1 = 5 0+ 2 −9 𝑠 𝑠−1 5 9 = 2− . 𝑠 𝑠−1

1 Solutions

43

3. { } ℒ 𝑒2𝑡 − 3𝑒−𝑡 (𝑠) ∫ ∞ = 𝑒−𝑠𝑡 (𝑒2𝑡 − 3𝑒−𝑡 ) 𝑑𝑡 0 ∫ ∞ ∫ ∞ −𝑠𝑡 2𝑡 = 𝑒 𝑒 𝑑𝑡 − 3 𝑒−𝑠𝑡 𝑒−𝑡 𝑑𝑡 0 0 ∫ ∞ ∫ ∞ = 𝑒−(𝑠−2)𝑡 𝑑𝑡 − 3 𝑒−(𝑠+1)𝑡 𝑑𝑡 0

0

1 3 = − . 𝑠−2 𝑠+1 4.

= = = =

{ } ℒ 𝑡𝑒−3𝑡 (𝑠) ∫ ∞ 𝑒−𝑠𝑡 𝑡𝑒−3𝑡 𝑑𝑡 ∫0 ∞ 𝑡𝑒−(𝑠+3)𝑡 𝑑𝑡 0 ∞ ∫ ∞ 𝑡𝑒−(𝑠+3)𝑡 1 + 𝑒−(𝑠+3)𝑡 𝑑𝑡 −(𝑠 + 3) 0 𝑠+3 0 1 . (𝑠 + 3)2

{ } { } 5. ℒ 5𝑒2𝑡 = 5ℒ 𝑒2𝑡 =

5 𝑠−2

{ } { } { } 6. ℒ 3𝑒−7𝑡 − 7𝑡3 = 3ℒ 𝑒−7𝑡 − 7ℒ 𝑡3 =

3 3! 3 42 −7 4 = − 𝑠+7 𝑠 𝑠 + 7 𝑠4

{ } { } 5 4 2 7. ℒ 𝑡2 − 5𝑡 + 4 = ℒ 𝑡2 − 5ℒ {𝑡} + 4ℒ {1} = 3 − 2 + 𝑠 𝑠 𝑠 {3 } { } { } 6 2 1 1 8. ℒ 𝑡 + 𝑡2 + 𝑡 + 1 = ℒ 𝑡3 + ℒ 𝑡2 + ℒ {𝑡} + ℒ {1} = 4 + 3 + 2 + 𝑠 𝑠 𝑠 𝑠 { } { } { } 1 7 9. ℒ 𝑒−3𝑡 + 7𝑡𝑒−4𝑡 = ℒ 𝑒−3𝑡 + 7ℒ 𝑡𝑒−4𝑡 = + 𝑠 + 3 (𝑠 + 4)2 { } 10. ℒ 𝑡2 𝑒4𝑡 (𝑠) =

2 (𝑠 − 4)3

11. ℒ {cos 2𝑡 + sin 2𝑡} = ℒ {cos 2𝑡} + ℒ {sin 2𝑡} =

𝑠 2 𝑠+2 + = 2 𝑠2 + 2 2 𝑠2 + 2 2 𝑠 +4

44

1 Solutions

12. ℒ {𝑒𝑡 (𝑡 − cos 4𝑡)} (𝑠) = ℒ {𝑡𝑒−𝑡 } (𝑠) − ℒ {𝑒−𝑡 cos 4𝑡} (𝑠) = 𝑠−1 . (𝑠 − 1)2 + 16

{ } { } 13. ℒ (𝑡𝑒−2𝑡 )2 (𝑠) = ℒ 𝑡2 𝑒−4𝑡 (𝑠) = √ } { 14. ℒ 𝑒−𝑡/3 cos 6𝑡 (𝑠) =

1 − (𝑠 − 1)2

2 (𝑠 + 4)3

𝑠 + 31 (𝑠 + 13 )2 + 6

{ } { } { } { } 15. ℒ (𝑡 + 𝑒2𝑡 )2 (𝑠) = ℒ 𝑡2 + 2𝑡𝑒2𝑡 + 𝑒4𝑡 (𝑠) = ℒ 𝑡2 (𝑠)+2ℒ 𝑡𝑒2𝑡 (𝑠)+ { } 2 2 1 ℒ 𝑒4𝑡 (𝑠) = 3 + + 2 𝑠 (𝑠 − 2) 𝑠−4

16. ℒ {5 cos 3𝑡 − 3 sin 3𝑡 + 4} (𝑠) = 5ℒ {cos 3𝑡} (𝑠)−3ℒ {sin 3𝑡} (𝑠)+ℒ {4} (𝑠) = 𝑠 3 4 5𝑠 − 9 4 5 2 −3 2 + = 2 + 𝑠 + 32 𝑠 + 32 𝑠 𝑠 +9 𝑠 { 4} { } 𝑡 4! 24 17. ℒ (𝑠) = ℒ 𝑡4 𝑒−4𝑡 (𝑠) = = 𝑒4𝑡 (𝑠 + 4)5 (𝑠 + 4)5 { } { } { } 18. ℒ 𝑒5𝑡 (8 cos 2𝑡 + 11 sin 2𝑡) (𝑠) = 8ℒ 𝑒5𝑡 cos 2𝑡 (𝑠)+11ℒ 𝑒5𝑡 sin 2𝑡 (𝑠) = 8(𝑠 − 5) 22 8𝑠 − 18 + = 2 (𝑠 − 5)2 + 4 (𝑠 − 5)2 + 4 𝑠 − 10𝑠 + 29 ( )′ { 3𝑡 } ( { 3𝑡 })′ 3 1 19. ℒ 𝑡𝑒 (𝑠) = − ℒ 𝑒 (𝑠) = − = 𝑠−3 (𝑠 − 3)2 ( )′ 𝑠 𝑠2 − 9 ′ 20. ℒ {𝑡 cos 3𝑡} (𝑠) = − (ℒ {cos 3𝑡}) = − 2 = 2 𝑠 +9 (𝑠 + 9)2 { } 21. Here we use the transform derivative principle twice to get ℒ 𝑡2 sin 2𝑡 (𝑠) = ( )′′ ( )′ 2 −4𝑠 12𝑠2 − 16 (ℒ {sin 2𝑡})′′ = = = 𝑠2 + 4 (𝑠2 + 4)2 (𝑠2 + 4)3 )′ ( ( )′ 𝑠 𝑠+1 ′ −𝑡 −𝑡 22. ℒ {𝑡𝑒 cos 𝑡} (𝑠) = −ℒ {𝑒 cos 𝑡} (𝑠) = − = − = 𝑠2 + 1 𝑠→𝑠+1 𝑠2 + 2𝑠 + 2 𝑠2 + 2𝑠 2 (𝑠 + 2𝑠 + 2)2 ( ( 2 ))′ 𝑠 2𝑠 2 ′ 23. ℒ {𝑡𝑓 (𝑡)} (𝑠) = −ℒ {𝑓 (𝑡)} (𝑠) = − ln = 2 − 2 𝑠 +1 𝑠 +1 𝑠 24. ℒ

{

1 − cos 5𝑡 𝑡

}

= 5ℒ

{

1 − cos 5𝑡 5𝑡

}

=

1 (𝑠/5)2 1 𝑠2 ln = ln . 2 (𝑠/5)2 + 1 2 𝑠2 + 25

1 Solutions

45

1 1 ln((𝑠/6) + 1) ln(𝑠 + 6) − ln 6 ℒ {Ei(𝑡)} (𝑠)∣𝑠→𝑠/6 = = 6 6 𝑠/6 𝑠 ( ) { } 1 1 1 𝑠 𝑠2 + 2𝑏2 + 2 = 26. ℒ cos2 𝑏𝑡 (𝑠) = ℒ {1 + cos 2𝑏𝑡} (𝑠) = 2 2 2 𝑠 𝑠 + 4𝑏 𝑠(𝑠2 + 4𝑏2 )

25. ℒ {Ei(6𝑡)} (𝑠) =

{ } 1 1 27. We use the identity sin2 𝜃 = (1−cos 2𝜃). ℒ sin2 𝑏𝑡 (𝑠) = ℒ {1 − cos 2𝑏𝑡} (𝑠) = 2 2 ( ) 1 1 𝑠 2𝑏2 − 2 = ; 2 2 2 𝑠 𝑠 + 4𝑏 𝑠(𝑠 + 4𝑏2 )

1 28. We use the identity sin 2𝜃 = 2 sin 𝜃 cos 𝜃. ℒ {sin 𝑏𝑡 cos 𝑏𝑡} (𝑠) = ℒ {sin 2𝑏𝑡} (𝑠) = 2 1 2𝑏 𝑏 = 2 ; 2 𝑠2 + 4𝑏2 𝑠 + 4𝑏2 29. We use the identity sin 𝑎𝑡 cos 𝑏𝑡 =

1 (sin(𝑎 + 𝑏)𝑡 + sin(𝑎 − 𝑏)𝑡). 2

1 (ℒ {sin(𝑎 + 𝑏)𝑡} + ℒ {sin(𝑎 − 𝑏)𝑡}) 2( ) 1 𝑎−𝑏 𝑎+𝑏 = + . 2 𝑠2 + (𝑎 − 𝑏)2 𝑠2 + (𝑎 + 𝑏)2

ℒ {sin 𝑎𝑡 cos 𝑏𝑡} =

30. ℒ {cosh 𝑏𝑡} = 31. ℒ {sinh 𝑏𝑡} =

}) 1 1 ( { 𝑏𝑡 ℒ 𝑒 + 𝑒−𝑏𝑡 = 2 2

}) 1 1 ( { 𝑏𝑡 ℒ 𝑒 − 𝑒−𝑏𝑡 = 2 2

(

1 1 + 𝑠+𝑏 𝑠−𝑏

)

=

(

1 1 − 𝑠+𝑏 𝑠−𝑏

)

=

𝑠2

𝑠 − 𝑏2

𝑏 𝑠2 − 𝑏 2

32. Let 𝑓 (𝑡) = 𝑒𝑎𝑡 . Then 𝑓 ′ (𝑡) = 𝑎𝑒𝑎𝑡 and 𝑓 (𝑡)∣𝑡=0 = 1. Thus 𝑎ℒ {𝑒𝑎𝑡 } = 1 ℒ {𝑓 ′ (𝑡)} = 𝑠ℒ {𝑒𝑎𝑡 } − 1. Solving for ℒ {𝑒𝑎𝑡 } gives ℒ {𝑒𝑎𝑡 } = . 𝑠−𝑎 33. Let 𝑓 (𝑡) = sinh 𝑏𝑡. Then 𝑓 ′ (𝑡) = 𝑏 cosh 𝑡 and 𝑓 ′′ (𝑡) = 𝑏2 sinh 𝑡. Further, 𝑓 (𝑡)∣𝑡=0 = 0 and 𝑓 ′ (𝑡)∣𝑡=0 = 𝑏. Thus 𝑏2 ℒ {sinh 𝑏𝑡} = ℒ {𝑓 ′′ (𝑡)} = 𝑠2 ℒ {𝑓 (𝑡)} − 𝑠𝑓 (0) − 𝑓 ′ (0) = 𝑠2 ℒ {𝑓 (𝑡)} − 𝑏. Solving for ℒ {𝑓 (𝑡)} gives 𝑏 ℒ {sinh 𝑏𝑡} = 2 . 𝑠 − 𝑏2 34. Let 𝑓 (𝑡) = cosh 𝑏𝑡. Then 𝑓 ′ (𝑡) = 𝑏 sinh 𝑡 and 𝑓 ′′ (𝑡) = 𝑏2 cosh 𝑡. Further, 𝑓 (𝑡)∣𝑡=0 = 1 and 𝑓 ′ (𝑡)∣𝑡=0 = 0. Thus 𝑏2 ℒ {cosh 𝑏𝑡} = ℒ {𝑓 ′′ (𝑡)} = 𝑠2 ℒ {𝑓 (𝑡)} − 𝑠𝑓 (0) − 𝑓 ′ (0) = 𝑠2 ℒ {𝑓 (𝑡)} − 𝑠. Solving for ℒ {𝑓 (𝑡)} gives 𝑠 ℒ {cosh 𝑏𝑡} = 2 . 𝑠 − 𝑏2 ∫𝑡 ∫0 35. Let 𝑔(𝑡) = 0 𝑓 (𝑢) 𝑑𝑢 and note that 𝑔 ′ (𝑡) = 𝑓 (𝑡) and 𝑔(0) = 0 𝑓 (𝑢) 𝑑𝑢 = 0. Now apply the input derivative formula to 𝑔(𝑡), to get

46

1 Solutions

𝐹 (𝑠) = ℒ {𝑓 (𝑡)} (𝑠) = ℒ {𝑔 ′ (𝑡)} (𝑠) = 𝑠ℒ {𝑔(𝑡)} (𝑠) − 𝑔(0) = 𝑠𝐺(𝑠). Solving for 𝐺(𝑠) gives 𝐺(𝑠) = 𝐹 (𝑠)/𝑠. 36. For 𝑡 ≥ 0, 𝑒𝑎𝑡 ≤ 𝑒𝑏𝑡 . Thus ∣𝑓 (𝑡)∣ ≤ 𝐾𝑒𝑎𝑡 ≤ 𝐾𝑒𝑏𝑡 . So 𝑓 is of exponential type of order 𝑏. 37. Suppose 𝑓 is on exponential type of order 𝑎 and 𝑔 is of exponential type of order 𝑏. Suppose 𝑎 ≤ 𝑏. Then there are numbers 𝐾 and 𝐿 so that ∣𝑓 (𝑡)∣ ≤ 𝐾𝑒𝑎𝑡 and ∣𝑔(𝑡)∣ ≤ 𝐿𝑒𝑏𝑡 . Now ∣𝑓 (𝑡) + 𝑔(𝑡)∣ ≤ ∣𝑓 (𝑡)∣ + ∣𝑔(𝑡)∣ ≤ 𝐾𝑒𝑎𝑡 + 𝐿𝑒𝑏𝑡 ≤ (𝐾 + 𝐿)𝑒𝑏𝑡 . If follows that 𝑓 + 𝑔 is of exponential type of order 𝑏. 38. Suppose 𝑓 is of exponential type of order 𝑎 and 𝑔 is of exponential type of order 𝑏. Then there are numbers 𝐾 and 𝐿 so that ∣𝑓 (𝑡)∣ ≤ 𝐾𝑒𝑎𝑡 and ∣𝑔(𝑡)∣ ≤ 𝐿𝑒𝑏𝑡 . Now ∣𝑓 (𝑡)𝑔(𝑡)∣ ≤ 𝐾𝑒𝑎𝑡 𝐿𝑒𝑏𝑡 = 𝐾𝐿𝑒(𝑎+𝑏)𝑡 . If follows that 𝑓 + 𝑔 is of exponential type of order 𝑎 + 𝑏. 39. If 𝑓 is bounded by 𝐾, say, then 𝑓 (𝑡) ≤ 𝐾𝑒0𝑡 . So 𝑓 is of exponential type of order 0. 40. Suppose 𝑓 is of exponential type of order 𝑎 in the sense given in the text. Then 𝑁 can be chosen to be 0 and 𝑓 satisfies the definition given in the statement of the problem. Now suppose 𝑓 satisfies the definition given in the statement of the problem. I.e. there is a 𝐾 ≥ 0 and 𝑁 ≥ 0 so that ∣𝑓 ∣ ≤ 𝐾𝑒𝑎𝑡 for 𝑡 ≥ 𝑁 . Since ∣𝑓 ∣ is continuous on the interval [0, 𝑁 ] it has a maximum, 𝐾1 , say. It follows that ∣𝑓 ∣ ≤ 𝐾1 ≤ 𝐾1 𝑒𝑎𝑡 on [0, 𝑁 ] and hence ∣𝑓 ∣ ≤ (𝐾 + 𝐾1 )𝑒𝑎𝑡 , for all 𝑡 ≥ 0. It follows that 𝑓 is of exponential type in the sense given in the text. 41. Suppose 𝑎 and 𝐾 are real and ∣𝑦(𝑡)∣ ≤ 𝐾𝑒𝑎𝑡 . Then 𝑦(𝑡)𝑒−𝑎𝑡 is bounded by 𝐾. But 2

𝑒𝑡 𝑒−𝑎𝑡 = 𝑒𝑡

2

2

−𝑎𝑡+ 𝑎4 𝑎 2

= 𝑒(𝑡− 2 ) 𝑒 2

= 𝑒𝑢 𝑒

−𝑎2 4

𝑒

−𝑎2 4

−𝑎2 4

, 2

where 𝑢 = 𝑡− 𝑎2 . As 𝑡 approaches infinity so does 𝑢. Since lim𝑢→∞ 𝑒𝑢 = ∞ 2 it is clear that lim𝑡→∞ 𝑒𝑡 𝑒−𝑎𝑡 = ∞, for all 𝑎 ∈ ℝ, and hence 𝑦(𝑡)𝑒−𝑎𝑡 is not bounded. It follows that 𝑦(𝑡) is not of exponential type. ∫∞ 2 2 42. First of all, fix 𝑎 ∈ ℝ. Since 𝑒𝑡 > 1 for all 𝑡 it follows that 𝑎 𝑒𝑡 𝑑𝑡 > ∫∞ ∫ ∞ 𝑡2 𝑎 1 𝑑𝑡 = ∞ by the comparison test. Thus 𝑎 𝑒 𝑑𝑡 does not exist for any real number 𝑎. Now let 𝑠 be any real number. Then

1 Solutions

{

ℒ 𝑒𝑡

} 2

(2𝑠) =

∫

∞

∫0 ∞

47

2

𝑒−2𝑠𝑡 𝑒𝑡 𝑑𝑡 2

2

2

𝑒𝑡 −2𝑠𝑡+𝑠 −𝑠 𝑑𝑡 0 ∫ ∞ 2 −𝑠2 =𝑒 𝑒(𝑡−𝑠) 𝑑𝑡 ∫0 ∞ 2 −𝑠2 =𝑒 𝑒𝑡 𝑑𝑡.

=

−𝑠

But this last integral does not exist. Since the Laplace transform does not exist at 2𝑠, for any 𝑠, the Laplace transform does not exist. 43. 𝑦(𝑡) is of exponential type because it is continuous and bounded. On the 2 2 other hand, 𝑦 ′ (𝑡) = cos(𝑒𝑡 )𝑒𝑡 (2𝑡). Suppose there is a 𝐾 and 𝑎 so that ∣𝑦 ′ (𝑡)∣ ≤ 𝐾𝑒𝑎𝑡 for all 𝑡 ≥ 0. We need only show that there are some 𝑡 for 2 which this inequality does not hold. Since cos 𝑒𝑡 oscillates between −1 2 2 and 1 let’s focus on those 𝑡 for which cos 𝑒𝑡 = 1. This happens when 𝑒𝑡 √ is a multiple of 2𝜋𝑛 or 𝑡 = 𝑡𝑛 = ln(2𝜋𝑛). If the inequality ∣𝑦 ′ (𝑡)∣ ≤ 𝐾𝑒𝑎𝑡 is valid for all 𝑡 ≥ 0 it is valid for 𝑡𝑛 for all 𝑛 > 0. We then get the 2 inequality 2𝑡𝑛 𝑒𝑡𝑛 ≤ 𝐾𝑒𝑎𝑡𝑛 . Now divide by 𝑒𝑎𝑡𝑛 , combine, complete the 2 2 square, and simplify to get the inequality 2𝑡𝑛 𝑒(𝑡𝑛 −𝑎/2) ≤ 𝐾𝑒𝑎 /4 . Choose 𝑛 so that 𝑡𝑛 > 𝐾 and 𝑡𝑛 > 𝑎. Then this last inequality is not satisfied. It follows that 𝑦 ′ (𝑡) is not of exponential type. Now consider the definite ∫𝑀 integral 0 𝑒−𝑠𝑡 𝑦 ′ (𝑡) 𝑑𝑡 and compute by parts: We get ∫

𝑀

0

𝑀 𝑒−𝑠𝑡 𝑦 ′ (𝑡) 𝑑𝑡 = 𝑦(𝑡)𝑒−𝑠𝑡 0 + 𝑠

∫

𝑀

𝑒−𝑠𝑡 𝑦(𝑡) 𝑑𝑡.

0

2

Since 𝑦(𝑡) = sin(𝑒𝑡 ) is bounded and 𝑦(0) = 0 it follows that 𝑀 lim 𝑦(𝑡)𝑒−𝑠𝑡 0 = 0.

𝑀 →∞

Taking limits as 𝑀 → ∞ in the equation above gives ℒ {𝑦 ′ (𝑡)} = 𝑠ℒ {𝑦(𝑡)}. The righthand side exists because 𝑦(𝑡) is bounded. (a) Show that 𝛤 (1) = 1. (b) Show that 𝛤 satisfies the recursion formula 𝛤 (𝛽 + 1) = 𝛽𝛤 (𝛽). (Hint: Integrate by parts.) (c) Show that 𝛤 (𝑛 + 1) = 𝑛! when 𝑛 is a nonnegative integer. ∫∞ ∞ 44. (a) 𝛤 (1) = 0 𝑒−𝑥 𝑑𝑥 = −𝑒−𝑥 ∣0 = 1. ∞ ∫∞ ∫ ∞ 𝛽 −𝑥 (b) 𝛤 (𝛽 + 1) = 0 𝑥 𝑒 𝑑𝑥 = −𝑥𝛽 𝑒−𝑥 0 + 𝛽 0 𝑥𝛽−1 𝑒−𝑥 𝑑𝑥 = 𝛽𝛤 (𝛽). The second equality is obtained by integration by parts using 𝑢 = 𝑥𝛽 , 𝑑𝑣 = 𝑒−𝑥 𝑑𝑥. (c) Repeated use of (b) gives

48

1 Solutions

𝛤 (𝑛+1) = 𝑛𝛤 (𝑛) = 𝑛(𝑛−1)𝛤 (𝑛−1) = ⋅ ⋅ ⋅ = 𝑛(𝑛−1) ⋅ ⋅ ⋅ 2⋅1𝛤 (1) = 𝑛!. 45. Using polar coordinates 𝑥 = 𝑟 cos 𝜃, 𝑦 = 𝑟 sin 𝜃. Then 𝑑𝑥 𝑑𝑦 = 𝑟 𝑑𝑟 𝑑𝜃 and the domain of integration is the first quadrant of the plane, which in polar coordinates is given by 0 ≤ 𝜃 ≤ 𝜋/2, 0 ≤ 𝑟 < ∞. Thus ∫ ∞∫ ∞ ∫ 𝜋/2 ∫ ∞ 2 2 2 𝑒−𝑟 𝑟 𝑑𝑟 𝑑𝜃 𝑒−(𝑥 +𝑦 ) 𝑑𝑥 𝑑𝑦 = 0 0 0 0 ∫ 𝜋 ∞ −𝑟2 = 𝑒 𝑟 𝑑𝑟 2 0 2 ∞ 𝜋 −𝑒−𝑟 𝜋 = = . 2 2 4 0

√

Hence, 𝐼 = 𝜋/2. ∫∞ 46. 𝛤 ( 21 ) = 0 𝑥−1/2 𝑒−𝑥 𝑑𝑥. Using the change of variables 𝑥 = 𝑢2 , so 𝑑𝑥 = 2𝑢 𝑑𝑢 it follows that ∫ ∞ ∫ ∞ √ 2 2 1 𝛤( ) = 𝑢−1 𝑒−𝑢 2𝑢 𝑑𝑢 = 2 𝑒−𝑢 𝑑𝑢 = 𝜋. 2 0 0 Then using the recursion formula√gives: √ (a) 𝛤 ( 32 ) = 𝛤 ( 21 + 1) = 12 𝛤 ( 21 ) = 𝜋/2, (b) 𝛤 ( 52 ) = 23 𝛤 ( 32 ) = 3 𝜋/4. Now apply the general power formula (Formula 11) to get √ √ {√ } 𝛤 ( 23 ) { } 𝛤 ( 52 ) 𝜋 3 𝜋 (c) ℒ 𝑡 = 3/2 = 3/2 , (d) ℒ 𝑡3/2 = 5/2 = 5/2 . 𝑠 2𝑠 𝑠 4𝑠

Section 2.2 The 𝑠 − 1 -chain 1.

5𝑠 + 10 (𝑠 − 1)(𝑠 + 4) 2 𝑠+4

3 𝑠−1

The 𝑠 − 2 -chain 2.

10𝑠 − 2 (𝑠 + 1)(𝑠 − 2) 4 (𝑠 + 1)

6 (𝑠 − 2)

1 Solutions

The 𝑠 − 5 -chain 3.

1 (𝑠 + 2)(𝑠 − 5) −1/7 (𝑠 + 2)

1/7 (𝑠 − 5)

The 𝑠 + 3 -chain 4.

5𝑠 + 9 (𝑠 − 1)(𝑠 + 3) 7/2 𝑠−1

3/2 𝑠+3

The 𝑠 − 1 -chain 5.

3𝑠 + 1 (𝑠 − 1)(𝑠2 + 1) −2𝑠 + 1 𝑠2 + 1

2 𝑠−1

The 𝑠 + 1 -chain 6.

3𝑠2 − 𝑠 + 6 (𝑠 + 1)(𝑠2 + 4) 𝑠−2 𝑠2 + 4

2 (𝑠 + 1)

The 𝑠 + 3 -chain 2

7.

𝑠 +𝑠−3 (𝑠 + 3)3 𝑠−2 (𝑠 + 3)2 1 𝑠+3 0

3 (𝑠 + 3)3 −5 (𝑠 + 3)2 1 𝑠+3

49

50

1 Solutions

The 𝑠 + 2 -chain 2

8.

5𝑠 − 3𝑠 + 10 (𝑠 + 1)(𝑠 + 2)2 5𝑠 + 23 (𝑠 + 1)(𝑠 + 2) 18 𝑠+1

−36 (𝑠 + 2)2 −13 𝑠+2

The 𝑠 + 1 -chain 𝑠 9.

(𝑠 +

2)2 (𝑠

1)2

+ 𝑠+4 (𝑠 + 2)2 (𝑠 + 1) −3𝑠 − 8 (𝑠 + 2)2

−1 (𝑠 + 1)2 3 𝑠+1

The 𝑠 − 1 -chain

10.

16𝑠 (𝑠 − 1)3 (𝑠 − 3)2 −4𝑠 + 36 (𝑠 − 1)2 (𝑠 − 3)2 −8𝑠 + 36 (𝑠 − 1)(𝑠 − 3)2 −7𝑠 + 27 (𝑠 − 3)2

4 (𝑠 − 1)3 8 (𝑠 − 1)2 7 (𝑠 − 1)

1 Solutions

The 𝑠 − 5 -chain

11.

1 (𝑠 − 5)5 (𝑠 − 6) 1 (𝑠 − 5)4 (𝑠 − 6) 1 (𝑠 − 5)3 (𝑠 − 6) 1 (𝑠 − 5)2 (𝑠 − 6) 1 (𝑠 − 5)(𝑠 − 6) 1 𝑠−6

−1 (𝑠 − 5)5 −1 (𝑠 − 5)4 −1 (𝑠 − 5)3 −1 (𝑠 − 5)2 −1 𝑠−5

12. Use the technique of distinct linear factors (Example 5). 7/2 3/2 + 𝑠+3 𝑠−1 13. Use the technique of distinct linear factors (Example 5). 13/8 5/8 − 𝑠−5 𝑠+3 14.

−1 1 + 𝑠−1 𝑠−2

15.

37 23 + 12(𝑠 − 5) 12(𝑠 + 7)

16.

1 2 + 𝑠 𝑠+1

17.

25 9 − 8(𝑠 − 7) 8(𝑠 + 1)

18.

25 9 15 + − 2(𝑠 − 3) 2(𝑠 − 1) 𝑠 − 2

19.

1 1 1 − + 2(𝑠 + 5) 2(𝑠 − 1) 𝑠 − 2

20. Use Theorem 1 to write

51

52

1 Solutions

𝑠2 𝐴1 = (𝑠 − 1)3 (𝑠 − 1)3

+

𝑝1 (𝑠) (𝑠 − 1)2

𝑠2 =1 1 𝑠=1 1 1 (𝑠2 − (1)(1)) = (𝑠2 − 1) = 𝑠 + 1 and 𝑝1 (𝑠) = 𝑠−1 𝑠−1 where 𝐴1 =

Continuing gives 𝑠+1 𝐴2 = (𝑠 − 1)2 (𝑠 − 1)2 𝑠 + 1 where 𝐴2 = 1

and 𝑝2 (𝑠) = Thus

+

𝑝2 (𝑠) 𝑠−1

=2 𝑠=1

1 1 (𝑠 + 1 − (2)(1)) = (𝑠 − 1) = 1 𝑠−1 𝑠−1

𝑠2 1 2 1 = + + . (𝑠 − 1)3 (𝑠 − 1)3 (𝑠 − 1)2 𝑠−1

Alternate Solution: Write 𝑠 = (𝑠 − 1) + 1 so that 𝑠2 ((𝑠 − 1) + 1)2 1 + 2(𝑠 − 1) + (𝑠 − 1)2 1 2 1 = = = + + 3 3 3 3 2 (𝑠 − 1) (𝑠 − 1) (𝑠 − 1) (𝑠 − 1) (𝑠 − 1) 𝑠 − 1 21.

7 (𝑠 + 4)4

22. Use Theorem 1 to write 𝑠 𝐴1 = (𝑠 − 3)3 (𝑠 − 3)3

+

𝑝1 (𝑠) (𝑠 − 3)2

𝑠 =3 1 𝑠=3 1 and 𝑝1 (𝑠) = (𝑠 − (3)(1)) = 1. 𝑠−3 where 𝐴1 =

Thus,

𝑠 3 1 = + (𝑠 − 3)3 (𝑠 − 3)3 (𝑠 − 3)2

Alternate Solution: Write 𝑠 = (𝑠 − 3) + 3 so that 𝑠 (𝑠 − 3) + 3 3 1 = = + . (𝑠 − 3)3 (𝑠 − 3)3 (𝑠 − 3)3 (𝑠 − 3)2

1 Solutions

53

23. Use Theorem 1 to write 𝑠2 + 𝑠 − 3 𝐴1 = (𝑠 + 3)3 (𝑠 + 3)3

𝑝1 (𝑠) (𝑠 + 3)2

+

𝑠2 + 𝑠 − 3 =3 1 𝑠=−3 1 1 and 𝑝1 (𝑠) = (𝑠2 + 𝑠 − 3 − (3)(1)) = (𝑠2 + 𝑠 − 6) = 𝑠 − 2 𝑠+3 𝑠+3 where 𝐴1 =

Continuing gives 𝑠−2 𝐴2 = (𝑠 + 3)2 (𝑠 + 3)2

+

𝑝2 (𝑠) 𝑠+3

𝑠 − 2 = −5 where 𝐴2 = 1 𝑠=−3 1 1 and 𝑝2 (𝑠) = (𝑠 − 2 − (−5)(1)) = (𝑠 + 3) = 1 𝑠+3 𝑠+3 Thus

𝑠2 + 𝑠 − 3 3 5 1 = − + (𝑠 + 3)3 (𝑠 + 3)3 (𝑠 + 3)2 𝑠+3

Alternate Solution: Write 𝑠 = (𝑠 + 3) − 3 so that 𝑠2 + 𝑠 − 3 ((𝑠 + 3) − 3)2 + ((𝑠 + 3) − 3) − 3 = 3 (𝑠 + 3) (𝑠 − 3)3 (𝑠 + 3)2 − 5(𝑠 + 3) + 3 = (𝑠 + 3)3 3 5 1 = − + . 3 2 (𝑠 + 3) (𝑠 + 3) 𝑠+3 24. Use Theorem 1 to compute the (𝑠 + 2)-chain: 5𝑠2 − 3𝑠 + 10 𝐴1 = (𝑠 + 1)(𝑠 + 2)2 (𝑠 + 2)2

+

𝑝1 (𝑠) (𝑠 + 2)(𝑠 + 1)

5𝑠2 − 3𝑠 + 10 where 𝐴1 = = −36 𝑠+1 𝑠=−2 1 1 and 𝑝1 (𝑠) = (5𝑠2 − 3𝑠 + 10 − (−36)(𝑠 + 1)) = (5𝑠2 + 33𝑠 + 46) = 5𝑠 + 23 𝑠+2 𝑠+2 Continuing gives

54

1 Solutions

5𝑠 + 23 𝐴2 = (𝑠 + 2)(𝑠 + 1) 𝑠+2

+

𝑝2 (𝑠) 𝑠+1

5𝑠 + 23 = −13 𝑠 + 1 𝑠=−2 1 1 (5𝑠 + 23 − (−13)(𝑠 + 1)) = (18𝑠 + 36) = 18 and 𝑝2 (𝑠) = 𝑠+2 𝑠+2 where 𝐴2 =

Thus 25.

5𝑠2 − 3𝑠 + 10 −36 13 18 = − + (𝑠 + 1)(𝑠 + 2)2 (𝑠 + 2)2 𝑠+2 𝑠+1

𝑠2 − 6𝑠 + 7 𝑠2 − 6𝑠 + 7 = , so use Theorem 1 to compute the (𝑠2 − 4𝑠 − 5)2 (𝑠 + 1)2 (𝑠 − 5)2 (𝑠 + 1)-chain: 𝑠2 − 6𝑠 + 7 𝐴1 = (𝑠 + 1)2 (𝑠 − 5)2 (𝑠 + 1)2

+

𝑠2 − 6𝑠 + 7 where 𝐴1 = (𝑠 − 5)2

𝑝1 (𝑠) (𝑠 + 1)(𝑠 − 5)2 =

𝑠=−1

7 18

1 and 𝑝1 (𝑠) = (𝑠2 − 6𝑠 + 7 − (7/18)(𝑠 − 5)2 ) 𝑠+1 1 1 = (11𝑠2 − 38𝑠 − 49)/18 = (11𝑠 − 49) 𝑠+1 18 Continuing gives 1 11𝑠 − 49 𝐴2 = 2 18 (𝑠 + 1)(𝑠 − 5) 𝑠+1

+

𝑝2 (𝑠) (𝑠 − 5)2

1 11𝑠 − 49 = −5/54 18 (𝑠 − 5)2 𝑠=−1 1 and 𝑝2 (𝑠) = ((11𝑠 − 49)/18 − (−5/54)(𝑠 − 5)2 ) = (5𝑠 − 22)/54 𝑠+1 where 𝐴2 =

𝑠2 − 6𝑠 + 7 1/18 5/54 (5𝑠 − 22)/54 = − + Now either con(𝑠 + 1)2 (𝑠 − 5)2 (𝑠 + 1)2 𝑠 + 1 (𝑠 − 5)2 tinue with Theorem 1 or replace 𝑠 with 𝑠 = (𝑠 − 5) + 5 in the numerator 𝑠2 − 6𝑠 + 7 of the last fraction to finish the calculation and get = (𝑠 + 1)2 (𝑠 − 5)2 ( ) 1 5 21 3 5 + + − 2 2 54 𝑠 − 5 (𝑠 + 1) (𝑠 − 5) 𝑠+1 Thus

26. Use Theorem 1 to compute the (𝑠 + 9)-chain:

1 Solutions

81 𝐴1 = + 9) 𝑠+9

𝑠3 (𝑠

+

55

𝑝1 (𝑠) 𝑠3

81 = −1/9 𝑠3 𝑠=−9 1 1 (81 − (−1/9)(𝑠3 )) = (𝑠3 + 93 )/9 = (𝑠2 − 9𝑠 + 81)/9 and 𝑝1 (𝑠) = 𝑠+9 𝑠+9 where 𝐴1 =

Thus

81 9 1 1 1 1 = 3− 2+ − + 9) 𝑠 𝑠 9𝑠 9 𝑠 + 1

𝑠3 (𝑠

27. Use Theorem 1 to compute the (𝑠 + 2)-chain: 𝑠 𝐴1 = (𝑠 + 2)2 (𝑠 + 1)2 (𝑠 + 2)2

+

𝑝1 (𝑠) (𝑠 + 2)(𝑠 + 1)2

𝑠 where 𝐴1 = = −2 2 (𝑠 + 1) 𝑠=−2 1 and 𝑝1 (𝑠) = (𝑠 − (−2)(𝑠 + 1)2 ) 𝑠+2 2𝑠2 + 5𝑠 + 2 (2𝑠 + 1)(𝑠 + 1) = = = 2𝑠 + 1 𝑠+2 𝑠+2 Continuing gives 2𝑠 + 1 𝐴2 = (𝑠 + 2)(𝑠 + 1)2 𝑠+2

+

𝑝2 (𝑠) (𝑠 + 1)2

2𝑠 + 1 where 𝐴2 = = −3 (𝑠 + 1)2 𝑠=−2 1 and 𝑝2 (𝑠) = (2𝑠 + 1 − (−3)(𝑠 + 1)2 ) = 3𝑠 + 2 𝑠+2 −2 3 3𝑠 + 2 𝑠 = − + . Now continue using (𝑠 + 2)2 (𝑠 + 1)2 (𝑠 + 2)2 𝑠 + 2 (𝑠 + 1)2 Theorem 1 or replace 𝑠 by (𝑠 + 1) − 1 in the numerator of the last fraction 𝑠 −2 3 1 3 to get = − − + (𝑠 + 2)2 (𝑠 + 1)2 (𝑠 + 2)2 𝑠 + 2 (𝑠 + 1)2 𝑠+1 Thus

28. Use Theorem 1 to compute the (𝑠 + 2)-chain:

56

1 Solutions

𝑠2 𝐴1 = (𝑠 + 2)2 (𝑠 + 1)2 (𝑠 + 2)2

𝑝1 (𝑠) (𝑠 + 2)(𝑠 + 1)2

+

𝑠2 =4 (𝑠 + 1)2 𝑠=−2 1 (𝑠2 − (4)(𝑠 + 1)2 ) and 𝑝1 (𝑠) = 𝑠+2 −3𝑠2 − 8𝑠 − 4 −(3𝑠 + 2)(𝑠 + 2) = = = −(3𝑠 + 2) 𝑠+2 𝑠+2 where 𝐴1 =

Continuing gives −3𝑠 − 2 𝐴2 = (𝑠 + 2)(𝑠 + 1)2 𝑠+2

𝑝2 (𝑠) (𝑠 + 1)2

+

−3𝑠 − 2 =4 (𝑠 + 1)2 𝑠=−2 1 and 𝑝2 (𝑠) = (−3𝑠 − 2 − (4)(𝑠 + 1)2 ) = −(4𝑠 + 3) 𝑠+2 where 𝐴2 =

𝑠2 4 −4𝑠 − 3 = + + . Now continue using (𝑠 + 2)2 (𝑠 + 1)2 (𝑠 + 2)2 𝑠 + 2 (𝑠 + 1)2 Theorem 1 or replace 𝑠 by (𝑠 + 1) − 1 in the numerator of the last fraction 𝑠2 4 4 1 4 to get = + + − (𝑠 + 2)2 (𝑠 + 1)2 (𝑠 + 2)2 𝑠 + 2 (𝑠 + 1)2 𝑠+1 Thus

29. Use Theorem 1 to compute the (𝑠 − 3)-chain: 8𝑠 𝐴1 = (𝑠 − 1)(𝑠 − 2)(𝑠 − 3)3 (𝑠 − 3)3

+

𝑝1 (𝑠) (𝑠 − 1)(𝑠 − 2)(𝑠 − 3)2

8𝑠 = 12 (𝑠 − 1)(𝑠 − 2) 𝑠=3 1 and 𝑝1 (𝑠) = (8𝑠 − (12)(𝑠 − 1)(𝑠 − 2)) 𝑠−3 −12𝑠2 + 44𝑠 − 24 (−12𝑠 + 8)(𝑠 − 3) = = = −12𝑠 + 8 𝑠−3 𝑠−3 where 𝐴1 =

For the second step in the (𝑠 − 3)-chain:

1 Solutions

−12𝑠 + 8 𝐴2 = 2 (𝑠 − 1)(𝑠 − 2)(𝑠 − 3) (𝑠 − 3)2

57

𝑝2 (𝑠) (𝑠 − 1)(𝑠 − 2)(𝑠 − 3)2

+

−12𝑠 + 8 = −14 (𝑠 − 1)(𝑠 − 2) 𝑠=3 1 and 𝑝2 (𝑠) = (−12𝑠 + 8 − (−14)(𝑠 − 1)(𝑠 − 2)) 𝑠−3 14𝑠2 − 54𝑠 + 36 (14𝑠 − 12)(𝑠 − 3) = = = 14𝑠 − 12 𝑠−3 𝑠−3 Continuing gives where 𝐴2 =

14𝑠 − 12 𝐴3 = (𝑠 − 1)(𝑠 − 2)(𝑠 − 3)2 𝑠−3

𝑝3 (𝑠) (𝑠 − 1)(𝑠 − 2)

+

14𝑠 − 12 where 𝐴3 = = 15 (𝑠 − 1)(𝑠 − 2) 𝑠=3 1 and 𝑝3 (𝑠) = (14𝑠 − 12 − (15)(𝑠 − 1)(𝑠 − 2)) = −15𝑠 + 14 𝑠−3 8𝑠 12 14 15 −15𝑠 + 14 Thus = − + + . (𝑠 − 1)(𝑠 − 3)(𝑠 − 3)3 (𝑠 − 3)3 (𝑠 − 3)2 𝑠 − 3 (𝑠 − 1)(𝑠 − 2) The last fraction has a denominator with distinct linear factors so we get 12 −14 15 −16 1 8𝑠 = + + + + 3 3 2 (𝑠 − 1)(𝑠 − 3)(𝑠 − 3) (𝑠 − 3) (𝑠 − 3) 𝑠−3 𝑠−2 𝑠−1 30. Use Theorem 1 to compute the 𝑠-chain: 25 𝐴1 = 2 𝑠2 (𝑠 − 5)(𝑠 + 1) 𝑠

+

𝑝1 (𝑠) 𝑠(𝑠 − 5)(𝑠 + 1)

25 = −5 (𝑠 − 5)(𝑠 + 1) 𝑠=0 1 and 𝑝1 (𝑠) = (25 − (−5)(𝑠 − 5)(𝑠 + 1)) 𝑠 5𝑠2 − 20𝑠 = = 5𝑠 − 20 𝑠 where 𝐴1 =

Continuing gives 5𝑠 − 20 𝐴2 = 𝑠(𝑠 − 5)(𝑠 + 1) 𝑠

+

𝑝2 (𝑠) (𝑠 − 5)(𝑠 + 1)

5𝑠 − 20 where 𝐴2 = =4 (𝑠 − 5)(𝑠 + 1) 𝑠=0 1 and 𝑝2 (𝑠) = (5𝑠 − 20 − (4)(𝑠 − 5)(𝑠 + 1)) = −4𝑠 + 16 𝑠

58

1 Solutions

25 5 4 −4𝑠 + 16 = 2 − + . The last fraction has 𝑠2 (𝑠 − 5)(𝑠 + 1) 𝑠 𝑠 (𝑠 − 5)(𝑠 + 1) 25 a denominator with distinct linear factors so we get 2 = 𝑠 (𝑠 − 5)(𝑠 + 1) 2 1 20 1 5 4 − − − + 3𝑠−5 3 𝑠 + 1 𝑠2 𝑠 Thus

31. Use Theorem 1 to compute the (𝑠 − 2)-chain: 𝑠 𝐴1 = (𝑠 − 2)2 (𝑠 − 3)2 (𝑠 − 2)2

+

𝑝1 (𝑠) (𝑠 − 2)(𝑠 − 3)2

𝑠 =2 2 (𝑠 − 3) 𝑠=2 1 (𝑠 − (2)(𝑠 − 3)2 ) and 𝑝1 (𝑠) = 𝑠−2 −2𝑠2 + 13𝑠 − 18 (−2𝑠 + 9)(𝑠 − 2) = = −2𝑠 + 9 = 𝑠−2 𝑠−2 where 𝐴1 =

Continuing gives −2𝑠 + 9 𝐴2 = (𝑠 − 2)(𝑠 − 3)2 𝑠−3

+

𝑝2 (𝑠) (𝑠 − 3)2

−2𝑠 + 9 where 𝐴2 = =5 (𝑠 − 3)2 𝑠=2 1 and 𝑝2 (𝑠) = (−2𝑠 + 9 − (5)(𝑠 − 3)2 ) = −5𝑠 + 18 𝑠−2 𝑠

2 5 −5𝑠 + 18 + + . Now continue using 2 (𝑠 − − (𝑠 − 2) 𝑠 − 2 (𝑠 − 3)2 Theorem 1 or replace 𝑠 by (𝑠 − 3) + 3 in the numerator of the last fraction 𝑠 2 5 3 5 to get = + + − (𝑠 − 2)2 (𝑠 − 3)2 (𝑠 − 2)2 𝑠 − 2 (𝑠 − 3)2 𝑠−3 Thus

2)2 (𝑠

3)2

=

32. Use Theorem 1 to compute the (𝑠 − 1)-chain: 16𝑠 𝐴1 = (𝑠 − 1)3 (𝑠 − 3)2 (𝑠 − 1)3

+

𝑝1 (𝑠) (𝑠 − 1)2 (𝑠 − 3)2

16𝑠 where 𝐴1 = =4 (𝑠 − 3)2 𝑠=1 1 and 𝑝1 (𝑠) = (16𝑠 − (4)(𝑠 − 3)2 ) 𝑠−1 −4𝑠2 + 40𝑠 − 36 (−4𝑠 + 36)(𝑠 − 1) = = = −4𝑠 + 36 𝑠−1 𝑠−1

1 Solutions

59

For the second step in the (𝑠 − 1)-chain: −4𝑠 + 36 𝐴2 = 2 2 (𝑠 − 1) (𝑠 − 3) (𝑠 − 1)2

+

𝑝2 (𝑠) (𝑠 − 1)(𝑠 − 3)2

−4𝑠 + 36 where 𝐴2 = =8 (𝑠 − 3)2 𝑠=1 1 and 𝑝2 (𝑠) = (−4𝑠 + 36 − (8)(𝑠 − 3)2 ) 𝑠−1 −8𝑠2 + 44𝑠 − 36 (−8𝑠 + 36)(𝑠 − 1) = = = −8𝑠 + 36 𝑠−1 𝑠−1 Continuing gives −8𝑠 + 36 𝐴3 = (𝑠 − 1)(𝑠 − 3)2 𝑠−1

+

𝑝3 (𝑠) (𝑠 − 3)2

−8𝑠 + 36 where 𝐴3 = =7 (𝑠 − 3)2 𝑠=1 1 and 𝑝3 (𝑠) = (−8𝑠 + 36 − (7)(𝑠 − 3)2 ) = −7𝑠 + 27 𝑠−1 4 8 7 −7𝑠 + 27 16𝑠 = + + + . Now (𝑠 − 1)3 (𝑠 − 3)2 (𝑠 − 1)3 (𝑠 − 1)2 𝑠−1 (𝑠 − 3)2 continue using Theorem 1 or replace 𝑠 by (𝑠 − 3) + 3 in the numerator of 16𝑠 4 8 7 the last fraction to get = + + + 3 2 3 2 (𝑠 − 1) (𝑠 − 3) (𝑠 − 1) (𝑠 − 1) 𝑠−1 6 7 − 2 (𝑠 − 3) 𝑠−3 Thus

Section 2.3 1. Note that 𝑠 = 𝑖 is a root of 𝑠2 + 1. Applying Theorem 1 gives (𝑠2

1 𝐵1 𝑠 + 𝐶 1 𝑝1 (𝑠) = 2 + 2 2 2 2 + 1) (𝑠 + 2) (𝑠 + 1) (𝑠 + 1)(𝑠2 + 2) 1 1 = 2 =1 2 (𝑠 + 2) 𝑠=𝑖 𝑖 +2 ⇒ 𝐵1 = 0 and 𝐶1 = 1 1 and 𝑝1 (𝑠) = 2 (1 − (1)(𝑠2 + 2)) 𝑠 +1 −𝑠2 − 1 = 2 = −1. 𝑠 +1

where 𝐵1 𝑖 + 𝐶1 =

60

1 Solutions

We now apply Theorem 1 on the remainder term

−1 . (𝑠2 + 1)(𝑠2 + 2)

−1 𝐵2 𝑠 + 𝐶 2 𝑝2 (𝑠) = + 2 (𝑠2 + 1)(𝑠2 + 2) (𝑠2 + 1) (𝑠 + 2) −1 = −1 (𝑠2 + 2) 𝑠=𝑖 ⇒ 𝐵2 = 0 and 𝐶2 = −1 1 and 𝑝2 (𝑠) = 2 (−1 − (−1)(𝑠2 + 2)) 𝑠 +1 𝑠2 + 1 = 1. = 2 𝑠 +1

where 𝐵2 𝑖 + 𝐶2 =

Thus the (𝑠2 + 1)-chain is The 𝑠2 + 1 -chain 1 + 1)2 (𝑠2 + 2) −1 (𝑠2 + 1)(𝑠2 + 2) 1 2 𝑠 +2

(𝑠2

2. Note that 𝑠 =

1 + 1)2 −1 (𝑠2 + 1)

(𝑠2

√ 2𝑖 is a root of 𝑠2 + 2. Applying Theorem 1 gives

𝑠3 (𝑠2 + 2)2 (𝑠2 + 3) √

where 𝐵1 2𝑖 + 𝐶1

=

= ⇒

and 𝑝1 (𝑠)

= =

𝐵1 𝑠 + 𝐶 1 𝑝1 (𝑠) + 2 (𝑠2 + 2)2 (𝑠 + 2)(𝑠2 + 3) √ √ 𝑠3 ( 2𝑖)3 = −2 2𝑖 = √ 2 √ 2 (𝑠 + 3) 𝑠= 2𝑖 ( 2𝑖) + 3 𝐵1 = −2 and 𝐶1 = 0 1 (𝑠3 − (−2𝑠)(𝑠2 + 3)) 2 𝑠 +1 3𝑠3 + 6𝑠 = 3𝑠. 𝑠2 + 2

We now apply Theorem 1 on the remainder term

3𝑠 . (𝑠2 + 2)(𝑠2 + 3)

1 Solutions

3𝑠 + 2)(𝑠2 + 3)

=

√ where 𝐵2 2𝑖 + 𝐶2

=

(𝑠2

⇒ and 𝑝2 (𝑠)

= =

61

𝐵2 𝑠 + 𝐶 2 𝑝2 (𝑠) + 2 2 (𝑠 + 2) (𝑠 + 3) √ 3𝑠 = 3 2𝑖 2 √ (𝑠 + 3) 𝑠= 2𝑖 𝐵2 = 3 and 𝐶2 = 0 1 (3𝑠 − (3𝑠)(𝑠2 + 3)) 𝑠2 + 2 −3𝑠(𝑠2 + 2) = −3𝑠. 𝑠2 + 2

Thus the (𝑠2 + 2)-chain is The 𝑠2 + 2 -chain 𝑠3 (𝑠2 + 2)2 (𝑠2 + 3) 3𝑠 (𝑠2 + 2)(𝑠2 + 3) −3𝑠 𝑠2 + 3 3. Note that 𝑠 =

−2𝑠 + 2)2 3𝑠 𝑠2 + 2

(𝑠2

√ 3𝑖 is a root of 𝑠2 + 3. Applying Theorem 1 gives

8𝑠 + 8𝑠2 (𝑠2 + 3)3 (𝑠2 + 1) √

where 𝐵1 3𝑖 + 𝐶1

=

= = ⇒

and 𝑝1 (𝑠)

= =

𝐵1 𝑠 + 𝐶 1 𝑝1 (𝑠) + 2 (𝑠2 + 3)3 (𝑠 + 3)2 (𝑠2 + 1) √ √ 8𝑠 + 8𝑠2 8 3𝑖 + 8( 3𝑖)2 √ = (𝑠2 + 1) 𝑠=√3𝑖 ( 3𝑖)2 + 1 √ −4 3𝑖 + 12 𝐵1 = −4 and 𝐶1 = 12 1 (8𝑠 + 8𝑠2 − (−4𝑠 + 12)(𝑠2 + 1)) 𝑠2 + 3 4𝑠3 − 4𝑠2 + 12𝑠 − 12 = 4(𝑠 − 1). 𝑠2 + 3

Apply Theorem 1 a second time on the remainder term

(𝑠2

4𝑠 − 4 . + 3)2 (𝑠2 + 1)

62

1 Solutions

4𝑠 − 4 + 3)2 (𝑠2 + 1)

=

√ where 𝐵2 3𝑖 + 𝐶2

=

(𝑠2

⇒ and 𝑝2 (𝑠)

= =

𝐵2 𝑠 + 𝐶 2 𝑝2 (𝑠) + 2 2 2 (𝑠 + 3) (𝑠 + 3)(𝑠2 + 1) √ 4𝑠 − 4 = −2 3𝑖 + 2 2 √ (𝑠 + 1) 𝑠= 3𝑖 𝐵2 = −2 and 𝐶2 = 2 1 (4𝑠 − 4 − (−2𝑠 + 2)(𝑠2 + 1)) 𝑠2 + 3 2𝑠3 − 2𝑠2 + 6𝑠 − 6 = 2𝑠 − 2. 𝑠2 + 3

A third application of Theorem 1 on the remainder term

2𝑠 − 2 (𝑠2 + 3)(𝑠2 + 1)

gives 2𝑠 − 2 (𝑠2 + 3)(𝑠2 + 1)

=

√

where 𝐵3 3𝑖 + 𝐶3

= ⇒

and 𝑝3 (𝑠)

= =

𝐵3 𝑠 + 𝐶 3 𝑝3 (𝑠) + 2 (𝑠2 + 3) (𝑠 + 1) √ 2𝑠 − 2 = − 3𝑖 + 1 2 √ (𝑠 + 1) 𝑠= 3𝑖 𝐵3 = −1 and 𝐶3 = 1 1 (2𝑠 − 2 − (−𝑠 + 1)(𝑠2 + 1)) 2 𝑠 +3 𝑠3 − 𝑠2 + 3𝑠 − 3 = 𝑠 − 1. 𝑠2 + 3

Thus the (𝑠2 + 3)-chain is The 𝑠2 + 3 -chain 8𝑠 + 8𝑠2 (𝑠+ 3)3 (𝑠2 + 1) 4(𝑠 − 1) 2 (𝑠 + 3)2 (𝑠2 + 1) 2(𝑠 − 1) (𝑠2 + 3)(𝑠2 + 1) 𝑠−1 𝑠2 + 1

12 − 4𝑠 (𝑠2 + 3)3 2 − 2𝑠 (𝑠2 + 3)2 1−𝑠 𝑠2 + 3

4. Note that 𝑠 = 2𝑖 is a root of 𝑠2 + 4. Applying Theorem 1 gives

1 Solutions

4𝑠4 + 4)3 (𝑠2 + 6)

=

where 𝐵1 2𝑖 + 𝐶1

=

(𝑠2

= ⇒ and 𝑝1 (𝑠)

= =

𝐵1 𝑠 + 𝐶 1 𝑝1 (𝑠) + 2 (𝑠2 + 4)4 (𝑠 + 4)3 (𝑠2 + 6) 4𝑠4 4(2𝑖)4 = 2 (𝑠 + 6) 𝑠=2𝑖 (2𝑖)2 + 6 32 𝐵1 = 0 and 𝐶1 = 32 1 (4𝑠4 − (32)(𝑠2 + 6)) 2 𝑠 +4 4𝑠4 − 32𝑠2 − 192 = 4𝑠2 − 48. 𝑠2 + 4

Apply Theorem 1 a second time on the remainder term 4𝑠2 − 48 (𝑠2 + 4)3 (𝑠2 + 6) where 𝐵2 2𝑖 + 𝐶2

4𝑠2 − 48 . (𝑠2 + 4)3 (𝑠2 + 6)

𝐵2 𝑠 + 𝐶 2 𝑝2 (𝑠) + 2 (𝑠2 + 4)3 (𝑠 + 4)2 (𝑠2 + 6)

=

4𝑠2 − 48 = −32 (𝑠2 + 6) 𝑠=2𝑖 𝐵2 = 0 and 𝐶2 = −32 1 (4𝑠2 − 48 − (−32)(𝑠2 + 6)) 𝑠2 + 4 36𝑠2 + 144 = 36. 𝑠2 + 4

= ⇒

and 𝑝2 (𝑠)

63

= =

A third application of Theorem 1 on the remainder term

36 (𝑠2 + 4)2 (𝑠2 + 6)

gives 36 (𝑠2 + 4)2 (𝑠2 + 6)

=

where 𝐵3 2𝑖 + 𝐶3

= ⇒

and 𝑝3 (𝑠)

= =

𝐵3 𝑠 + 𝐶 3 𝑝3 (𝑠) + 2 (𝑠2 + 4)2 (𝑠 + 4)(𝑠2 + 6) 36 = 18 (𝑠2 + 6) 𝑠=2𝑖 𝐵3 = 0 and 𝐶3 = 18 1 (36 − (18)(𝑠2 + 6)) 𝑠2 + 4 −18𝑠2 − 72 = −18. 𝑠2 + 4

A fourth (and final) application of Theorem 1 on the remainder term −18 gives (𝑠2 + 4)(𝑠2 + 6)

64

1 Solutions

−18 + 4)(𝑠2 + 6)

=

where 𝐵4 2𝑖 + 𝐶4

=

(𝑠2

⇒ and 𝑝4 (𝑠)

= =

𝐵4 𝑠 + 𝐶 4 𝑝4 (𝑠) + 2 2 (𝑠 + 4) (𝑠 + 6) −18 = −9 (𝑠2 + 6) 𝑠=2𝑖 𝐵4 = 0 and 𝐶4 = −9 1 (−18 − (−9)(𝑠2 + 6)) 2 𝑠 +4 9𝑠2 + 36 = 9. 𝑠2 + 4

Thus the (𝑠2 + 4)-chain is The 𝑠2 + 4 -chain 4𝑠4 (𝑠2 + 4)4 (𝑠2 + 6) 4𝑠2 − 48 2 (𝑠 + 4)3 (𝑠2 + 6) 36 (𝑠2 + 4)2 (𝑠2 + 6) −18 (𝑠2 + 4)(𝑠2 + 6) 9 𝑠2 + 6

32 (𝑠2 + 4)4 −32 (𝑠2 + 4)3 18 (𝑠2 + 4)2 −9 𝑠2 + 4

5. Note that 𝑠2 +2𝑠+2 = (𝑠+1)2 +1 so 𝑠 = −1±𝑖 are the roots of 𝑠2 +2𝑠+2. We will use the root 𝑠 = −1+𝑖 for the partial fraction algorithm. Applying Theorem 1 gives

1 Solutions

1 (𝑠2

+ 2𝑠 +

2)2 (𝑠2

+ 2𝑠 +

3)2

𝐵1 𝑠 + 𝐶 1 + 2𝑠 + 2)2

=

(𝑠2

+

where 𝐵1 (−1 + 𝑖) + 𝐶1

=

⇒ = = =

Now apply Theorem 1 to the remainder term

(𝑠2

−(𝑠2 + 2𝑠 + 4) + 2𝑠 + 2)(𝑠2 + 2𝑠 + 3)2

where 𝐵2 (−1 + 𝑖) + 𝐶2

−(𝑠2 + 2𝑠 + 4) . (𝑠2 + 2𝑠 + 2)(𝑠2 + 2𝑠 + 3)2

𝐵2 𝑠 + 𝐶 2 𝑝2 (𝑠) + (𝑠2 + 2𝑠 + 2) (𝑠2 + 2𝑠 + 3)2

=

−(𝑠2 + 2𝑠 + 4) = −2 (𝑠2 + 2𝑠 + 3) 𝑠=−1+𝑖 𝐵2 = 0 and 𝐶2 = −2 −(𝑠2 + 2𝑠 + 4) − (−2)(𝑠2 + 2𝑠 + 3)2 𝑠2 + 2𝑠 + 2 2 (2(𝑠 + 1) + 5)((𝑠 + 1)2 + 1) 𝑠2 + 2𝑠 + 2 2 2𝑠 + 4𝑠 + 7.

= ⇒

and 𝑝2 (𝑠)

𝑝1 (𝑠) (𝑠2 + 2𝑠 + 2)(𝑠2 + 2𝑠 + 3)2

1 2 2 (𝑠 + 2𝑠 + 3) 𝑠=−1+𝑖 1 =1 ((−1 + 𝑖)2 + 2)2 𝐵1 = 0 and 𝐶1 = 1 1 − (1)(𝑠2 + 2𝑠 + 3)2 𝑠2 + 2𝑠 + 2 2 −(𝑠 + 2𝑠 + 4)(𝑠2 + 2𝑠 + 2) 𝑠2 + 2𝑠 + 2 2 −(𝑠 + 2𝑠 + 4).

=

and 𝑝1 (𝑠)

= = =

Thus the (𝑠2 + 2𝑠 + 2)-chain is The 𝑠2 + 2𝑠 + 2 -chain 1 (𝑠2 + 2𝑠 + 2)2 (𝑠2 + 2𝑠 + 3)2 −(𝑠2 + 2𝑠 + 4) 2 (𝑠 + 2𝑠 + 2)(𝑠2 + 2𝑠 + 3)2 2𝑠2 + 4𝑠 + 7 (𝑠2 + 2𝑠 + 3)2

65

1 (𝑠2 + 2𝑠 + 2)2 −2 𝑠2 + 2𝑠 + 2

66

1 Solutions

6. Note that 𝑠2 +2𝑠+2 = (𝑠+1)2 +1 so 𝑠 = −1±𝑖 are the roots of 𝑠2 +2𝑠+2. We will use the root 𝑠 = −1+𝑖 for the partial fraction algorithm. Applying Theorem 1 gives 1 (𝑠2 + 2𝑠 + 2)2 (𝑠2 + 4𝑠 + 5)

=

𝐵1 𝑠 + 𝐶 1 (𝑠2 + 2𝑠 + 2)2 +

where 𝐵1 (−1 + 𝑖) + 𝐶1

= = ⇒

and 𝑝1 (𝑠)

= = =

5𝑠 − 5 (𝑠2 + 4𝑠 + 5) 𝑠=−1+𝑖 5𝑖 − 10 = 5𝑖 2𝑖 + 1 𝐵1 = 5 and 𝐶1 = 5 5𝑠 − 5 − (5𝑠 + 5)(𝑠2 + 4𝑠 + 5) 𝑠2 + 2𝑠 + 2 −(5𝑠 + 15)(𝑠2 + 2𝑠 + 2) 𝑠2 + 2𝑠 + 2 −(5𝑠 + 15).

Now apply Theorem 1 to the remainder term −(5𝑠 + 15) (𝑠2 + 2𝑠 + 2)(𝑠2 + 4𝑠 + 5)

=

where 𝐵2 (−1 + 𝑖) + 𝐶2

= ⇒

and 𝑝2 (𝑠)

= = =

Thus the (𝑠2 + 2𝑠 + 2)-chain is

𝑝1 (𝑠) (𝑠2 + 2𝑠 + 2)(𝑠2 + 4𝑠 + 5)

−(5𝑠 + 15) . (𝑠2 + 2𝑠 + 2)(𝑠2 + 4𝑠 + 5)

𝐵2 𝑠 + 𝐶 2 𝑝2 (𝑠) + (𝑠2 + 2𝑠 + 2) (𝑠2 + 4𝑠 + 5) −(5𝑠 + 15) = −4 + 3𝑖 (𝑠2 + 4𝑠 + 5) 𝑠=−1+𝑖 𝐵2 = 3 and 𝐶2 = −1 −(5𝑠 + 15) − (3𝑠 − 1)(𝑠2 + 4𝑠 + 5) 𝑠2 + 2𝑠 + 2 (−3𝑠 − 5)(𝑠2 + 2𝑠 + 2) 𝑠2 + 2𝑠 + 2 −3𝑠 − 5.

1 Solutions

67

The 𝑠2 + 2𝑠 + 2 -chain 5𝑠 − 5 + 2𝑠 + 2)2 (𝑠2 + 4𝑠 + 5) −5𝑠 − 15 (𝑠2 + 2𝑠 + 2)(𝑠2 + 4𝑠 + 5) −3𝑠 − 5 2 𝑠 + 4𝑠 + 5

5𝑠 + 5 + 2𝑠 + 2)2 3𝑠 − 1 𝑠2 + 2𝑠 + 2

(𝑠2

(𝑠2

7. Use Theorem 1 of Section 2.2 to compute the (𝑠 − 3)-chain: (𝑠2

𝑠 𝐴1 = + 1)(𝑠 − 3) 𝑠−3

+

𝑝1 (𝑠) 𝑠2 + 1

𝑠 3 = 2 𝑠 + 1 𝑠=3 10 1 1 and 𝑝1 (𝑠) = (𝑠 − (3/10)(𝑠2 + 1)) = (−3𝑠2 + 10𝑠 − 3) 𝑠−3 10(𝑠 − 3) −3𝑠 + 1 = 10 where 𝐴1 =

Since the remainder term

−3𝑠 + 1 is already a simple partial fraction, 10(𝑠2 + 1)

we conclude 1 𝑠 = (𝑠2 + 1)(𝑠 − 3) 10

(

3 1 − 3𝑠 + 𝑠 − 3 𝑠2 + 1

)

8. Use Theorem of Section 2.2 1 to compute the (𝑠 + 1)-chain: (𝑠2

4𝑠 𝐴1 = 2 + 1) (𝑠 + 1) 𝑠+1 where 𝐴1 =

+

𝑝1 (𝑠) (𝑠2 + 1)2

4𝑠 = −1 2 2 (𝑠 + 1) 𝑠=−1

1 𝑠4 + 2𝑠2 + 4𝑠 + 1 (4𝑠 − (−1)(𝑠2 + 1)2 ) = 𝑠+1 𝑠+1 3 2 = 𝑠 − 𝑠 + 3𝑠 + 1

and 𝑝1 (𝑠) =

𝑠3 − 𝑠2 + 3𝑠 + 1 (𝑠2 + 1)2 2 using Theorem 1. Since 𝑠 = 𝑖 is a root of 𝑠 + 1, an application of this theorem gives Now compute the 𝑠2 + 1-chain for the remainder term

68

1 Solutions

𝑠3 − 𝑠2 + 3𝑠 + 1 𝐵1 𝑠 + 𝐶 1 𝑝1 (𝑠) = 2 + 2 (𝑠2 + 1)2 (𝑠 + 1)2 (𝑠 + 1) 𝑠3 − 𝑠2 + 3𝑠 + 1 = 2𝑖 + 2 1 𝑠=𝑖 ⇒ 𝐵1 = 2 and 𝐶1 = 2 1 and 𝑝1 (𝑠) = 2 (𝑠3 − 𝑠2 + 3𝑠 + 1 − (2𝑠 + 2)(1)) 𝑠 +1 𝑠3 − 𝑠 2 + 𝑠 − 1 = = 𝑠 − 1. 𝑠2 + 1

where 𝐵1 𝑖 + 𝐶1 =

𝑠−1 is a simple partial fraction, we conclude 𝑠2 + 1 that the complete partial fraction decomposition is Since the remainder term

(𝑠2

4𝑠 2𝑠 + 2 𝑠−1 1 = 2 + 2 − 2 2 + 1) (𝑠 + 1) (𝑠 + 1) 𝑠 +1 𝑠+1

9. Use Theorem of Section 2.2 1 to compute the (𝑠 − 3)-chain: 2 𝐴1 = (𝑠2 − 6𝑠 + 10)(𝑠 − 3) 𝑠−3

+

𝑝1 (𝑠) (𝑠2 − 6𝑠 + 10)

2 where 𝐴1 = 2 =2 (𝑠 − 6𝑠 + 10) 𝑠=3

1 −2𝑠2 + 12𝑠 − 18 (2 − (2)(𝑠2 − 6𝑠 + 10)) = 𝑠−3 𝑠−3 = −2𝑠 + 6

and 𝑝1 (𝑠) =

−2𝑠 + 6 is a simple partial fraction, we con− 6𝑠 + 10 2 clude that the complete partial fraction decomposition is 2 = (𝑠 − 6𝑠 + 10)(𝑠 − 3) 2 6 − 2𝑠 + 𝑠 − 3 (𝑠 − 3)2 + 1 Since the remainder term

𝑠2

10. Use Theorem of Section 2.2 1 to compute the (𝑠 − 1)-chain: (𝑠2

30 𝐴1 = − 4𝑠 + 13)(𝑠 − 1) 𝑠−1 where 𝐴1 =

+

(𝑠2

𝑝1 (𝑠) − 4𝑠 + 13)

30 =3 2 (𝑠 − 4𝑠 + 13) 𝑠=1

−3𝑠2 + 12𝑠 − 9 1 (30 − (3)(𝑠2 − 4𝑠 + 13)) = 𝑠−1 𝑠−1 = −3𝑠 + 9

and 𝑝1 (𝑠) =

1 Solutions

69

−3𝑠 + 9 is a simple partial fraction, we con𝑠2 − 4𝑠 + 13 30 clude that the complete partial fraction decomposition is 2 = (𝑠 − 4𝑠 + 13)(𝑠 − 1) 9 − 3𝑠 3 + 2 ((𝑠 − 2) + 9) 𝑠 − 1 Since the remainder term

11. Note that 𝑠2 −4𝑠+8 = (𝑠−2)2 +2 so 𝑠 = 2±2𝑖 are the roots of 𝑠2 −4𝑠+8. We will use the root 𝑠 = 2+2𝑖 to compute the (𝑠2 −4𝑠+8)-chain. Applying Theorem 1 gives (𝑠2

25 𝐵1 𝑠 + 𝐶 1 = 2 2 − 4𝑠 + 8) (𝑠 − 1) (𝑠 − 4𝑠 + 8)2 𝑝1 (𝑠) + 2 (𝑠 − 4𝑠 + 8)(𝑠 − 1)

where 𝐵1 (2 + 2𝑖) + 𝐶1 = = ⇒ and 𝑝1 (𝑠) = = =

25 𝑠 − 1) 𝑠=2+2𝑖 25 = 5 − 10𝑖 2𝑖 + 1 𝐵1 = −5 and 𝐶1 = 15 25 − (−5𝑠 + 15)(𝑠 − 1) 𝑠2 − 4𝑠 + 8 2 (5)(𝑠 − 4𝑠 + 8) 𝑠2 − 4𝑠 + 8 5.

Now apply Theorem 1 to the remainder term

(𝑠2

5 . − 4𝑠 + 8)(𝑠 − 1)

5 𝐵2 𝑠 + 𝐶 2 𝑝2 (𝑠) = 2 + − 4𝑠 + 8)(𝑠 − 1) (𝑠 − 4𝑠 + 8) 𝑠 − 1

where 𝐵2 (2 + 2𝑖) + 𝐶2 = ⇒ and 𝑝2 (𝑠) = = = Thus the partial fraction expansion is 𝑠2

(𝑠2

−𝑠 + 3 1 + − 4𝑠 + 8 𝑠 − 1

5) = 1 − 2𝑖 𝑠 − 1 𝑠=2+2𝑖 𝐵2 = −1 and 𝐶2 = 3 5 − (3 − 𝑠)(𝑠 − 1) 𝑠2 − 4𝑠 + 8 (1)(𝑠2 − 4𝑠 + 8) 𝑠2 − 4𝑠 + 8 1.

25 −5𝑠 + 15 = 2 + (𝑠2 − 4𝑠 + 8)2 (𝑠 − 1) (𝑠 − 4𝑠 + 8)2

70

1 Solutions

12. Note that 𝑠2 + 6𝑠 + 10 = (𝑠 + 3)2 + 1 so 𝑠 = −3 ± 𝑖 are the roots of 𝑠2 + 6𝑠 + 10. We will use the root 𝑠 = −3 + 𝑖 to compute the (𝑠2 + 6𝑠 + 10)chain. Applying Theorem 1 gives 𝑠 𝐵1 𝑠 + 𝐶 1 = 2 (𝑠2 + 6𝑠 + 10)2 (𝑠 + 3)2 (𝑠 + 6𝑠 + 10)2 𝑝1 (𝑠) + (𝑠2 + 6𝑠 + 10)(𝑠 + 3)2 𝑠 2 (𝑠 + 3) ) 𝑠=−3+𝑖 = 3−𝑖 ⇒ 𝐵1 = −1 and 𝐶1 = 0 𝑠 − (−𝑠)(𝑠 + 3)2 and 𝑝1 (𝑠) = 𝑠2 + 6𝑠 + 10 (𝑠)(𝑠2 + 6𝑠 + 10) = 𝑠2 + 6𝑠 + 10 = 𝑠. 𝑠 Now apply Theorem 1 to the remainder term 2 . (𝑠 + 6𝑠 + 10)(𝑠 + 3)2 ) where 𝐵1 (−3 + 𝑖) + 𝐶1 =

𝐵2 𝑠 + 𝐶 2 𝑝2 (𝑠) 𝑠 = 2 + (𝑠2 + 6𝑠 + 10)(𝑠 + 3)2 (𝑠 + 6𝑠 + 10) (𝑠 + 3)2 where 𝐵2 (−3 + 𝑖) + 𝐶2 = ⇒ and 𝑝2 (𝑠) = = =

𝑠) =3−𝑖 (𝑠 + 3)2 𝑠=−3+𝑖 𝐵2 = −1 and 𝐶2 = 0 𝑠 − (−𝑠)(𝑠 + 3)2 𝑠2 + 6𝑠 + 10 (𝑠)(𝑠2 + 6𝑠 + 10) 𝑠2 + 6𝑠 + 10 𝑠.

The remainder term is 𝑠 (𝑠 + 3) − 3 −3 1 = = + , 2 2 2 (𝑠 + 3) (𝑠 + 3) (𝑠 + 3) 𝑠+3 so the partial fraction expansion of the entire rational function is 𝑠 −𝑠 𝑠 = 2 − 2 (𝑠2 + 6𝑠 + 10)2 (𝑠 + 3)2 (𝑠 + 6𝑠 + 10)2 𝑠 + 6𝑠 + 10 3 1 − + (𝑠 + 3)2 𝑠+3

1 Solutions

71

13. Note that 𝑠2 +4𝑠+5 = (𝑠+2)2 +1 so 𝑠 = −2±𝑖 are the roots of 𝑠2 +4𝑠+5. We will use the root 𝑠 = −2+𝑖 to compute the (𝑠2 +4𝑠+5)-chain. Applying Theorem 1 gives 𝑠+1 𝐵1 𝑠 + 𝐶 1 = 2 (𝑠2 + 4𝑠 + 5)2 (𝑠2 + 4𝑠 + 6)2 (𝑠 + 4𝑠 + 5)2 +

where 𝐵1 (−2 + 𝑖) + 𝐶1 = = ⇒ and 𝑝1 (𝑠) = = = =

𝑝1 (𝑠) (𝑠2 + 4𝑠 + 5)(𝑠2 + 4𝑠 + 6)2

𝑠+1 (𝑠2 + 4𝑠 + 6)2 ) 𝑠=−2+𝑖 −1 + 𝑖 𝐵1 = 1 and 𝐶1 = 1 𝑠 + 1 − (𝑠 + 1)(𝑠2 + 4𝑠 + 6)2 𝑠2 + 4𝑠 + 5 −(𝑠 + 1)((𝑠2 + 4𝑠 + 6)2 − 1) 𝑠2 + 4𝑠 + 5 −(𝑠 + 1)(𝑠2 + 4𝑠 + 7)(𝑠2 + 4𝑠 + 5) 𝑠2 + 4𝑠 + 5 2 −(𝑠 + 1)(𝑠 + 4𝑠 + 7).

Now apply Theorem 1 to the remainder term

−(𝑠 + 1)(𝑠2 + 4𝑠 + 7) . (𝑠2 + 4𝑠 + 5)(𝑠2 + 4𝑠 + 6)2 )

𝐵2 𝑠 + 𝐶 2 𝑝2 (𝑠) −(𝑠 + 1)(𝑠2 + 4𝑠 + 7) = 2 + 2 2 2 2 (𝑠 + 4𝑠 + 5)(𝑠 + 4𝑠 + 6) (𝑠 + 4𝑠 + 5) (𝑠 + 4𝑠 + 6)2 where 𝐵2 (−2 + 𝑖) + 𝐶2 = ⇒ and 𝑝2 (𝑠) = = =

−(𝑠 + 1)(𝑠2 + 4𝑠 + 7) = 2 − 2𝑖 (𝑠2 + 4𝑠 + 6)2 𝑠=−2+𝑖 𝐵2 = −2 and 𝐶2 = −2 −(𝑠 + 1)(𝑠2 + 4𝑠 + 7) − (−2𝑠 − 2)(𝑠2 + 4𝑠 + 6)2 𝑠2 + 4𝑠 + 5 (𝑠 + 1)(2(𝑠2 + 4𝑠 + 6) + 1)(𝑠2 + 4𝑠 + 5) 𝑠2 + 4𝑠 + 5 2 (𝑠 + 1)(2(𝑠 + 4𝑠 + 6) + 1).

The remainder term is (𝑠 + 1)(2(𝑠2 + 4𝑠 + 6) + 1) 𝑠+1 2𝑠 + 2 = 2 + 2 2 2 2 (𝑠 + 4𝑠 + 6) (𝑠 + 4𝑠 + 6) 𝑠 + 4𝑠 + 6 so the partial fraction expansion of the entire rational function is

72

1 Solutions

𝑠+1 𝑠+1 2𝑠 + 2 = 2 + 2 (𝑠2 + 4𝑠 + 5)2 (𝑠2 + 4𝑠 + 6)2 (𝑠 + 4𝑠 + 6)2 𝑠 + 4𝑠 + 6 𝑠+1 2𝑠 + 2 + 2 − 2 (𝑠 + 4𝑠 + 5)2 𝑠 + 4𝑠 + 5 𝑢 , (𝑢 + 5)3 (𝑢 + 6)2 which can be put in partial fraction form by the technique of Section 2.2. Use Theorem 1 of Section 2.2 to compute the (𝑢 + 5)-chain:

14. Using the hint, let 𝑢 = 𝑠2 . Then the rational function becomes

𝑢 𝐴1 = (𝑢 + 5)3 (𝑢 + 6)2 (𝑢 + 5)3

𝑝1 (𝑢) (𝑢 + 5)2 (𝑢 + 6)2

+

𝑢 = −5 where 𝐴1 = 2 (𝑢 + 6) 𝑢=−5 1 and 𝑝1 (𝑢) = (𝑢 − (−5)(𝑢 + 6)2 ) 𝑢+5 5𝑢2 + 61𝑢 + 180 (5𝑢 + 36)(𝑢 + 5) = = = 5𝑢 + 36. 𝑢+6 𝑢+5 Continuing gives 5𝑢 + 36 𝐴2 = (𝑢 + 5)2 )(𝑢 + 6)2 (𝑢 + 5)2

+

𝑝2 (𝑢) (𝑢 + 5)(𝑢 + 6)2

5𝑢 + 36 where 𝐴2 = = 11 (𝑢 + 6)2 𝑢=−5 1 and 𝑝2 (𝑢) = (5𝑢 + 36 − (11)(𝑢 + 6)2 ) 𝑢+5 −11𝑢2 − 127𝑢 − 360 −(11𝑢 + 72)(𝑢 + 5) = = 𝑢+5 𝑢+5 = −(11𝑢 + 72). Another recursion gives −11𝑢 − 72 𝐴3 = (𝑢 + 5)(𝑢 + 6)2 (𝑢 + 5)

+

𝑝3 (𝑢) (𝑢 + 6)2

−11𝑢 − 72 where 𝐴3 = = −17 (𝑢 + 6)2 𝑢=−5 1 and 𝑝3 (𝑢) = (−11𝑢 − 72 − (−17)(𝑢 + 6)2 ) 𝑢+5 (17𝑢 + 108)(𝑢 + 5) 17𝑢2 + 193𝑢 + 540 = = 𝑢+5 𝑢+5 = 17𝑢 + 108.

1 Solutions

73

Thus we get 𝑢 −5 11 17 = + − (𝑢 + 5)3 (𝑢 + 6)2 (𝑢 + 5)3 (𝑢 + 5)2 𝑢+5 17𝑢 + 108 + (𝑢 + 6)2 −5 11 17 = + − (𝑢 + 5)3 (𝑢 + 5)2 𝑢+5 17 6 + + . (𝑢 + 6)2 𝑢+6 Replacing 𝑢 by 𝑠2 in the above expression, gives the following expression 𝑠2 : for the partial fraction decomposition of 2 (𝑠 + 5)3 (𝑠2 + 6)2 (𝑠2

11 17 6 17 −5 + 2 − 2 + 2 + 2 3 2 2 + 5) (𝑠 + 5) 𝑠 + 5 (𝑠 + 6) 𝑠 +6

Section 2.4 1. ℒ−1 {−5/𝑠} = −5ℒ−1 {1/𝑠} = −5 2. ℒ−1 {3/(𝑠 − 4)} = 3ℒ−1 {1/(𝑠 − 4)} = 3𝑒4𝑡 { } { } { } 3 4 3. ℒ−1 − = 3ℒ−1 1/𝑠2 − 2ℒ−1 2/𝑠3 = 3𝑡 − 2𝑡2 2 3 𝑠 𝑠 } { } { 1 4 4. ℒ−1 = 2ℒ−1 = 2𝑒−3𝑡/2 2𝑠 + 3 𝑠 + (3/2) { } { } 3𝑠 𝑠 −1 −1 5. ℒ = 3ℒ = 3 cos 2𝑡 𝑠2 + 4 𝑠2 + 2 2 { } √ } { √ 2 −1 2 3 2 −1 √ 6. ℒ =√ ℒ = √ sin 3 𝑡 2 2 2 𝑠 +3 3 𝑠 + ( 3) 3 2𝑠 − 5 = 7. First, we have 𝑠2 + 6𝑠 + 9 = (𝑠 + 3)2 . Partial fractions gives 2 𝑠 + 6𝑠 + 9 { } −11 2 2𝑠 − 5 + . So ℒ−1 = −11𝑡𝑒−3𝑡 + 2𝑒−3𝑡 (𝑠 + 3)2 𝑠+3 (𝑠 + 3)2 { } 2𝑠 − 5 −11 2 2𝑠 − 5 −1 8. Partial fractions gives = + . Thus ℒ = (𝑠 + 3)3 (𝑠 + 3)3 (𝑠 + 3)2 (𝑠 + 3)3 −11 2 −3𝑡 𝑡 𝑒 + 2𝑡𝑒−3𝑡 2

74

1 Solutions

{ } 6 6 −1 1 6 −1 = = + . So ℒ = 𝑠2 + 2𝑠 − 8 (𝑠 − 2)(𝑠 + 4) 𝑠+4 𝑠−2 𝑠2 + 2𝑠 − 8 𝑒2𝑡 − 𝑒−4𝑡 { } 𝑠 𝑠 −2 3 𝑠 −1 10. 2 = = + . So ℒ = 𝑠 − 5𝑠 + 6 (𝑠 − 2)(𝑠 − 3) 𝑠−2 𝑠−3 (𝑠 − 2)(𝑠 + 3) 3𝑒3𝑡 − 2𝑒2𝑡 { 2 } 2𝑠2 − 5𝑠 + 1 −1 3 2 2𝑠 − 5𝑠 + 1 −1 = + + . So ℒ = 11. (𝑠 − 2)4 (𝑠 − 2)4 (𝑠 − 2)3 (𝑠 − 2)2 (𝑠 − 2)4 −1 3 2𝑡 3 2 2𝑡 𝑡 𝑒 + 𝑡 𝑒 + 2𝑡𝑒2𝑡 6 2 { } 2𝑠 + 6 2𝑠 + 6 −2 4 2𝑠 + 6 −1 12. 2 = = + . So ℒ = 𝑠 − 6𝑠 + 5 (𝑠 − 1)(𝑠 − 5) 𝑠−1 𝑠−5 𝑠62 − 6𝑠 + 5 5𝑡 𝑡 4𝑒 − 2𝑒 { } 1 1 1 1 4𝑠2 4𝑠2 −1 13. = + + − . So ℒ = (𝑠 − 1)2 (𝑠 + 1)2 (𝑠 − 1)2 𝑠 − 1 (𝑠 + 1)2 𝑠 + 1 (𝑠 − 1)2 (𝑠 + 1)2 𝑡𝑒𝑡 + 𝑒𝑡 + 𝑡𝑒−𝑡 − 𝑒−𝑡 { } 27 9 3 1 1 27 9𝑡2 14. 2 = 3− 2+ − . So ℒ−1 −3𝑡+1−𝑒−3𝑡 = 3 𝑠 (𝑠 + 3) 𝑠 𝑠 𝑠 𝑠−3 𝑠 (𝑠 + 3) 2 { } 8𝑠 + 16 4 1 𝑠 2 8𝑠 + 16 −1 15. 2 = − + − . So ℒ = (𝑠 + 4)(𝑠 − 2)2 (𝑠 − 2)2 𝑠 − 2 𝑠2 + 4 𝑠2 + 4 (𝑠2 + 4)(𝑠 − 2)2 2𝑡 2𝑡 4𝑡𝑒 − 𝑒 + cos 2𝑡 − sin 2𝑡 { } 5𝑠 + 15 2 𝑠 3 5𝑠 + 15 −1 16. 2 = −2 2 + . So ℒ = (𝑠 + 9)(𝑠 − 1) 𝑠−1 𝑠 + 9 𝑠2 + 9 (𝑠2 + 9)(𝑠 − 1) 𝑡 −2 cos 3𝑡 + sin 3𝑡 + 2𝑒 { } 12 −6 3 4 1 12 = 2 + − + . So ℒ−1 17. 2 = 𝑠 (𝑠 + 1)(𝑠 − 2) 𝑠 𝑠 𝑠+1 𝑠−2 𝑠2 (𝑠 + 1)(𝑠 − 2) 3 − 6𝑡 + 𝑒2𝑡 − 4𝑒−𝑡 9.

18.

2𝑠 6 14 22 8 22 = + + + − . So 3 2 3 2 2 (𝑠 −{3) (𝑠 − 4) (𝑠}− 3) (𝑠 − 3) 𝑠−3 (𝑠 − 4) 𝑠−4 2𝑠 ℒ−1 = 3𝑡2 𝑒3𝑡 + 14𝑡𝑒3𝑡 + 22𝑒3𝑡 + 8𝑡𝑒4𝑡 − 22𝑒4𝑡 (𝑠 − 3)3 (𝑠 − 4)2

19. First we have 𝑠2 + 2𝑠 + 5 = (𝑠 + 1)2 + 4. So

2𝑠 2𝑠 = = 𝑠2 + 2𝑠 + 5 (𝑠 + 1)2 + 4

2(𝑠 + 1) − 2 𝑠+1 2 =2 − . The First Translation princi2 (𝑠 + 1)2 + 4 { (𝑠 + 1)2 +}4 (𝑠 + 1) { +4 } { } 2𝑠 𝑠+1 2 −1 −1 −1 ple gives ℒ = 2ℒ −ℒ = 𝑠2 + 2𝑠 + 5 (𝑠 + 1)2 + 4 (𝑠 + 1)2 + 4 −𝑡 −𝑡 2𝑒 cos 2𝑡 − 𝑒 sin 2𝑡

1 Solutions

20.

𝑠2

1 1 = . Thus ℒ−1 + 6𝑠 + 10 (𝑠 + 3)2 + 1

{

1 (𝑠 + 3)2 + 1

}

75

= 𝑒−3𝑡 sin 𝑡

{ } 𝑠−1 𝑠−4 1 𝑠−1 −1 = +3 . Thus ℒ = 𝑠2 − 8𝑠 + 17 (𝑠 − 4)2 + 1 (𝑠 − 4)2 + 1 𝑠2 − 8𝑠 + 17 𝑒4𝑡 cos 𝑡 + 3𝑒4𝑡 sin 𝑡 √ { } √ 2𝑠 + 4 8 𝑠−2 2𝑠 + 4 −1 √ + 8 √ . Thus ℒ 22. 2 =2 = 2 + ( 8)2 𝑠 − 4𝑠 + 12 𝑠2 − 4𝑠 + 12 (𝑠 − 2)√ (𝑠 − 2)2 + ( 8)2 √ √ 2𝑒2𝑡 cos 8 𝑡 + 8 𝑒2𝑡 sin 8 𝑡 { } 𝑠−1 𝑠−1 𝑠−1 −1 = . Thus ℒ = 𝑒𝑡 cos 3𝑡 23. 2 𝑠 − 2𝑠 + 10 (𝑠 − 1)2 + 32 𝑠2 − 2𝑠 + 10 { } 𝑠−5 𝑠−3 2 𝑠−5 −1 24. 2 = − . Thus ℒ = 𝑠 − 6𝑠 + 13 (𝑠 − 3)2 + 22 (𝑠 − 3)2 + 22 𝑠2 − 6𝑠 + 13 𝑒3𝑡 cos 2𝑡 − 𝑒3𝑡 sin 2𝑡 } { } { 𝑠 8 8𝑠 −1 = 8ℒ = (2𝑡 sin 2𝑡) = 2𝑡 sin 2𝑡 25. ℒ−1 (𝑠2 + 4)2 (𝑠2 + 22 )2 2 ⋅ 22 { } { } 9 3 3 −1 −1 26. ℒ = 3ℒ = (sin 3𝑡 − 3𝑡 cos 3𝑡) = (𝑠2 + 9)2 (𝑠2 + 32 )2 2 ⋅ 32 1 1 sin 3𝑡 − 𝑡 cos 3𝑡 6 2 21.

2 27. We first complete the square = (𝑠 + 2)2{+ 1. By the transla{ 𝑠 + 4𝑠 + 5 } } 2𝑠 (𝑠 + 2) − 2 −1 tion principle we get ℒ−1 = 2ℒ = 2 4𝑠 + 5)2 }) ((𝑠(+ 2)2 + 1)2 ( { } (𝑠 + { ) 𝑠 1 1 1 −1 −2𝑡 2𝑒−2𝑡 ℒ−1 − 2ℒ = 2𝑒 𝑡 sin 𝑡 − 2( (sin 𝑡 − 𝑡 cos 𝑡) = (𝑠2 + 1)2 (𝑠2 + 1)2 2 2 2𝑡𝑒−2𝑡 cos 𝑡 + (𝑡 − 2)𝑒−2𝑡 sin 𝑡

28. We first complete the square 𝑠2 −6𝑠+10}= (𝑠−3)2 +1. { { By the translation } 2𝑠 + 2 (𝑠 − 3) + 3 + 1 −1 −1 principle we get ℒ = 2ℒ = 2 2 2 2 ((𝑠 ( { } (𝑠 − 6𝑠{+ 10) }) ( − 3) + 1) ) 𝑠 1 1 1 3𝑡 −1 −1 3𝑡 2𝑒 ℒ + 4ℒ = 2𝑒 𝑡 sin 𝑡 + 4( (sin 𝑡 − 𝑡 cos 𝑡) = (𝑠2 + 1)2 (𝑠2 + 1)2 2 2 −4𝑡𝑒3𝑡 cos 𝑡 + (𝑡 + 4)𝑒3𝑡 sin 𝑡 29. We first complete the square 𝑠2 +8𝑠+17}= (𝑠+4)2 +1. { { By the translation } 2𝑠 (𝑠 + 4) − 4 −1 −1 principle we get ℒ = 2ℒ = 2 2 ((𝑠 ( + 4)2 + 1)2 ( { }(𝑠 + 8𝑠 + { 17) }) ) 𝑠 1 1 1 −1 −4𝑡 2𝑒−4𝑡 ℒ−1 − 4ℒ = 2𝑒 𝑡 sin 𝑡 − 4( (sin 𝑡 − 𝑡 cos 𝑡) = (𝑠2 + 1)2 (𝑠2 + 1)2 2 2 4𝑡𝑒−4𝑡 cos 𝑡 + (𝑡 − 4)𝑒−4𝑡 sin 𝑡

76

1 Solutions

2 30. We first complete the square = (𝑠 + 1){2 + 1. By the trans{ 𝑠 + 2𝑠 + 2 } } 𝑠+1 𝑠+1 −1 lation principle we get ℒ−1 = ℒ = (𝑠2 + 2𝑠 + 2)3 ((𝑠 + 1)2 + 1)3 }) ( { ) 1 𝑠 1( = 𝑒−𝑡 𝑡 sin 𝑡 − 𝑡2 cos 𝑡 = (𝑡𝑒−𝑡 sin 𝑡−𝑡2 𝑒−𝑡 cos 𝑡) 𝑒−𝑡 ℒ−1 (𝑠2 + 1)3 8 8 2 2 2 31. We first complete the square { 𝑠 − 2𝑠 + 5 = } (𝑠 − 1) {+ 2 . By the transla} 1 1 −1 −1 tion principle we get ℒ =ℒ = 2 2 3 (𝑠2 ( − 2𝑠 +{5)3 }) }) ((𝑠 − 1) + 2 ) ( { 1 1 2 = 𝑒𝑡 ℒ−1 𝑒𝑡 ℒ−1 2 2 3 2 (𝑠 + 2 ) 2 (𝑠 + 22 )3 ( ) 1 = 𝑒𝑡 (3 − (2𝑡)2 ) sin 2𝑡 − 6𝑡 cos 2𝑡 4 2⋅8⋅2 ) 1 ( = (3 − 4𝑡2 )𝑒𝑡 sin 2𝑡 − 6𝑡𝑒𝑡 cos 2𝑡 256

32. We first complete the square 𝑠2 −6𝑠+10}= (𝑠−3)2 +1. { { By the translation } 8𝑠 (𝑠 − 3) + 3 −1 −1 principle we get ℒ = 8ℒ = 2 3 ((𝑠 − 3)2 + 1)3 ( { } (𝑠 − 6𝑠{+ 10) }) 𝑠 1 8𝑒3𝑡 ℒ−1 + 3ℒ−1 2 3 2 (𝑠 + 1) (𝑠 + 1)3 ( ) ( ) 1 1 = 8𝑒3𝑡 𝑡 sin 𝑡 − 𝑡2 cos 𝑡 + 3 ((3 − 𝑡2 ) sin 𝑡 − 3𝑡 cos 𝑡) 8 8 = (−3𝑡2 + 𝑡 + 9)𝑒3𝑡 sin 𝑡 − (𝑡2 + 9𝑡)𝑒3𝑡 cos 𝑡

2 2 33. We first complete the square { 𝑠 − 8𝑠 + 17 = } (𝑠 − 4) {+ 1. By the transla} 𝑠−4 𝑠−4 −1 −1 tion principle we get ℒ =ℒ = (𝑠2 − 8𝑠 + 17)4 ((𝑠 − 4)2 + 1)4 ( { }) ) 𝑠 1 ( 𝑒4𝑡 ℒ−1 = 𝑒4𝑡 (3𝑡 − 𝑡3 ) sin 𝑡 − 3𝑡2 cos 𝑡 2 4 (𝑠 + 1) 48 ) 1 ( 3 4𝑡 2 4𝑡 = (−𝑡 + 3𝑡)𝑒 cos 𝑡 − 3𝑡 𝑒 cos 𝑡 48

2 2 2 34. We first complete the square { 𝑠 + 4𝑠 + 8 = } (𝑠 + 2) {+ 2 . By the transla} 2 2 −1 tion principle we get ℒ−1 = ℒ = 2 3 ((𝑠 + 2)2 + 22 )3 ( { }) (𝑠 + 4𝑠 + 8) ) 2 1 ( 𝑒−2𝑡 ℒ−1 = 𝑒−2𝑡 (3 − (2𝑡)2 ) sin 2𝑡 − 6𝑡 cos 2𝑡 (𝑠2 + 22 )3 8 ⋅ 24 ) 1 ( = (3 − 4𝑡2 )𝑒−2𝑡 sin 2𝑡 − 6𝑡𝑒−2𝑡 cos 2𝑡 128

35. Compute the partial fraction Then { ℒ−1

1 (𝑠 − 𝑎)(𝑠 − 𝑏)

}

= ℒ−1

{

1 1/(𝑎 − 𝑏) −1/(𝑏 − 𝑎) = + . (𝑠 − 𝑎)(𝑠 − 𝑏) 𝑠−𝑎 𝑠−𝑏 1/(𝑎 − 𝑏) −1/(𝑏 − 𝑎) + 𝑠−𝑎 𝑠−𝑏

}

=

𝑒𝑎𝑡 𝑒𝑏𝑡 + . 𝑎−𝑏 𝑏−𝑎

1 Solutions

77

36. Apply the inverse Laplace transform to the partial fraction expansion 𝑠 𝑎/(𝑎 − 𝑏) 𝑏/(𝑏 − 𝑎) = + . (𝑠 − 𝑎)(𝑠 − 𝑏) 𝑠−𝑎 𝑠−𝑏 37. Apply the inverse Laplace transform to the partial fraction expansion 1 1 1 1 1 1 1 = + + . (𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) (𝑎 − 𝑏)(𝑎 − 𝑐) 𝑠 − 𝑎 (𝑏 − 𝑎)(𝑏 − 𝑐) 𝑠 − 𝑏 (𝑐 − 𝑎)(𝑐 − 𝑏) (𝑠 − 𝑐) 38. Apply the inverse Laplace transform to the partial fraction expansion 𝑠 𝑎 1 𝑏 1 𝑐 1 = + + . (𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) (𝑎 − 𝑏)(𝑎 − 𝑐) 𝑠 − 𝑎 (𝑏 − 𝑎)(𝑏 − 𝑐) 𝑠 − 𝑏 (𝑐 − 𝑎)(𝑐 − 𝑏) (𝑠 − 𝑐) 39. Apply the inverse Laplace transform to the partial fraction expansion 𝑠2 𝑎2 1 𝑏2 1 𝑐2 1 = + + . (𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) (𝑎 − 𝑏)(𝑎 − 𝑐) 𝑠 − 𝑎 (𝑏 − 𝑎)(𝑏 − 𝑐) 𝑠 − 𝑏 (𝑐 − 𝑎)(𝑐 − 𝑏) (𝑠 − 𝑐) 40. Apply the inverse Laplace transform to the partial fraction expansion 𝑠𝑘 𝐴1 𝐴𝑛 = +⋅⋅⋅+ (𝑠 − 𝑟1 ) ⋅ ⋅ ⋅ (𝑠 − 𝑟𝑛 ) 𝑠 − 𝑟1 𝑠 − 𝑟𝑛 where 𝐴𝑖 =

𝑠𝑘 𝑟𝑘 = ′ 𝑖 . 𝑞(𝑠)/(𝑠 − 𝑟𝑖 ) 𝑠=𝑟𝑖 𝑞 (𝑟𝑖 )

The last equality is true since the product rule for derivatives implies that 𝑞 ′ (𝑟𝑖 ) = (𝑟1 − 𝑟1 ) ⋅ ⋅ ⋅ (𝑟𝑖 − 𝑟𝑖−1 )(𝑟𝑖 − 𝑟𝑖+1 ) ⋅ ⋅ ⋅ (𝑟𝑖 − 𝑟𝑛 ), that is, the derivative of 𝑞(𝑠) evaluated at one of the roots 𝑟𝑖 is obtained by deleting the term 𝑠 − 𝑟𝑖 from 𝑞(𝑠) and then evaluating at 𝑟𝑖 and this is the same expression which is evaluated to put in the denominator of the coefficient 𝐴𝑖 . 41. This is directly from Table 2.3. 42. Apply the inverse Laplace transform to the partial fraction expansion 𝑠 (𝑠 − 𝑎) + 𝑎 1 𝑎 = = + . (𝑠 − 𝑎)2 (𝑠 − 𝑎)2 𝑠 − 𝑎 (𝑠 − 𝑎)2 43. This is directly from Table 2.3.

78

1 Solutions

44. Apply the inverse Laplace transform to the partial fraction expansion 𝑠 (𝑠 − 𝑎) + 𝑎 1 𝑎 = = + . (𝑠 − 𝑎)3 (𝑠 − 𝑎)3 (𝑠 − 𝑎)2 (𝑠 − 𝑎)3 45. Apply the inverse Laplace transform to the partial fraction expansion 𝑠2 ((𝑠 − 𝑎) + 𝑎)2 1 2𝑎 𝑎2 = = + + . 3 3 2 (𝑠 − 𝑎) (𝑠 − 𝑎) 𝑠 − 𝑎 (𝑠 − 𝑎) (𝑠 − 𝑎)3 46. To compute the partial fraction write 𝑠 = (𝑠 − 𝑎) + 𝑎 and compute 𝑠𝑘 = ((𝑠 − 𝑎) + 𝑎)𝑘 by the binomial theorem: 𝑠𝑘 = ((𝑠 − 𝑎) + 𝑎)𝑘 =

𝑘 ( ) ∑ 𝑘 𝑙=0

𝑙

𝑎𝑘−𝑙 (𝑠 − 𝑎)𝑙 .

Thus, 𝑘 ( ) ∑ 𝑠𝑘 𝑘 𝑘−𝑙 1 𝑎 = . (𝑠 − 𝑎)𝑛 𝑙 (𝑠 − 𝑎)𝑛−𝑙 𝑙=0

Now apply the inverse Laplace transform.

Section 2.5 1. Yes. 2. No; 𝑡−2 is not a polynomial. 3. Yes;

𝑡 = 𝑡𝑒−𝑡 . 𝑒𝑡

1 is not a polynomial. 𝑡 √ √ 𝜋 2 2 5. Yes; 𝑡 sin(4𝑡 − ) = 𝑡( sin 4𝑡 − cos 4𝑡). 4 2 2 4. No;

6. Yes; (𝑡 + 𝑒𝑡 )2 = 𝑡2 + 2𝑡𝑒𝑡 + 𝑒2𝑡 . 7. No. 8. Yes. 1

9. No; 𝑡 2 is not a polynomial. 10. Yes;

sin 2𝑡 = 𝑒−2𝑡 sin 2𝑡. 𝑒2𝑡

1 Solutions

11. No. { } 12. ℬ𝑞 = 𝑒4𝑡 { } 13. ℬ𝑞 = 𝑒−6𝑡 { } 14. ℬ𝑞 = 1, 𝑒−5𝑡 { } 15. ℬ𝑞 = 𝑒4𝑡 , 𝑒−𝑡 { } 16. ℬ𝑞 = 𝑒3𝑡 , 𝑡𝑒3𝑡 { } 17. ℬ𝑞 = 𝑒2𝑡 , 𝑒7𝑡 { } 18. ℬ𝑞 = 𝑒3𝑡 , 𝑒−2𝑡 { } 19. ℬ𝑞 = 𝑒−3𝑡 , 𝑒−6𝑡 { } 20. ℬ𝑞 = 𝑒𝑡/2 , 𝑒4𝑡/3 { √ } √ 21. ℬ𝑞 = 𝑒(−1+ 2)𝑡 , 𝑒(−1− 2)𝑡 { √ √ } 22. ℬ𝑞 = 𝑒(2+ 3)𝑡 , 𝑒(2− 3)𝑡 { } 23. ℬ𝑞 = 𝑒5𝑡 , 𝑡𝑒5𝑡 { } 24. ℬ𝑞 = 𝑒−3𝑡/2 , 𝑡𝑒−3𝑡/2 25. ℬ𝑞 = {cos 3𝑡, sin 3𝑡}

26. ℬ𝑞 = {cos(5𝑡/2), sin(5𝑡/2)} { } 27. ℬ𝑞 = 𝑒−2𝑡 cos 3𝑡, 𝑒−2𝑡 sin 3𝑡

28. ℬ𝑞 = {𝑒𝑡 cos 2𝑡, 𝑒𝑡 sin 2𝑡} { √ √ } 29. ℬ𝑞 = 𝑒𝑡/2 cos 23 𝑡, 𝑒𝑡/2 sin 23 𝑡

30. ℬ𝑞 = {𝑒𝑡 , cos 𝑡, sin 𝑡} { } 31. ℬ𝑞 = 1, 𝑡, 𝑡2 √ √ } { 32. ℬ𝑞 = 𝑒−2𝑡 , 𝑒𝑡 cos 3𝑡, 𝑒𝑡 sin 3𝑡 { } 33. ℬ𝑞 = 𝑒𝑡/2 , 𝑒𝑡 , 𝑡𝑒𝑡 { } 34. ℬ𝑞 = 𝑒−2𝑡 , 𝑒3𝑡 𝑡, 𝑒−3𝑡

79

80

1 Solutions

35. ℬ𝑞 = {𝑒𝑡 , 𝑒−𝑡 , cos 𝑡, sin 𝑡} √ √ √ √ } { 36. ℬ𝑞 = cos 3𝑡, sin 3𝑡, cos 2𝑡, sin 2𝑡 } { 37. ℬ𝑞 = 𝑒2𝑡 , 𝑒−2𝑡 , 𝑡𝑒2𝑡 , 𝑡𝑒−2𝑡 { } 38. ℬ𝑞 = 𝑒𝑡 , 𝑡𝑒𝑡 , 𝑡2 𝑒𝑡 , 𝑒−7𝑡 , 𝑡𝑒−7𝑡 { } 39. ℬ𝑞 = cos 𝑡, 𝑡 cos 𝑡, 𝑡2 cos 𝑡, sin 𝑡, 𝑡 sin 𝑡, 𝑡2 sin 𝑡

𝑝1 (𝑠) 𝑝2 (𝑠) with deg 𝑝1 (𝑠) < deg 𝑞1 (𝑠) and 𝑟2 (𝑠) = with 𝑞1 (𝑠) 𝑞2 (𝑠) 𝑝1 (𝑠)𝑝2 (𝑠) deg 𝑝2 (𝑠) < deg 𝑞2 (𝑠). Thus, 𝑟1 (𝑠)𝑟2 (𝑠) = and 𝑞1 (𝑠)𝑞2 (𝑠)

40. 𝑟1 (𝑠) =

deg(𝑝1 (𝑠)𝑝2 (𝑠) = deg 𝑝1 (𝑠)+deg 𝑝2 (𝑠) < deg 𝑞1 (𝑠)+deg 𝑞2 (𝑠) = deg(𝑞1 (𝑠)𝑞2 (𝑠)). 𝑝(𝑠) with deg 𝑝(𝑠) < deg 𝑞(𝑠). But deg 𝑝(𝑠) = deg 𝑝(𝑠−𝑎) since, for 𝑞(𝑠) any natural number 𝑛, (𝑠−𝑎)𝑛 = 𝑠𝑛 +lower degree terms. Thus translating the highest power of 𝑠 found in 𝑝(𝑠) will produce the same highest power of 𝑠 in 𝑝(𝑠−𝑎). Since the same is also true for 𝑞(𝑠), it follows that deg 𝑝(𝑠− 𝑝(𝑠 − 𝑎) 𝑎) = deg 𝑝(𝑠) < deg 𝑞(𝑠) = deg 𝑞(𝑠 − 𝑎). Thus, 𝑟(𝑠 − 𝑎) = is a 𝑞(𝑠 − 𝑎) proper rational function.

41. 𝑟(𝑠) =

42. If 𝑟(𝑠) ∈ ℛ then 𝑟(𝑠) =

𝑝(𝑠) where deg 𝑝(𝑠) = 𝑚 < 𝑛 = deg 𝑞(𝑠). Then 𝑞(𝑠)

𝑟′ (𝑠) =

𝑞(𝑠)𝑝′ (𝑠) − 𝑞 ′ (𝑠)𝑝(𝑠) , (𝑞(𝑠))2

and deg(𝑞(𝑠)𝑝′ (𝑠) − 𝑞 ′ (𝑠)𝑝(𝑠)) ≤ max(deg(𝑞(𝑠)𝑞 ′ (𝑠)), deg(𝑞 ′ (𝑠)𝑝(𝑠))) = max(𝑛 + (𝑚 − 1), (𝑛 − 1) + 𝑚) = 𝑛 + 𝑚 − 1 < 2𝑛 = deg(𝑞(𝑠))2 . Hence 𝑟′ (𝑠) is a proper rational function. 43. Suppose 𝑟(𝑠) ∈ ℛ𝑞𝑛 but not in ℛ𝑞𝑛−1 . Then we can write 𝑟(𝑠) = 𝑞𝑝(𝑠) 𝑛 (𝑠) 2 2 where 𝑝(𝑠) does not have a factor of 𝑞(𝑠) = (𝑠 − 𝑎) + 𝑏 . By the quotient rule we have 𝑝′ 𝑞 𝑛 − 𝑛𝑞 𝑛−1 2(𝑠 − 𝑎)𝑝 𝑞 2𝑛 𝑝′ 𝑞 − 2𝑛(𝑠 − 𝑎)𝑝 = . 𝑞 𝑛+1

𝑟′ (𝑠) =

Thus 𝑟′ (𝑠) ∈ ℛ𝑞𝑛+1 . Suppose 𝑟′ (𝑠) ∈ ℛ𝑞𝑛 . Then the numerator 𝑝′ 𝑞 − 2𝑛(𝑠 − 𝑎)𝑝 has a factor of 𝑞. But this implies 𝑝 has a factor of 𝑞. But this is impossible since 𝑟(𝑠) ∈ / ℛ𝑞𝑛−1 . Hence 𝑟′ (𝑠) ∈ / ℛ𝑞 𝑛 .

1 Solutions

81

44. By exercise 43 this is true for 𝑛 = 1. Now apply induction. If 𝑛 is given and we assume the result is true for derivatives of order 𝑛−1, then 𝑟 (𝑛−1) ∈ ℛ𝑞𝑛 but not in ℛ𝑞𝑛−1 . Another application of exercise 43 then shows that ( )′ 𝑟(𝑛) = 𝑟(𝑛−1) ∈ ℛ𝑞𝑛+1 but not in ℛ𝑞𝑛 .

45. First note the following trigonometric identities:

1 (cos(𝑏 + 𝑑)𝑡 + cos(𝑏 − 𝑑)𝑡) 2 1 sin 𝑏𝑡 sin 𝑑𝑡 = (cos(𝑏 − 𝑑)𝑡 − cos(𝑏 + 𝑑)𝑡) 2 1 sin 𝑏𝑡 cos 𝑑𝑡 = (sin(𝑏 + 𝑑)𝑡 + sin(𝑏 − 𝑑)𝑡) . 2

cos 𝑏𝑡 cos 𝑑𝑡 =

Thus if 𝑡1 (𝑡) and 𝑡2 (𝑡) are two basic trig functions ( sin or cos) then 𝑡1 (𝑡)𝑡2 (𝑡) is a linear combination of basic trig functions. Now if 𝑓 (𝑡) = 𝑡𝑛 𝑒𝑎𝑡 𝑡1 (𝑡) and 𝑔(𝑡) = 𝑡𝑚 𝑒𝑐𝑡 𝑡2 (𝑡) then 𝑓 (𝑡)𝑔(𝑡) = 𝑡𝑛+𝑚 𝑒(𝑎+𝑐)𝑡 𝑡1 (𝑡)𝑡2 (𝑡). Since 𝑡1 (𝑡)𝑡2 (𝑡) is a linear combinations of basic trig functions then 𝑓 (𝑡)𝑔(𝑡) is a linear combination of exponential polynomial. Now if 𝑓 (𝑡) and 𝑔(𝑡) is any exponential polynomial then they are each made up of linear combinations of simple exponential polynomials and their product is a sum of terms of products of simple exponential polynomials, which we have shown is the sum of possibly two simple exponential polynomials. It now follows that 𝑓 (𝑡)𝑔(𝑡) is an exponential polynomial. 46. Observe that 𝑒𝑡−𝑡0 = 𝑒−𝑡0 𝑒𝑡 . So the translate of an exponential function is a multiple of an exponential function. Also, if 𝑓 (𝑡) is a polynomial so is 𝑓 (𝑡−𝑡0 ). By the addition rule for cos we have cos 𝑏(𝑡−𝑡0 ) = cos 𝑏𝑡 cos 𝑏𝑡0 − sin 𝑏𝑡 sin 𝑏𝑡0 and similarly for sin. It follows that all these translates remain in ℰ. By Theorem ?? and Exercise ?? the result follows. 47. Since 𝑓 ∈ ℰ is a linear combination of terms of the form 𝑡𝑛 𝑒𝑎𝑡 cos 𝑏𝑡 and 𝑡𝑛 𝑒𝑎𝑡 sin 𝑏𝑡 it is enough to show that the derivative of each of these terms is again in ℰ. But, by the product rule we have (𝑡𝑛 𝑒𝑎𝑡 cos 𝑏𝑡)′ = 𝑛𝑡𝑛−1 𝑒𝑎𝑡 cos 𝑏𝑡 + 𝑎𝑡𝑛 𝑒𝑎𝑡 cos 𝑏𝑡 − 𝑏𝑡𝑛 𝑒𝑎𝑡 sin 𝑏𝑡, a linear combination of simple exponential polynomials and hence back in ℰ. A similar calculation holds for 𝑡𝑛 𝑒𝑎𝑡 sin 𝑏𝑡. 48. All exponential polynomials are continuous functions on ℝ. If 𝑓 (𝑡) = 𝑡 then 𝑓 is an exponential polynomial. However, 𝑓 −1 (𝑡) = 1/𝑡 is not continuous at 0. Thus 𝑓 −1 is not an exponential polynomial.

Section 2.6 1.

82

1 Solutions

𝑡∗𝑡 = = =

∫

∫

𝑡

𝑡 0

(

=𝑡

𝑥(𝑡 − 𝑥) 𝑑𝑥

0

(𝑡𝑥 − 𝑥2 ) 𝑑𝑥

) 𝑥=𝑡 𝑥2 𝑥3 𝑡 − 2 3 𝑥=0

𝑡2 𝑡3 𝑡3 − = 2 3 6

2. 𝑡 ∗ 𝑡3 = 𝑡3 ∗ 𝑡 = = =

∫ ∫

𝑡 0 𝑡 0

(

𝑥3 (𝑡 − 𝑥) 𝑑𝑥 (𝑡𝑥3 − 𝑥4 ) 𝑑𝑥

) 𝑥=𝑡 𝑥5 𝑥4 𝑡 − 4 5 𝑥=0

𝑡4 𝑡5 − 4 5 𝑡5 = . 20

=𝑡

3. 3 ∗ sin 𝑡 = sin 𝑡 ∗ 3 =

∫

𝑡

(sin 𝑥)(3) 𝑑𝑥 0 𝑥=𝑡

= −3 cos 𝑥∣𝑥=0 = −3(cos 𝑡 − cos 0) = −3 cos 𝑡 + 3

4. (3𝑡 + 1) ∗ 𝑒

4𝑡

∫

𝑡

(3𝑥 + 1)𝑒4(𝑡−𝑥) 𝑑𝑥 ( ∫ 𝑡 ) ∫ 𝑡 4𝑡 −4𝑥 −4𝑥 =𝑒 3 𝑥𝑒 𝑑𝑥 + 𝑒 𝑑𝑥 0 ) ) ( (0 1 −4𝑡 1 −4𝑡 1 1 −4𝑡 1 4𝑡 − 𝑒 + − 𝑒 + =𝑒 3 − 𝑡𝑒 4 16 16 4 4 7𝑒4𝑡 − 12𝑡 − 7 = . 16

=

0

1 Solutions

5. From the Convolution table we get 1 (2𝑒3𝑡 − 2 cos 2𝑡 − 3 sin 2𝑡) 32 + 2 2 1 = (2𝑒3𝑡 − 2 cos 2𝑡 − 3 sin 2𝑡). 13

sin 2𝑡 ∗ 𝑒3𝑡 =

6. From linearity and the Convolution table we get (2𝑡 + 1) ∗ cos 2𝑡 = 2𝑡 ∗ cos 2𝑡 + 1 ∗ cos 2𝑡 ∫ 𝑡 1 − cos 2𝑡 =2 + ( cos 2𝑥 𝑑𝑥) 22 0 1 − cos 2𝑡 sin 2𝑡 = + 2 2 7. From the Convolution table we get 2 (𝑒−6𝑡 − (−6 − 6𝑡 + (36𝑡2 )/2)) (−6)3 1 = (18𝑡2 − 6𝑡 − 6 + 𝑒−6𝑡 ). 108

𝑡2 ∗ 𝑒−6𝑡 =

8. From the Convolution table we get 1 (sin 𝑡 − 2 sin 2𝑡) 12 − 2 2 1 = (2 sin 2𝑡 − sin 𝑡). 3

cos 𝑡 ∗ cos 2𝑡 =

9. From the Convolution table we get 𝑒2𝑡 ∗ 𝑒−4𝑡 =

𝑒2𝑡 − 𝑒−4𝑡 2 − (−4)

1 2𝑡 (𝑒 − 𝑒−4𝑡 ). 6 10. ℒ {𝑡 ∗ 𝑡𝑛 } (𝑠) = ℒ {𝑡} (𝑠)ℒ {𝑡𝑛 } (𝑠) 1 𝑛! = 2 𝑛+1 𝑠 𝑠 𝑛! = 𝑛+3 𝑠 1 (𝑛 + 2)! = (𝑛 + 1)(𝑛 + 2) 𝑠𝑛+3 { } 1 = ℒ 𝑡𝑛+2 . (𝑛 + 1)(𝑛 + 2)

83

84

1 Solutions

Thus 𝑡 ∗ 𝑡𝑛 =

𝑡𝑛+2 . (𝑛 + 1)(𝑛 + 2)

11. 1 𝑏 𝑠 − 𝑎 𝑠2 + 𝑏 2 ( ) 1 𝑏 𝑏𝑠 + 𝑏𝑎 − 2 = 2 𝑠 + 𝑏2 𝑠 − 𝑎 𝑠 + 𝑏2 ( { 𝑎𝑡 } ) 1 = 2 𝑏ℒ 𝑒 − (𝑏ℒ {cos 𝑏𝑡} + 𝑎ℒ {sin 𝑏𝑡}) 𝑠 + 𝑏2

{ } ℒ 𝑒𝑎𝑡 ∗ sin 𝑏𝑡 (𝑠) =

Thus

𝑒𝑎𝑡 ∗ sin 𝑏𝑡 =

1 (𝑏𝑒𝑎𝑡 − 𝑏 cos 𝑏𝑡 − 𝑎 sin 𝑏𝑡). 𝑎2 + 𝑏 2

12. 1 𝑠 𝑠 − 𝑎 𝑠2 + 𝑏 2 ( ) 1 𝑎 𝑎𝑠 − 𝑏2 = 2 − 𝑎 + 𝑏 2 𝑠 − 𝑎 𝑠2 + 𝑏 2 ( { 𝑎𝑡 } ) 1 = 2 𝑎ℒ 𝑒 − (𝑎ℒ {cos 𝑏𝑡} − 𝑏ℒ {sin 𝑏𝑡}) 𝑎 + 𝑏2

{ } ℒ 𝑒𝑎𝑡 ∗ cos 𝑏𝑡 (𝑠) =

Thus

𝑒𝑎𝑡 ∗ sin 𝑏𝑡 =

𝑎2

1 (𝑎𝑒𝑎𝑡 − 𝑎 cos 𝑏𝑡 + 𝑏 sin 𝑏𝑡). + 𝑏2

13. First assume 𝑎 ∕= 𝑏. Then 𝑎 𝑏 2 2 + 𝑎 𝑠 + 𝑏2 ( ) 𝑎𝑏 𝑎𝑏 1 = 2 − 2 𝑏 − 𝑎 2 𝑠2 + 𝑎 2 𝑠 + 𝑏2

ℒ {sin 𝑎𝑡 ∗ sin 𝑏𝑡} =

𝑠2

From this it follows that sin 𝑎𝑡 ∗ sin 𝑏𝑡 =

1 (𝑏 sin 𝑎𝑡 − 𝑎 sin 𝑏𝑡). 𝑏2 − 𝑎 2

Now assume 𝑎 = 𝑏. Then ℒ {sin 𝑎𝑡 ∗ sin 𝑎𝑡} = By Table 2.4 in Section 2.4 we get

(𝑠2

𝑎2 + 𝑎 2 )2

1 Solutions

ℒ {sin 𝑎𝑡 ∗ sin 𝑎𝑡} = 14. First assume 𝑎 ∕= 𝑏. Then

85

1 (sin 𝑎𝑡 − 𝑎𝑡 cos 𝑎𝑡). 2𝑎

𝑎 𝑠 𝑠2 + 𝑎 2 𝑠2 + 𝑏 2 ( ) 1 𝑎𝑠 𝑎𝑠 − = 2 𝑏 − 𝑎 2 𝑠2 + 𝑎 2 𝑠2 + 𝑏 2

ℒ {sin 𝑎𝑡 ∗ cos 𝑏𝑡} =

From this it follows that sin 𝑎𝑡 ∗ cos 𝑏𝑡 =

𝑏2

1 (𝑎 cos 𝑎𝑡 − 𝑎 cos 𝑏𝑡). − 𝑎2

Now assume 𝑎 = 𝑏. Then ℒ {sin 𝑎𝑡 ∗ cos 𝑎𝑡} =

𝑎𝑠 (𝑠2 + 𝑎2 )2

By Table 2.4 in Section 2.4 we get ℒ {sin 𝑎𝑡 ∗ cos 𝑎𝑡} = 15. First assume 𝑎 ∕= 𝑏. Then

1 (𝑎𝑡 sin 𝑎𝑡). 2𝑎

𝑠 𝑠 𝑠2 + 𝑎 2 𝑠2 + 𝑏 2 ( ) 1 −𝑎2 𝑏2 = 2 + 2 𝑏 − 𝑎 2 𝑠2 + 𝑎 2 𝑠 + 𝑏2

ℒ {cos 𝑎𝑡 ∗ cos 𝑏𝑡} =

From this it follows that cos 𝑎𝑡 ∗ cos 𝑏𝑡 =

1 (−𝑎 sin 𝑎𝑡 + 𝑏 sin 𝑏𝑡). 𝑏2 − 𝑎 2

Now assume 𝑎 = 𝑏. Then 𝑠2 (𝑠2 + 𝑎2 )2 1 𝑎2 = 2 − 2 . 2 𝑠 +𝑎 (𝑠 + 𝑎2 )2

ℒ {cos 𝑎𝑡 ∗ cos 𝑎𝑡} =

By Table 2.4 we get ℒ {cos 𝑎𝑡 ∗ cos 𝑎𝑡} =

1 1 1 sin 𝑎𝑡− (sin 𝑎𝑡−𝑎𝑡 cos 𝑎𝑡) = (sin 𝑎𝑡+𝑎𝑡 cos 𝑎𝑡). 𝑎 2𝑎 2𝑎

86

1 Solutions

16. The key is to recognize the integral defining 𝑓 (𝑡) as the convolution integral of two functions. Thus 𝑓 (𝑡) = (cos 2𝑡) ∗ 𝑡 so that 𝐹 (𝑠) = 𝑠 1 1 ℒ {(cos 2𝑡) ∗ 𝑡} = ℒ {cos 2𝑡} ℒ {𝑡} = 2 = . 2 2 𝑠 +4𝑠 𝑠(𝑠 + 4) 2 2 4 = 3 2 . 3 2 𝑠 𝑠 +4 𝑠 (𝑠 + 4)

17. 𝑓 (𝑡) = 𝑡2 ∗ sin 2𝑡 so 𝐹 (𝑠) = 18. 𝑓 (𝑡) = 𝑡3 ∗ 𝑒−3𝑡 so 𝐹 (𝑠) =

6 1 6 = 4 4 𝑠 𝑠+3 𝑠 (𝑠 + 3)

19. 𝑓 (𝑡) = 𝑡3 ∗ 𝑒−3𝑡 so 𝐹 (𝑠) =

6 1 6 = 4 4 𝑠 𝑠+3 𝑠 (𝑠 + 3)

20. 𝑓 (𝑡) = sin 2𝑡 ∗ cos 𝑡 so 𝐹 (𝑠) = 21. 𝑓 (𝑡) = sin 2𝑡 ∗ sin 2𝑡 so 𝐹 (𝑠)

𝑠2

2 𝑠 2𝑠 = 2 2 +4𝑠 +1 (𝑠 + 4)(𝑠2 + 1)

2 2 4 = 2 𝑠2 + 2 2 𝑠2 + 2 2 (𝑠 + 4)2

22. ℒ

−1

{

1 (𝑠 − 2)(𝑠 + 4)

}

=ℒ

−1

{

1 𝑠−2

}

= 𝑒2𝑡 ∗ 𝑒−4𝑡 𝑒2𝑡 − 𝑒−4𝑡 = 2 − −4 1 2𝑡 = (𝑒 − 𝑒−4𝑡 ). 6

∗ℒ

−1

{

1 𝑠+4

}

23. ℒ

−1

{

1 2 𝑠 − 6𝑠 + 5

}

} 1 =ℒ (𝑠 − 1)(𝑠 − 5) { } { } 1 1 −1 −1 =ℒ ∗ℒ 𝑠−1 𝑠−5 −1

{

= 𝑒𝑡 ∗ 𝑒5𝑡 𝑒𝑡 − 𝑒5𝑡 = 1−5 1 = (−𝑒𝑡 + 𝑒5𝑡 ) 4 24.

1 Solutions

ℒ−1

{

(𝑠2

1 + 1)2

}

{

𝑠 (𝑠2 + 1)2

}

{

}

1 ∗ ℒ−1 +1 = sin 𝑡 ∗ sin 𝑡 1 = (sin 𝑡 − 𝑡 cos 𝑡) 2 = ℒ−1

𝑠2

{

𝑠2

1 +1

}

{

𝑠 𝑠2 + 1

}

25. ℒ

−1

{

1 =ℒ 𝑠2 + 1 = sin 𝑡 ∗ cos 𝑡 1 = 𝑡 sin 𝑡 2 −1

}

∗ℒ

−1

26. ℒ

−1

{

1 (𝑠 + 6)𝑠3

}

} { } 1 1 −1 =ℒ ∗ℒ 𝑠+6 𝑠3 2 𝑡 = 𝑒−6𝑡 ∗ 2 1 = (−𝑒−6𝑡 + 1 − 6𝑡 + 18𝑡2 ) 216 −1

{

27. ℒ−1

{

2 (𝑠 − 3)(𝑠2 + 4)

}

{

𝑠 (𝑠 − 4)(𝑠2 + 1)

}

= ℒ−1

{

1 𝑠−3

}

{

1 𝑠−4

}

{

1 𝑠−𝑎

}

∗ ℒ−1

{

2 2 𝑠 +4

{

𝑠 2 𝑠 +1

{

1 𝑠−𝑏

= 𝑒3𝑡 ∗ sin 2𝑡 1 = (2𝑒3𝑡 − 2 cos 2𝑡 − 3 sin 2𝑡) 13

}

28. ℒ

−1

=ℒ

−1

∗ℒ

−1

= 𝑒4𝑡 ∗ cos 𝑡 1 = (4𝑒4𝑡 − 4 cos 𝑡 + sin 𝑡) 17

}

29. ℒ

−1

{

1 (𝑠 − 𝑎)(𝑠 − 𝑏)

}

=ℒ

−1

= 𝑒𝑎𝑡 ∗ 𝑒𝑏𝑡 𝑒𝑎𝑡 − 𝑒𝑏𝑡 = . 𝑎−𝑏

∗ℒ

−1

}

87

88

1 Solutions

30. ℒ−1

{

𝐺(𝑠) 𝑠+2

}

= ℒ−1 {𝐺(𝑠)} ∗ ℒ−1

{

1 𝑠+2

= 𝑔(𝑡) ∗ 𝑒−2𝑡 ∫ 𝑡 = 𝑔(𝑥)𝑒−2(𝑡−𝑥) 𝑑𝑥

}

0

31. ℒ

−1

{

𝐺(𝑠) 𝑠2 + 2

}

=ℒ

−1

{𝐺(𝑠)} ∗ ℒ

−1

{

𝑠 √ 2 2 𝑠 + 2

√ = 𝑔(𝑡) ∗ cos( 2)𝑡 ∫ 𝑡 √ = 𝑔(𝑥) cos 2(𝑡 − 𝑥) 𝑑𝑥

}

0

32. Hint: use the Input Integral Principle repeatedly. 33. We apply the input integral principle twice: } ∫ 𝑡 { 1 −1 ℒ = sin 𝑥 𝑑𝑥 𝑠(𝑠2 + 1) 0 𝑡

= − cos 𝑥∣0 = − cos 𝑡 + 1

ℒ−1

{

1 𝑠2 (𝑠2 + 1)

}

34. The inverse Laplace transform of 𝑒2𝑡 − 𝑒−2𝑡 . Thus 4 ℒ−1

{

1 𝑠(𝑠2 − 4)

}

=

∫

𝑡 0

1 − cos 𝑥 𝑑𝑥

= 𝑡 − sin 𝑡 𝑑𝑥

1 1 1 = is 𝑒2𝑡 ∗ 𝑒−2𝑡 = 𝑠2 − 4 𝑠−2𝑠+2

1 4

∫

𝑡

(𝑒2𝑥 − 𝑒−2𝑥 ) 𝑑𝑥 (0 𝑡 𝑡 ) 1 𝑒2𝑥 𝑒−2𝑥 = + 4 2 0 2 0 =

1 2𝑡 (𝑒 − 1 + 𝑒−2𝑡 − 1) 8 1 1 1 = − + 𝑒2𝑡 + 𝑒−2𝑡 4 8 8 =

1 Solutions

Repeating the input integral principle we get { } ∫ 𝑡( ) 1 1 1 2𝑥 1 −2𝑥 ℒ−1 = − + 𝑒 + 𝑒 𝑑𝑥 𝑠2 (𝑠2 − 4) 4 8 8 0 𝑡 1 1 = − + (𝑒2𝑡 − 1) − (𝑒−2𝑡 − 1) 4 16 16 −𝑡 1 1 = + 𝑒2𝑡 − 𝑒−2𝑡 4 16 16 35. We apply the input integral principle three times: { } ∫ 𝑡 1 −1 ℒ = 𝑒−3𝑥 𝑑𝑥 𝑠(𝑠 + 3) 0 𝑡 𝑒−3𝑥 = −3 0

1 = (1 − 𝑒−3𝑡 ). 3

ℒ−1

ℒ

−1

{

{

1 2 𝑠 (𝑠 + 3)

1 𝑠3 (𝑠 + 3)

}

}

∫ 1 𝑡 1 − 𝑒−3𝑥 𝑑𝑥 3 0 ( ) 1 − 𝑒−3𝑡 1 𝑡− = 3 3 1 = (3𝑡 − 1 + 𝑒−3𝑡 ). 9 =

∫ 1 𝑡 = 3𝑥 − 1 + 𝑒−3𝑥 𝑑𝑥 9 0 ( 2 ) 1 𝑡 𝑒−3𝑡 − 1 = 3 −𝑡− 9 2 3 1 = (2 − 6𝑡 + 9𝑡2 − 2𝑒−3𝑡 ) 54

36. We apply the input integral principle twice: { } ∫ 𝑡 1 −1 ℒ = 𝑥𝑒2𝑥 𝑑𝑥 𝑠(𝑠 − 2)2 0

𝑡 1 2𝑥 2𝑥 = (2𝑥𝑒 − 𝑒 ) 4 0 1 2𝑡 2𝑡 = (2𝑡𝑒 − 𝑒 + 1) 4

89

90

1 Solutions

ℒ−1

37. First, ℒ

−1

ℒ−1

{

{

{

1 𝑠2 (𝑠 − 2)2

1 2 (𝑠 + 9)2

}

=

1 2 𝑠(𝑠 + 9)2

}

= = = =

}

=

1 4

∫

𝑡 0

2𝑥𝑒2𝑥 − 𝑒2𝑥 + 1 𝑑𝑥

𝑡 1 2𝑥 2𝑥 = (𝑥𝑒 − 𝑒 + 𝑥) 4 0 1 2𝑡 = (𝑡𝑒 − 𝑒2𝑡 + 𝑡 + 1) 4

1 (sin 3𝑡 − 3𝑡 cos 3𝑡). Thus 54 ∫ 𝑡 1 (sin 3𝑥 − 3𝑥 cos 3𝑥) 𝑑𝑥 54 0 ( ( )) 𝑡 1 cos 3𝑥 cos 3𝑥 − − 𝑥 sin 3𝑥 + 54 3 3 0 ( ) 2 cos 3𝑡 2 1 − − 𝑡 sin 3𝑡 + 54 3 3 1 (−2 cos 3𝑡 − 3𝑡 sin 3𝑡 + 2). 162

1 1 = 2 . We apply the input integral 𝑠3 + 𝑠 2 𝑠 (𝑠 + 1)

38. First, from Table 2.4, principle twice: ℒ

−1

{

1 𝑠(𝑠 + 1)

}

=

∫

𝑡

𝑒−𝑥 𝑑𝑥

0

𝑡 = −𝑒−𝑥 0

= 1 − 𝑒−𝑡 .

ℒ−1

{

1 𝑠2 (𝑠 + 1)

}

=

∫

𝑡 0

(1 − 𝑒−𝑥 ) 𝑑𝑥

( ) 𝑡 = 𝑥 + 𝑒−𝑥 0 = 𝑡 + 𝑒−𝑡 − 1.

Section 2.7 1 384

) (𝑡4 − 45𝑡2 + 105) sin 𝑡 + (10𝑡3 − 105𝑡) cos 𝑡 ( 4 ) 1 2. 384 (𝑡 − 15𝑡2 ) cos 𝑡 − (6𝑡3 − 15𝑡) sin 𝑡 ( ) 1 3. 3840 (945 − 420𝑡2 + 15𝑡4 ) sin 𝑡 − (945𝑡 − 105𝑡3 + 𝑡4 ) cos 𝑡 ( ) 1 4. 3840 (105𝑡 − 45𝑡3 + 𝑡5 ) sin 𝑡 − (105𝑡2 − 10𝑡4 ) cos 𝑡 1.

(

1 Solutions

91

Section 3.1 12 + 1. 𝑌 (𝑠) = (𝑠−2)(𝑠+1)(𝑠+2) −2𝑡 −𝑡 𝑒 −𝑒

𝑠+4 (𝑠+1)(𝑠+2)

2. 𝑌 (𝑠) =

−𝑠+4 (𝑠+1)(𝑠−5)

3. 𝑌 (𝑠) =

2𝑠+1 𝑠2 +4

4. 𝑌 (𝑠) =

1 (𝑠+2)2

+

1 (𝑠+2)3

5. 𝑌 (𝑠) =

1 (𝑠+2)2

+

4𝑠 (𝑠2 +4)(𝑠+2)2

6. 𝑌 (𝑠) =

2𝑠−3 (𝑠−1)(𝑠−2)

+

+

25 𝑠2 (𝑠+1)(𝑠−5)

8 𝑠(𝑠2 +4)

=

1 𝑠2 +4

+

=

1 (𝑠−1)2 (𝑠−2)

𝑠+12 2(𝑠+4)2

12. 𝑌 (𝑠) = 2𝑡2 𝑒2𝑡

−𝑠 (𝑠−2)2

13. 𝑌 (𝑠) =

and 𝑦(𝑡) =

2 𝑠

and 𝑦(𝑡) =

4 𝑠

+

1 2

+

1 𝑠+1

and 𝑦(𝑡) = 𝑒2𝑡 +

and 𝑦(𝑡) = 4 − 5𝑡 − 5𝑒−𝑡 sin 2𝑡 + 2

1 𝑠2 +4

=

−1 5

2 𝑠

sin 2𝑡

3 3 + 𝑠−3 − 𝑠−1 and 𝑦(𝑡) = 2 + 3𝑒2𝑡 − 3𝑒𝑡

−7 𝑠−1

+

4 𝑠−2

13 1 10 𝑠−1

+

3 𝑠+3

1 2

and 𝑦(𝑡) =

+

+

−3𝑠+4 𝑠2 +1

1 (𝑠−1)2

−

1 −4𝑠−14 5 𝑠2 +4

+

7 (𝑠+3)2

and 𝑦(𝑡) = −𝑡𝑒𝑡 + and 𝑦(𝑡) =

𝑠+1

−

and 𝑦(𝑡) = 3𝑒−3𝑡 +

sin 5𝑡 + cos 5𝑡

and 𝑦(𝑡) =

√ 3 −𝑡 2 3 𝑒

64𝑠 (𝑠2 +4)2 (𝑠+2)2

3 + (𝑠2 +9)(𝑠 2 +4) =

=

sin

(√ ) 3𝑡 2

16 (𝑠2 +4)2

8 1 5 𝑠2 +4

+

+𝑒

−𝑡 2

𝑠 𝑠2 +4

cos

(√ ) 3𝑡 2

and 𝑦(𝑡) = −2𝑡 cos 2𝑡 +

− 35 𝑠21+9 and 𝑦(𝑡) =

4 5

Section 3.2 1. no, not linear. 2. yes; (𝑫 2 − 3𝑫)(𝑦) = 𝑒𝑡 , 𝑞(𝑠) = 𝑠2 − 3𝑠, nonhomogeneous 3. no, third order. 4. no, not linear.

13 𝑡 20 𝑒

4 1 2 4 2𝑡 + (𝑠−2) +2𝑡𝑒2𝑡 + 3 = − 𝑠−2 + (𝑠−2)2 + (𝑠−2)3 and 𝑦(𝑡) = −𝑒

𝑠2 +𝑠+1

1 𝑠2 +4

−

5 𝑠+1

1 (𝑠+2)

and 𝑦(𝑡) = 12 𝑒−4𝑡 + 4𝑡𝑒−4𝑡

𝑠+4 14. 𝑌 (𝑠) = (𝑠+2) 2 + cos 2𝑡 + sin 2𝑡

15. 𝑌 (𝑠) =

−5 𝑠2

−1 1 2 𝑠+3

2 50 9. 𝑌 (𝑠) = (𝑠+3) 2 + (𝑠2 +1)(𝑠+3)2 = 7𝑡𝑒−3𝑡 − 3 cos 𝑡 + 4 sin 𝑡

11. 𝑌 (𝑠) =

=

+

and 𝑦(𝑡) = 𝑡𝑒−2𝑡 + 21 𝑡2 𝑒−2𝑡

26 8. 𝑌 (𝑠) = (𝑠2 +4)(𝑠+3)(𝑠−1) = 4 7 1 −3𝑡 𝑒 − cos 2𝑡 − sin 2𝑡 2 5 5

𝑠−1 𝑠2 +25

1 𝑠−2

4 + 𝑠(𝑠−1)(𝑠−2) =

−3𝑠+9 7. 𝑌 (𝑠) = (𝑠−1)(𝑠−2) + 4𝑒2𝑡 − 7𝑒𝑡

10. 𝑌 (𝑠) =

=

sin 2𝑡 − 15 sin 3𝑡

92

1 Solutions

5. no, not constant coefficient. 6. yes; (𝑫 2 + 2𝑫 + 3)(𝑦) = 𝑒−𝑡 , 𝑞(𝑠) = 𝑠2 + 2𝑠 + 3, nonhomogeneous 7. yes; (𝑫 2 − 7𝑫 + 10)(𝑦) = 0, 𝑞(𝑠) = 𝑠2 − 7𝑠 + 10, homogeneous 8. no, first order. 9. yes; 𝑫 2 (𝑦) = −2 + cos 𝑡, 𝑞(𝑠) = 𝑠2 , nonhomogeneous 10. yes; (2𝑫 2 − 12𝑫 + 18)(𝑦) = 0, 𝑞(𝑠) = 2𝑠2 − 12𝑠 + 18, homogeneous 11. (a) 6𝑒𝑡 (b) 0 (c) sin 𝑡 − 3 cos 𝑡 12. (a) 0 (b) 2 sin 𝑡 (c) −𝑒2𝑡 13. (a) −4𝑒−2𝑡 (b) 0 (c) sec2 𝑡 − 2 tan 𝑡 14. (a) 0 (b) 0 (c) 1 15. (a) 4𝑒2𝑡 (b) 0 (c) 0 16. 𝑦(𝑡) = cos 2𝑡 + 𝑐1 𝑒𝑡 + 𝑐2 𝑒4𝑡 where 𝑐1 , 𝑐2 are arbitrary constants. 17. 𝑦(𝑡) = 𝑡𝑒3𝑡 + 𝑐1 𝑒3𝑡 + 𝑐2 𝑒−2𝑡 where 𝑐1 , 𝑐2 are arbitrary constants. 18. 𝑦(𝑡) = cos 2𝑡 + 𝑒𝑡 − 𝑒4𝑡 19. 𝑦(𝑡) = 𝑡𝑒3𝑡 + 𝑒3𝑡 − 2𝑒−2𝑡

Section 3.3 1. 𝑦(𝑡) = 𝑐1 𝑒2𝑡 + 𝑐2 𝑒−𝑡 2. 𝑦(𝑡) = 𝑐1 𝑒−4𝑡 + 𝑐2 𝑒3𝑡 3. 𝑦(𝑡) = 𝑐1 𝑒−4𝑡 + 𝑐2 𝑒−6𝑡 4. 𝑦(𝑡) = 𝑐1 𝑒6𝑡 + 𝑐2 𝑒−2𝑡

1 Solutions

93

5. 𝑦(𝑡) = 𝑐1 𝑒−4𝑡 + 𝑐2 𝑡𝑒−4𝑡 6. 𝑦(𝑡) = 𝑐1 𝑒5𝑡 + 𝑐2 𝑒−2𝑡 7. 𝑦(𝑡) = 𝑐1 𝑒−𝑡 cos 2𝑡 + 𝑐2 𝑒−𝑡 sin 2𝑡 8. 𝑦(𝑡) = 𝑐1 𝑒3𝑡 + 𝑐2 𝑡𝑒3𝑡 9. 𝑦(𝑡) = 𝑐1 𝑒−9𝑡 + 𝑐2 𝑒−4𝑡 10. 𝑦(𝑡) = 𝑐1 𝑒−4𝑡 cos 3𝑡 + 𝑐2 𝑒−4𝑡 sin 3𝑡 11. 𝑦(𝑡) = 𝑐1 𝑒−5𝑡 + 𝑐2 𝑡𝑒−5𝑡 12. 𝑦(𝑡) = 𝑐1 𝑒7𝑡 + 𝑐2 𝑒−3𝑡 13. 𝑦 =

𝑒𝑡 −𝑒−𝑡 2

14. 𝑦 = 2𝑒5𝑡 + 3𝑒−2𝑡 15. 𝑦 = 𝑡𝑒5𝑡 16. 𝑦 = 𝑒−2𝑡 cos 3𝑡 − 𝑒−2𝑡 sin 3𝑡

Section 3.4 { } 1. 𝑞(𝑠)𝑣(𝑠) − 2)(𝑠 − 3) so ℬ𝑞𝑣 = 𝑒−𝑡 , 𝑒2𝑡 , 𝑒3𝑡 while ℬ𝑞 = { −𝑡 2𝑡 }= (𝑠 + 1)(𝑠 𝑒 , 𝑒 . Since 𝑒3𝑡 is the only function in the first set but not in the second 𝑦𝑝 (𝑡) = 𝑐1 𝑒3𝑡 . { } 2. 𝑞(𝑠)𝑣(𝑠) = }(𝑠 + 2)(𝑠 + 4)(𝑠 + 3) so ℬ𝑞𝑣 = 𝑒−2𝑡 , 𝑒−4𝑡 , 𝑒−3𝑡 while ℬ𝑞 = { −2𝑡 −4𝑡 𝑒 ,𝑒 . Since 𝑒−3𝑡 is the only function in the first set but not in the second 𝑦𝑝 (𝑡) = 𝑐1 𝑒−3𝑡 . { } { } 3. 𝑞(𝑠)𝑣(𝑠) = (𝑠 − 2)2 (𝑠 − 3) so ℬ𝑞𝑣 = 𝑒2𝑡 , 𝑡𝑒2𝑡 , 𝑒3𝑡 while ℬ𝑞 = 𝑒2𝑡 , 𝑒3𝑡 . Since 𝑡𝑒2𝑡 is the only function in the first set but not in the second 𝑦𝑝 (𝑡) = 𝑐1 𝑡𝑒2𝑡 . { } 3 4. 𝑞(𝑠)𝑣(𝑠) ℬ𝑞𝑣 = 𝑒4𝑡 , 𝑡𝑒4𝑡 , 𝑡2 𝑒4𝑡 , 𝑒3𝑡 while ℬ𝑞 = { 4𝑡 3𝑡 }= (𝑠 − 4)4𝑡(𝑠 − 3)2 so 𝑒 , 𝑒 . Since 𝑡𝑒 and 𝑡 𝑒4𝑡 are the only functions in the first set but not in the second 𝑦𝑝 (𝑡) = 𝑐1 𝑡𝑒4𝑡 + 𝑐2 𝑡2 𝑒4𝑡 . { 5𝑡 5𝑡 } 2 2 5. 𝑞(𝑠)𝑣(𝑠) { 5𝑡 5𝑡= } (𝑠 − 5) (𝑠 + 25) so ℬ𝑞𝑣 = 𝑒 , 𝑡𝑒 , cos 5𝑡, sin 5𝑡 while ℬ𝑞 = 𝑒 , 𝑡𝑒 . Since cos 5𝑡 and sin 5𝑡 are the only functions in the first set that are not in the second 𝑦𝑝 (𝑡) = 𝑐1 cos 5𝑡 + 𝑐2 sin 5𝑡. 6. 𝑞(𝑠)𝑣(𝑠) = (𝑠2 + 1)(𝑠2 + 4) so ℬ𝑞𝑣 = {cos 𝑡, sin 𝑡, cos 2𝑡, sin 2𝑡} while ℬ𝑞 = {cos 𝑡, sin 𝑡}. Since cos 2𝑡 and sin 2𝑡 are the only functions in the first set that are not in the second 𝑦𝑝 (𝑡) = 𝑐1 cos 2𝑡 + 𝑐2 sin 2𝑡.

94

1 Solutions

7. 𝑞(𝑠)𝑣(𝑠) = (𝑠2 + 4)2 so ℬ𝑞𝑣 = {cos 2𝑡, sin 2𝑡, 𝑡 cos 2𝑡, 𝑡 sin 2𝑡} while ℬ𝑞 = {cos 2𝑡, sin 2𝑡}. Since 𝑡 cos 2𝑡 and 𝑡 sin 2𝑡 are the only functions in the first set that are not in the second 𝑦𝑝 (𝑡) = 𝑐1 𝑡 cos 2𝑡 + 𝑐2 𝑡 sin 2𝑡. { } 8. 𝑞(𝑠)𝑣(𝑠) = { (𝑠2 + 4𝑠 + 5)(𝑠 − 1)}2 so ℬ𝑞𝑣 = 𝑒𝑡 , 𝑡𝑒𝑡 , 𝑒−2𝑡 cos 𝑡, 𝑒−2𝑡 sin 𝑡 while ℬ𝑞 = 𝑒−2𝑡 cos 𝑡, 𝑒−2𝑡 sin 𝑡 . Since 𝑒𝑡 and 𝑡𝑒𝑡 are the only functions in the first set that are not in the second 𝑦𝑝 (𝑡) = 𝑐1 𝑒𝑡 + 𝑐2 𝑡𝑒𝑡 . { } 9. 𝑞(𝑠)𝑣(𝑠) = (𝑠2 + 4𝑠 + 5)(𝑠 − 1)2 so ℬ𝑞𝑣 = 𝑒𝑡 , 𝑡𝑒𝑡 , 𝑒−2𝑡 cos 𝑡, 𝑒−2𝑡 sin 𝑡 while ℬ𝑞 = {𝑒𝑡 , 𝑡𝑒𝑡 }. Since 𝑒−2𝑡 cos 𝑡 and 𝑒−2𝑡 sin 𝑡 are the only functions in the first set that are not in the second 𝑦𝑝 (𝑡) = 𝑐1 𝑒−2𝑡 cos 𝑡 + 𝑐2 𝑒−2𝑡 sin 𝑡. 10. 𝑦 = 61 𝑒2𝑡 + 𝐴𝑒𝑡 + 𝐵𝑒−4𝑡 11. 𝑦 = −𝑡𝑒−2𝑡 + 𝐴𝑒−2𝑡 + 𝐵𝑒5𝑡 12. 𝑦 = 41 𝑒𝑡 + 𝐴𝑒−𝑡 + 𝐵𝑡𝑒−𝑡 13. 𝑦 = 2 + 𝐴𝑒−𝑡 + 𝐵𝑒−2𝑡 14. 𝑦 = 12 𝑒−3𝑡 + 𝐴𝑒−2𝑡 cos 𝑡 + 𝐵𝑒−2𝑡 sin 𝑡 15. 𝑦 =

1 4

+ 15 𝑒𝑡 + 𝐴 cos(2𝑡) + 𝐵 sin(2𝑡)

16. 𝑦 = −𝑡2 − 2 + 𝐴𝑒𝑡 + 𝐵𝑒−𝑡 17. 𝑦 = 𝑒𝑡 + 𝐴𝑒2𝑡 + 𝐵𝑡𝑒2𝑡 18. 𝑦 = 21 𝑡2 𝑒2𝑡 + 𝐴𝑒2𝑡 + 𝐵𝑡𝑒2𝑡 19. 𝑦 = −𝑡 cos 𝑡 + 𝐴 sin 𝑡 + 𝐵 cos 𝑡 20. 𝑦 = 𝑡𝑒2𝑡 − 52 𝑒2𝑡 + 𝐴𝑒−3𝑡 + 𝐵𝑡𝑒−3𝑡 21. 𝑦 =

25 3 −3𝑡 6 𝑡 𝑒

+ 𝐴𝑒−3𝑡 + 𝐵𝑡𝑒−3𝑡

22. 𝑦 = 14 𝑡𝑒−3𝑡 sin(2𝑡) + 𝐴𝑒−3𝑡 cos(2𝑡) + 𝐵𝑒−3𝑡 sin(2𝑡) 23. 𝑦 = −𝑡2 𝑒4𝑡 cos(3𝑡) + 𝑡𝑒4𝑡 sin(3𝑡) + 𝐴𝑒4𝑡 cos(3𝑡) + 𝐵𝑒4𝑡 sin 3𝑡 24. 𝑦 =

−1 3𝑡 12 𝑒

+

10 6𝑡 21 𝑒

+

135 −𝑡 84 𝑒

25. 𝑦 = 2𝑒−𝑡 − 2𝑒−𝑡 cos 2𝑡 + 4𝑒−𝑡 sin 2𝑡 26. 𝑦 = 2𝑒2𝑡 − 2 cos 𝑡 − 4 sin 𝑡 27. 𝑦 =

−1 −2𝑡 2 𝑒

+ 𝑒2𝑡 + 2𝑡 −

1 2

28. 𝑫 − 5. 29. (𝑫 + 6)2 = 𝑫2 + 12𝑫 + 36.

1 Solutions

30. 𝑫2 + 9. 31. (𝑫 − 2)2 + 16 = 𝑫 2 − 4𝑫 + 20. 32. (𝑫2 + 1)3 = 𝑫3 + 3𝑫2 + 3𝑫 + 1. 33. (𝑫 − 𝑎)𝑛+1 .

Section 3.5 1. 𝑦 =

1 −6𝑡 32 𝑒

+ 𝐴𝑒2𝑡 + 𝐵𝑒−2𝑡

2. 𝑦 = −𝑒𝑡 + 𝐴𝑒−3𝑡 + 𝐵𝑒5𝑡 3. 𝑦 = 𝑡𝑒−2𝑡 + 𝐴𝑒−2𝑡 + 𝐵𝑒−3𝑡 4. 𝑦 = 2 + 𝐴𝑒−𝑡 + 𝐵𝑒−2𝑡 5. 𝑦 = −𝑡𝑒−4𝑡 + 𝐴𝑒2𝑡 + 𝐵𝑒−4𝑡 3 6. 𝑦 = − 130 cos 𝑡 −

11 130

sin 𝑡

7. 𝑦 = −𝑡2 + 𝐴 + 𝐵𝑒−3𝑡 + 𝐶𝑒3𝑡 8. 𝑦 = 𝑡𝑒2𝑡 − 52 𝑒2𝑡 + 𝐴𝑒−3𝑡 + 𝐵𝑡𝑒−3𝑡 9. 𝑦 = −𝑡𝑒4𝑡 −

3 4𝑡 10 𝑒

+ 𝐴𝑒−𝑡 + 𝐵𝑒6𝑡

10. 𝑦 = −𝑡2 𝑒4𝑡 cos(3𝑡) + 𝑡𝑒4𝑡 sin(3𝑡) + 𝐴𝑒4𝑡 cos(3𝑡) + 𝐵𝑒4𝑡 sin 3𝑡 11. 𝑦 = 61 𝑡3 𝑒2𝑡 + 𝐴𝑒2𝑡 + 𝐵𝑡𝑒2𝑡 12. 𝑦 =

1 2

sin 𝑡 + 𝐴𝑒−𝑡 + 𝐵𝑡𝑒−𝑡

13. 𝑦 = 12 𝑡𝑒𝑡 sin 𝑡

Section 4.1 1. no 2. yes, yes, homogeneous 3. yes, yes, nonhomogeneous 4. no 5. yes, yes, nonhomogeneous 6. yes, yes, nonhomogeneous

95

96

1 Solutions

7. no 8. yes, yes, nonhomogeneous 9. yes, no, homogeneous 10. no 11. yes, no, homogeneous 12. yes, no, homogeneous 13. 𝑳(1) = 1′′ + 1 = 1 𝑳(𝑡) = 𝑡′′ + 𝑡 = 𝑡 𝑳(𝑒−𝑡 ) = (𝑒−𝑡 )′′ + 𝑒−𝑡 = 2𝑒−𝑡 𝑳(cos 2𝑡) = (cos 2𝑡)′′ + cos 2𝑡 = −4 cos 2𝑡 + cos 2𝑡 = −3 cos 2𝑡. 14. 𝑳(1) = 1 𝑳(𝑡) = 𝑡 𝑳(𝑒−𝑡 ) = (𝑡 + 1)𝑒−𝑡 𝑳(cos 2𝑡) = (−4𝑡 + 1) cos 2𝑡 15. 𝑳(1) = −3 𝑳(𝑡) = 1 − 3𝑡 𝑳(𝑒−𝑡 ) = −2𝑒−𝑡 𝑳(cos 2𝑡) = −11 cos 2𝑡 − 2 sin 2𝑡 16. 𝑳(1) = 5 𝑳(𝑡) = 5𝑡 + 6 𝑳(𝑒−𝑡 ) = 0 𝑳(cos 2𝑡) = cos 2𝑡 − 12 sin 2𝑡 17. 𝑳(1) = −4 𝑳(𝑡) = −4𝑡 𝑳(𝑒−𝑡 ) = −3𝑒−𝑡 𝑳(cos 2𝑡) = −8 cos 2𝑡 18. 𝑳(1) = −1 𝑳(𝑡) = 0 𝑳(𝑒−𝑡 ) = (𝑡2 − 𝑡 − 1)𝑒−𝑡 𝑳(cos 2𝑡) = (−4𝑡2 − 1) cos 2𝑡 − 2𝑡 sin 2𝑡 19. 𝑳(𝑒𝑟𝑡 ) = 𝑎(𝑒𝑟𝑡 )′′ + 𝑏(𝑒𝑟𝑡 )′ + 𝑐𝑒𝑟𝑡 = 𝑎𝑟2 𝑒𝑟𝑡 + 𝑏𝑟𝑒𝑟𝑡 + 𝑐𝑒𝑟𝑡 = (𝑎𝑟2 + 𝑏𝑟 + 𝑐)𝑒𝑟𝑡 20. 𝐶 = 21. 𝐶1 = 22. no

−4 3 −3 4

and 𝐶2 =

1 2

1 Solutions

97

23. yes, 𝐶 = 1. 24. Parts (1) and (2) are done by computing 𝑦 ′′ + 𝑦 where 𝑦(𝑡) = 𝑡2 − 2, 𝑦(𝑡) = cos 𝑡, or 𝑦(𝑡) = sin 𝑡. Then by Theorem 3, every function of the form 𝑦(𝑡) = 𝑡2 − 2 + 𝑐1 cos 𝑡 + 𝑐2 sin 𝑡 is a solution to 𝑦 ′′ + 𝑦 = 𝑡2 , where 𝑐1 and 𝑐2 are constants. If we want a solution to 𝑳(𝑦) = 𝑡2 with 𝑦(0) = 𝑎 and 𝑦 ′ (0) = 𝑏, then we need to solve for 𝑐1 and 𝑐2 : 𝑎 = 𝑦(0) = −2 + 𝑐1 𝑏 = 𝑦 ′ (0) = 𝑐2 These equations give 𝑐1 = 𝑎 + 2, 𝑐2 = 𝑏. Particular choices of 𝑎 and 𝑏 give the answers for Part (3). (3)a. 𝑦 = 21 𝑒𝑡 + 2𝑒2𝑡 − 32 𝑒3𝑡

(3)b. 𝑦 = 21 𝑒𝑡 − 2𝑒2𝑡 + 32 𝑒3𝑡 (3)c. 𝑦 = 21 𝑒𝑡 − 7𝑒2𝑡 +

11 3𝑡 2 𝑒

(3)d. 𝑦 = 21 𝑒𝑡 + (−1 + 3𝑎 − 𝑏)𝑒2𝑡 + 25.(3)a. 𝑦 = 21 𝑒𝑡 + 2𝑒2𝑡 − 32 𝑒3𝑡 (3)b. (3)c. (3)d.

1 𝑡 2𝑒 1 𝑡 2𝑒 1 𝑡 2𝑒

(1 2

) − 2𝑎 + 𝑏 𝑒3𝑡

− 2𝑒2𝑡 + 32 𝑒3𝑡 +

11 2𝑡 2 𝑒

− 7𝑒3𝑡

+ (3𝑎 − 𝑏 − 1)𝑒2𝑡 + (𝑏 − 2𝑎 + 21 )𝑒3𝑡

26.(3)a. 𝑦 = 61 𝑡5 + (3)b. 𝑦 = 16 𝑡5 − (3)c. 𝑦 = 16 𝑡5 −

(3)d. 𝑦 = 16 𝑡5 +

10 2 5 3 3 𝑡 − 2𝑡 2 2 1 3 3𝑡 + 2𝑡 17 2 9 3 3 𝑡 + 2𝑡 (1 ) 2 3 + 3𝑎 − 𝑏 𝑡

( ) + − 12 − 2𝑎 + 𝑏 𝑡3

27. Write the equation in the standard form:

3 1 𝑦 ′′ + 𝑦 ′ − 2 𝑦 = 𝑡2 . 𝑡 𝑡 3 1 , 𝑎2 (𝑡) = − 2 , and 𝑓 (𝑡) = 𝑡2 . These three functions are 𝑡 𝑡 all continuous on the intervals (0, ∞) and (−∞, 0). Thus, Theorem 6 shows that if 𝑡0 ∈ (0, ∞) then the unique solution is also defined on the interval (0, ∞), and if 𝑡0 ∈ (−∞, 0), then the unique solution is defined on (−∞, 0).

Then 𝑎1 (𝑡) =

28. Maximal intervals are (−∞, −1), (−1, 1), (1, ∞) 29. (𝑘𝜋, (𝑘 + 1)𝜋) where 𝑘 ∈ ℤ

98

1 Solutions

30. (−∞, ∞) 31. (3, ∞) 32. (−∞, −2), (−2, 0), (0 2), (2, ∞) 33. The initial condition occurs at 𝑡 = 0 which is precisely where 𝑎2 (𝑡) = 𝑡2 has a zero. Theorem 6 does not apply. 34. In this case 𝑦(𝑡0 ) = 0 and 𝑦 ′ (𝑡0 ) = 0. The function 𝑦(𝑡) = 0, 𝑡 ∈ 𝐼 is a solution to the initial value problem. By the uniqueness part of Theorem 6 𝑦 = 0 is the only solution. 35. The assumptions say that 𝑦1 (𝑡0 ) = 𝑦2 (𝑡0 ) and 𝑦1′ (𝑡0 ) = 𝑦2′ (𝑡0 ). Both 𝑦1 and 𝑦2 therefore satisfies the same initial conditions. By the uniqueness part of Theorem 6 𝑦1 = 𝑦2 .

Section 4.2 1. dependent; 2𝑡 and 5𝑡 are multiples of each other. 2. independent 3. independent 4. dependent; 𝑒2 𝑡 + 1 = 𝑒1 𝑒2𝑡 and 𝑒2𝑡−3 = 𝑒−3 𝑒2𝑡 , they are multiples of each other. 5. independent 6. dependent; ln 𝑡2 = 2 ln 𝑡 and ln 𝑡5 = 5 ln 𝑡, they are multiples of each other. 7. dependent; sin 2𝑡 = 2 sin 𝑡 cos 𝑡, they are multiples of each other. 8. dependent; cosh 𝑡 = 21 (𝑒𝑡 + 𝑒−𝑡 ) and 3𝑒𝑡 (1 + 𝑒−2𝑡 ) = 3(𝑒𝑡 + 𝑒−𝑡 ), they are multiples of each other. 9. 1. Suppose 𝑎𝑡3 + 𝑏 𝑡3 = 0 on (−∞, ∞). Then for 𝑡 = 1 and 𝑡 = −1 we get 𝑎+𝑏 = 0 −𝑎 + 𝑏 = 0.

These equations imply 𝑎 = 𝑏 = 0. So 𝑦1 and 𝑦2 are linearly independent. { −3𝑡2 if 𝑡 < 0 ′ 2 ′ 2. Observe that 𝑦1 (𝑡) = 3𝑡 and 𝑦2 (𝑡) = If 𝑡 < 0 2 3𝑡 if 𝑡 ≥ 0. ( 3 ) 𝑡 −𝑡3 then 𝑤(𝑦1 , 𝑦2 )(𝑡) = = 0. If 𝑡 ≥ 0 then 𝑤(𝑦1 , 𝑦2 )(𝑡) = 3𝑡2 −3𝑡2

1 Solutions

99

𝑡3 𝑡3 = 0. It follows that the Wronskian is zero for all 𝑡 ∈ 3𝑡2 3𝑡2 (−∞, ∞). 3. The condition that the coefficient function 𝑎2 (𝑡) be nonzero in Theorem 4 and Proposition 6 is essential. Here the coefficient function, 𝑡2 , of 𝑦 ′′ is zero at 𝑡 = 0, so Proposition 6 does not apply on (−∞, ∞). The largest open intervals on which 𝑡2 is nonzero are (−∞, 0) and (0, ∞). On each of these intervals 𝑦1 and 𝑦2 are linearly dependent. 4. Consider the cases 𝑡 < 0 and 𝑡 ≥ 0. The verification is then straightforward. 5. Again the condition that the coefficient function 𝑎2 (𝑡) be nonzero is essential. The Uniqueness and Existence theorem does not apply. (

)

Section 4.3 1. The indicial polynomial is 𝑞(𝑠) = 𝑠2 + 𝑠 − 2 = (𝑠 +{2)(𝑠 −}1). There are two distinct roots 1 and −2. The fundamental set is 𝑡, 𝑡−2 . The general solution is 𝑦(𝑡) = 𝑐1 𝑡 + 𝑐2 𝑡−2 . 2. The indicial polynomial is 𝑞(𝑠) = 2𝑠2 − 7𝑠 + 3 = (2𝑠 − { 1)(𝑠 − } 3). There are two distinct roots

1 2

and 3. The fundamental set is

1

𝑡 2 , 𝑡3 . The general

1

solution is 𝑦(𝑡) = 𝑐1 𝑡 2 + 𝑐2 𝑡3 .

3. The indicial polynomial is 𝑞(𝑠) = 9𝑠2 − 6𝑠 + 1 = (3𝑠 −{1)2 . There}is one 1 1 root, 1/3, with multiplicity 2. The fundamental set is 𝑡 3 , 𝑡 3 ln 𝑡 . The 1

1

general solution is 𝑦(𝑡) = 𝑐1 𝑡 3 + 𝑐2 𝑡 3 ln 𝑡.

√ √ are 4. The indicial polynomial is 𝑞(𝑠) = 𝑠2 − 2 = (𝑠 − 2)(𝑠 + { √ 2). There √ } √ √ two distinct roots 2 and − 2. The fundamental set is 𝑡 2 , 𝑡− 2 . The general solution is 𝑦(𝑡) = 𝑐1 𝑡

√ 2

+ 𝑐 2 𝑡−

√ 2

.

5. The indicial polynomial is 𝑞(𝑠) = 4𝑠2 − 𝑠 + 1 {= (2𝑠 − 1)}2 . The root is 1 2

with multiplicity 2. The fundamental set is 1

1

1

𝑡 2 , 𝑡 2 ln 𝑡 . The general

1

solution is 𝑦(𝑡) = 𝑐1 𝑡 2 + 𝑐2 𝑡 2 ln 𝑡.

6. The indicial polynomial is 𝑞(𝑠) = 𝑠2 − 4𝑠 { − 21 =} (𝑠 − 7)(𝑠 + 3). The roots are 7 and −3. The fundamental set is 𝑡7 , 𝑡−3 . The general solution is 𝑦(𝑡) = 𝑐1 𝑡7 + 𝑐2 𝑡−3 . 8. The indicial polynomial is 𝑞(𝑠) = 𝑠2 − 𝑠 + 1 = 𝑠2 − 𝑠 + 1/4 + 3/4√= √ √ 1+𝑖 3 2 2 (𝑠 − 1/2) + ( 3/2) . There are two complex}roots, and 1−𝑖2 3 . 2 { 1

The fundamental set is 1 2

𝑦(𝑡) = 𝑐1 𝑡 sin

√ 3 2 𝑡

𝑡 2 sin

1 2

+ 𝑐2 𝑡 cos

√ 1 3 2 2 𝑡, 𝑡

√ 3 2 𝑡.

cos

√ 3 2 𝑡

. The general solution is

100

1 Solutions

9. The indicial polynomial is 𝑞(𝑠) = 𝑠2 − 4 = (𝑠 − 2)(𝑠 { + 2). }There are two distinct roots, 2 and −2. The fundamental set is 𝑡2 , 𝑡−2 . The general solution is 𝑦(𝑡) = 𝑐1 𝑡2 + 𝑐2 𝑡−2 . 10. The indicial polynomial is 𝑞(𝑠) = 𝑠2 + 4. There are two complex roots, 2𝑖 and −2𝑖. The fundamental set is {cos(2 ln 𝑡), sin(2 ln 𝑡)}. The general solution is 𝑦(𝑡) = 𝑐1 cos(2 ln 𝑡) + 𝑐2 sin(2 ln 𝑡). 11. The indicial polynomial is 𝑞(𝑠) = 𝑠2 − 4𝑠 + 13 = (𝑠 − 2)2 + 9. There are roots, }2 + 3𝑖 and 2 − 3𝑖. The fundamental set is { 2 two complex 𝑡 cos(3 ln 𝑡), 𝑡2 sin(3 ln 𝑡) . The general solution is 𝑦(𝑡) = 𝑐1 𝑡2 cos(3 ln 𝑡)+ 𝑐2 𝑡2 sin(3 ln 𝑡). 12. 𝑦 =

1 (𝑡 − 𝑡−2 ) 3

13. 𝑦 = 2𝑡1/2 − 𝑡1/2 ln 𝑡 14. 𝑦 = −3 cos(2 ln 𝑡) + 2 sin(2 ln 𝑡) 15. No solution is possible.

Section 4.4 1. ln

(

𝑠−𝑎 𝑠−𝑏

)

) 𝑠2 + 𝑏 2 2. ln 2 𝑠 + 𝑎2 ( 2 ) ( ) ( ) 𝑠 + 𝑎2 𝑏 −1 −1 𝑎 3. 𝑠 ln + 2𝑏 tan − 2𝑎 tan 𝑠2 + 𝑏 2 𝑠 𝑠 (𝑎) 4. tan−1 𝑠 (

5. 𝑦 = 𝑎𝑡 + 𝑏𝑒𝑡

6. 𝑦 = 𝑎(𝑡 + 1) + 𝑏𝑒−𝑡 7. 𝑌 ′ (𝑠) +

1 𝐶 𝑌 (𝑠) = 0 and 𝑌 (𝑠) = and 𝑦(𝑡) = 𝐶𝑒−𝑡 𝑠+1 𝑠+1

8. 𝑌 ′ (𝑠) =

−𝑦0 𝑦0 , 𝑌 (𝑠) = , and 𝑦(𝑡) = 𝑒−2𝑡 . (𝑠 + 2)2 𝑠+2

4𝑠 3𝑦0 𝑦0 (𝑠3 + 3𝑠) + 𝐶 𝑌 (𝑠) = 2 . Solving gives 𝑌 (𝑠) = and +1 𝑠 +1 (𝑠2 + 1)2 𝑦(𝑡) = 𝐴(𝑡 cos 𝑡 − sin 𝑡) + 𝐵(𝑡 sin 𝑡 + cos 𝑡).

9. 𝑌 ′ (𝑠) +

𝑠2

1 Solutions

101

2𝑠2 + 𝑠 − 2 2(𝑦0 𝑠 + 𝑦1 ) 2𝑠2 − 2𝑠 + 1 2𝑠 − 1 = 𝑌 (𝑠) = 𝑦0 + 2 𝑦1 𝑠(𝑠 − 1) 𝑠(𝑠 − 1) 2𝑠2 (𝑠 − 1) 1𝑠 (𝑠 − 1) 𝑦(𝑡) = (−𝑡 + 1 + 𝑒𝑡 ) + (𝑡 − 1 + 𝑒𝑡 )𝑦1 The general solution can be written 𝐴(𝑡 − 1) + 𝐵𝑒𝑡

10. 𝑌 ′ (𝑠) +

11. 𝑌 ′ (𝑠) = 12. 𝑌 ′ (𝑠)+

−𝑦0 𝑦0 , 𝑌 (𝑠) = , and 𝑦(𝑡) = 𝑦0 𝑒−𝑡 . 2 (𝑠 + 1) 𝑠+1 ( ) 6𝑠 𝐶 𝑌 (𝑠) = 0, 𝑌 (𝑠) = 2 , and 𝑦(𝑡) = 𝐶 (3 − 𝑡2 ) sin 𝑡 − 3𝑡 cos 𝑡 3 +1 (𝑠 + 1)

𝑠2

𝑠−2 −𝑠𝑦0 − 𝑦1 𝑠2 + 3𝑠 + 3 𝑠+2 𝑌 (𝑠) = , 𝑌 (𝑠) = 𝑦0 + 𝑦1 , and 2 𝑠+1 (𝑠 + 1) (𝑠 + 1)3 (𝑠 + 1)3 𝑡2 𝑡2 𝑦(𝑡) = 𝑦0 (𝑒−𝑡 + 𝑡𝑒−𝑡 + 𝑒−𝑡 ) + 𝑦1 (𝑡𝑒−𝑡 + 𝑒−𝑡 ). The general solution can 2 2 2 𝑡 be written 𝑦(𝑡) = 𝐴𝑒−𝑡 + 𝐵(𝑡𝑒−𝑡 + 𝑒−𝑡 ). 2 ( ) 1 1 𝑠 14. 𝑌 ′ (𝑠) = 𝑦0 ( − ) Then 𝑌 (𝑠) = 𝑦0 ln + 𝐶. Take 𝐶 = 0. 𝑠 𝑠−1 𝑠−1 1 − 𝑒𝑡 . Hence 𝑦(𝑡) = 𝑦0 𝑡 ( ) −𝑦0 𝑦0 𝑠+1 15. 𝑌 ′ (𝑠) = 2 and 𝑌 (𝑠) = ln + 𝐶. Take 𝐶 = 0 Then 𝑦(𝑡) = 𝑠 −1 2 𝑠−1 𝑦0 𝑒𝑡 − 𝑒−𝑡 . 2 𝑡 ( ) −𝑦0 𝑠−2 16. 𝑌 ′ (𝑠) = 2 and 𝑌 (𝑠) = 𝑦0 ln + 𝐶. Take 𝐶 = 0. Then (𝑠 − 5𝑠 + ) 6) 𝑠−3 ( 3𝑡 𝑒 − 𝑒2𝑡 𝑦(𝑡) = 𝑦0 . 𝑡 13. 𝑌 ′ (𝑠) −

−𝑦0 −𝑦0 so 𝑌 (𝑠) = (tan−1 (𝑠/3) + 𝐶). Since lim𝑠→∞ 𝑌 (𝑠) = 0 𝑠2 + 9 3 𝜋 𝑦0 3 𝑦0 sin 3𝑡 we have 𝐶 = − . Hence 𝑌 (𝑠) = tan−1 ( ) and 𝑦(𝑡) = . 2 3 𝑠 3 𝑡 ( ) −𝑦0 𝑠+1 ′ 18. 𝑌 (𝑠) = 2 and hence 𝑦(𝑠) = 𝑦0 ln + 𝐶. But 𝐶 = 0 and 𝑠 +𝑠 𝑠 −𝑡 1−𝑒 𝑦(𝑡) = 𝑦0 . 𝑡 17. 𝑌 ′ (𝑠) =

19. We use the formula 𝑛 ( ) 𝑘 ∑ 𝑑𝑛 𝑛 𝑑 𝑑𝑛−𝑘 (𝑓 (𝑡)𝑔(𝑡)) = 𝑓 (𝑡) ⋅ 𝑔(𝑡). 𝑑𝑡𝑛 𝑚 𝑑𝑡𝑘 𝑑𝑡𝑛−𝑘 𝑘=0

Observe that

102

1 Solutions

𝑑𝑘 −𝑡 𝑒 = (−1)𝑘 𝑒−𝑡 𝑑𝑡𝑘 and

𝑑𝑛−𝑘 𝑛 𝑡 = 𝑛(𝑛 − 1) ⋅ ⋅ ⋅ (𝑘 + 1)𝑡𝑘 . 𝑑𝑡𝑛−𝑘

It now follows that 1 𝑡 𝑑𝑛 −𝑡 𝑛 𝑒 (𝑒 𝑡 ) 𝑛! 𝑑𝑡𝑛 𝑛 ( ) 1 𝑡 ∑ 𝑛 𝑑𝑘 −𝑡 𝑑𝑛−𝑘 𝑛 𝑒 𝑡 = 𝑒 𝑛! 𝑘 𝑑𝑡𝑘 𝑑𝑡𝑛−𝑘 𝑘=0 𝑛 ( ) ∑ 𝑛 𝑛(𝑛 − 1) ⋅ ⋅ ⋅ (𝑘 + 1) 𝑘 𝑡 =𝑒 (−1)𝑘 𝑒−𝑡 𝑡 𝑘 𝑛! 𝑘=0 ( ) 𝑘 𝑛 ∑ 𝑛 𝑡 = (−1)𝑘 𝑘 𝑘! 𝑘=0

= ℓ𝑛 (𝑡).

20. This follows in a similar manner to the proof of Equation (2) given in Theorem 11. 22. ℒ {ℓ𝑛 (𝑎𝑡)} (𝑠) { 𝑘} 𝑛 ( ) ∑ 𝑛 𝑡 = (−1)𝑘 𝑎𝑘 ℒ 𝑘 𝑘! 𝑘=0 ( ) 𝑛 ∑ 𝑛 1 = (−1)𝑘 𝑎𝑘 𝑘+1 𝑠 𝑘 𝑘=0 𝑛 ( ) 1 ∑ 𝑛 = 𝑛+1 (−𝑎)𝑘 𝑠𝑛−𝑘 𝑠 𝑘 𝑘=0

(𝑠 − 𝑎)𝑛 = . 𝑠𝑛+1

The last line follows from the binomial theorem. Note: the dilation principle gives the same formula for 𝑎 > 0. 23. Hint: Take the Laplace transform of each side. Use the previous exercise and the binomial theorem. ∫𝑡 24. We have that ℓ𝑛 ∗ 1(𝑡) = 0 ℓ𝑛 (𝑥) 𝑑𝑥. By the convolution theorem

1 Solutions

103

𝑛

1 (𝑠 − 1) 𝑠 𝑠𝑛+1 ) ( 𝑠 − 1 (𝑠 − 1)𝑛 = 1− 𝑠 𝑠𝑛+1 (𝑠 − 1)𝑛 (𝑠 − 1)𝑛+1 − = 𝑛+1 𝑠 𝑠𝑛+2 = ℒ {ℓ𝑛 } (𝑠) − ℒ {ℓ𝑛+1 } (𝑠).

ℒ {ℓ𝑛 ∗ 1} (𝑠) =

The result follows by inversion. 25. We compute the Laplace transform of both sides. We’ll do a piece at a time. ℒ {(2𝑛 + 1)ℓ𝑛 } (𝑠) (𝑠 − 1)𝑛 = (2𝑛 + 1) 𝑛+1 𝑠 (𝑠 − 1)𝑛−1 = (2𝑛 + 1)(𝑠(𝑠 − 1)). 𝑠𝑛+2 ℒ {−𝑡ℓ𝑛 } (𝑠) ( )′ (𝑠 − 1)𝑛 = 𝑠𝑛+1 (𝑠 − 1)𝑛−1 = (𝑛 + 1 − 𝑠). 𝑠𝑛+2 −𝑛ℒ {ℓ𝑛−1 } (𝑠) (𝑠 − 1)𝑛−1 = −𝑛 𝑠𝑛 𝑛−1 (𝑠 − 1) = (−𝑛𝑠2 ). 𝑠𝑛+2 We have written each so that the common factor is efficients are

(𝑠 − 1)𝑛−1 . The co𝑠𝑛+2

𝑛 + 1 − 𝑠 + (2𝑛 + 1)(𝑠(𝑠 − 1)) − 𝑛𝑠2

= (𝑛 + 1)(𝑠2 − 2𝑠 + 1) = (𝑛 + 1)(𝑠 − 1)2

The right hand side is now ( ) 1 (𝑠 − 1)𝑛−1 (𝑛 + 1)(𝑠 − 1)2 𝑛+1 𝑠𝑛+2 (𝑠 − 1)𝑛+1 = 𝑠𝑛+2 = ℒ {ℓ𝑛+1 } (𝑠).

104

1 Solutions

Taking the inverse Laplace transform completes the verification. ∫𝑡 26. We have that ℓ𝑛 ∗ ℓ𝑚 (𝑡) = 0 ℓ𝑛 (𝑥)ℓ𝑚 (𝑡 − 𝑥) 𝑑𝑥. By the convolution theorem ℒ {ℓ𝑛 ∗ ℓ𝑚 } (𝑠) (𝑠 − 1)𝑚 (𝑠 − 1)𝑛 = 𝑛+1 𝑠𝑚+1 𝑠( ) (𝑠 − 1)𝑚+𝑛 𝑠−1 = 1 − 𝑠𝑚+𝑛+1 𝑠 𝑚+𝑛 (𝑠 − 1) (𝑠 − 1)𝑚+𝑛+1 = − 𝑚+𝑛+1 𝑠 𝑠𝑚+𝑛+2 = ℒ {ℓ𝑚+𝑛 (𝑠)} − ℒ {ℓ𝑚+𝑛+1 } (𝑠). The result follows by inversion. ∫∞ 27. First of all 0 𝑒−𝑡 ℓ𝑛 (𝑡) 𝑑𝑡 = ℒ {ℓ𝑛 } (1) = 0. Thus ∫

∞

𝑒−𝑥 ℓ𝑛 (𝑥) 𝑑𝑡

𝑡

∫

𝑡

𝑒−𝑥 ℓ𝑛 (𝑥) 𝑑𝑥 ∫ ∞ 𝑒𝑡−𝑥 ℓ𝑛 (𝑥) 𝑑𝑥 = −𝑒−𝑡

=−

0

0

= −𝑒−𝑡 (𝑒𝑡 ∗ ℓ𝑛 (𝑡)).

By the convolution theorem { } ℒ 𝑒𝑡 ∗ ℓ𝑛 (𝑠) 1 (𝑠 − 1)𝑛 = 𝑠 − 1 𝑠𝑛+1 (𝑠 − 1)𝑛−1 = 𝑠𝑛+1 ( ) (𝑠 − 1)𝑛−1 𝑠−1 = 1 − 𝑠𝑛 𝑠 𝑛−1 (𝑠 − 1) (𝑠 − 1)𝑛 = − 𝑛 𝑠 𝑠𝑛+1 −1 = ℒ {ℓ𝑛−1 (𝑡)} − ℒ−1 {ℓ𝑛 (𝑡)} . It follows by inversion that 𝑒𝑡 ∗ℓ𝑛 = ℓ𝑛−1 −ℓ𝑛 and substituting this formula into the previous calculation gives the needed result.

1 Solutions

105

Section 4.5 1. Let 𝑦2 (𝑡) = 𝑡2 𝑢(𝑡). Then 𝑡4 𝑢′′ + 𝑡3 𝑢′ = 0, which gives 𝑢′ = 𝑡−1 and 𝑢(𝑡) = ln 𝑡. Substituting gives 𝑦2 (𝑡) = 𝑡2 ln 𝑡. The general solution can be written 𝑦(𝑡) = 𝑐1 𝑡2 + 𝑐2 𝑡2 ln 𝑡. 2. 𝑦2 (𝑡) =

−1 3𝑡2 .

The general solution can be written 𝑦(𝑡) = 𝑐1 𝑡 + 𝑐2 𝑡12 .

3. Let 𝑦2 (𝑡) = (1 − 𝑡2 )𝑢(𝑡). Substitution gives (1 − 𝑡2 )2 𝑢′′ − 4𝑡(1 − 𝑡2 )𝑢′ = 0 ′′ −2𝑡 1 ′ ′ and hence 𝑢𝑢′ = −2 1−𝑡 2 . From this we get 𝑢 = (1−𝑡)2 . Integrating 𝑢 by ( ) 𝑡 1 1+𝑡 partial fractions give 𝑢 = −1 and hence 2 1−𝑡2 + 4 ln 1−𝑡 𝑦2 (𝑡) =

4. 𝑦2 (𝑡) = −1 + ( 𝑐1 𝑡 + 𝑐2 −1 + 1

(

1+𝑡 1−𝑡

)

.

) . The general solution can be written 𝑦(𝑡) = ( )) 𝑡 1+𝑡 . 2 ln 1−𝑡 𝑡 2

ln

(

−1 1 𝑡 + (1 − 𝑡2 ) ln 2 4

1+𝑡 1−𝑡

5

3

2 𝑢′′ + 4𝑡 2 𝑢′ = 0 leads to 𝑢′ = 1/𝑡 and hence 5. Let 𝑦2 (𝑡) = 𝑡 2 𝑢(𝑡). Then 4𝑡√ 𝑢(𝑡) = ln√𝑡. Thus 𝑦√ 𝑡 ln 𝑡. The general solution can be written 2 (𝑡) = 𝑦(𝑡) = 𝑐1 𝑡 + 𝑐 + 2 𝑡 ln 𝑡.

6. 𝑦2 (𝑡) = 1. The general solution can be written 𝑦(𝑡) = 𝑐1 𝑡 + 𝑐2 . 7. 𝑦2 (𝑡) = 𝑡𝑒𝑡 . The general solution can be written 𝑦(𝑡) = 𝑐1 𝑡 + 𝑐2 𝑡𝑒𝑡 . 8. Let 𝑦2 (𝑡) = 𝑡2 cos 𝑡 𝑢(𝑡). Then 𝑡4 cos 𝑡 𝑢′′ − 2𝑡4 sin 𝑡 𝑢′ = 0 which gives 𝑢′ (𝑡) = sec2 𝑡 and hence 𝑢(𝑡) = tan 𝑡. Thus 𝑦2 (𝑡) = 𝑡2 sin 𝑡. The general solution can be written 𝑦(𝑡) = 𝑐1 𝑡2 cos 𝑡 + 𝑐2 𝑡2 sin 𝑡. 2 2 9. 𝑦2 (𝑡) = −1 2 cos 𝑡 . The general solution can be written 𝑦(𝑡) = 𝑐1 sin 𝑡 + 2 𝑐2 cos 𝑡 .

10. 𝑦2 (𝑡) = 𝑡𝑒2𝑡 . The general solution can be written 𝑦(𝑡) = 𝑐1 𝑒2𝑡 + 𝑐2 𝑡𝑒2𝑡 . 11. 𝑦2 (𝑡) = −1 − 𝑡 tan 𝑡. The general solution can be written 𝑦(𝑡) = 𝑐1 tan 𝑡 + 𝑐2 (1 + 𝑡 tan 𝑡). 12. 𝑦2 (𝑡) = 𝑡 − 1. The general solution can be written 𝑦(𝑡) = 𝑐1 𝑒−𝑡 + 𝑐2 (𝑡 − 1). 13. 𝑦2 (𝑡) = − sec 𝑡. The general solution can be written 𝑦(𝑡) = 𝑐1 tan 𝑡 + 𝑐2 sec 𝑡. 2

14. Let 𝑦2 (𝑡) = 𝑡𝑢(𝑡). Then (𝑡3 + 𝑡)𝑢′′ + 2𝑢′ = 0 which gives 𝑢′ = 𝑡 𝑡+1 and 2 𝑢 = 𝑡 − 1𝑡 . Thus 𝑦2 (𝑡) = 𝑡2 − 1. The general solution can be written 𝑦(𝑡) = 𝑐1 𝑡 + 𝑐2 (𝑡2 − 1).

106

1 Solutions

sin 2𝑡 𝑡 sin 2𝑡 . The general solution can be written 𝑦(𝑡) = 𝑐1 1+cos 15. 𝑦2 = −1− 1+cos 2𝑡 + ( )2𝑡 𝑡 sin 2𝑡 𝑐2 1 + 1+cos 2𝑡 .

16. 𝑦2 (𝑡) = 𝑡 sin 𝑡. The general solution can be written 𝑦(𝑡) = 𝑐1 𝑡 cos 𝑡 + 𝑐2 𝑡 sin 𝑡.

Section 4.6 1. sin 𝑡 and cos 𝑡 form a fundamental set for the homogeneous solutions. Let 𝑦(𝑝 (𝑡) = 𝑢1 cos)𝑡( + 𝑢) 2 sin 𝑡.(Then )the matrix equation cos 𝑡 sin 𝑡 𝑢′1 0 = implies 𝑢′1 (𝑡) = − sin2 𝑡 = 21 (cos 2𝑡 − 1) − sin 𝑡 cos 𝑡 𝑢′2 sin 𝑡 and 𝑢′2 (𝑡) = cos 𝑡 sin 𝑡 = 12 (sin 2𝑡). Integration give 𝑢1 (𝑡) = 14 (sin(2𝑡) − −1 2𝑡) = 12 (sin 𝑡 cos 𝑡−𝑡) and 𝑢2 (𝑡) = −1 4 cos 2𝑡 = 4 (2 cos 2𝑡−1). This implies 1 1 1 𝑦𝑝 (𝑡) = 4 sin 𝑡 − 2 𝑡 cos 𝑡. Since 4 sin 𝑡 is a homogeneous solution we can write the general solution in the form 𝑦(𝑡) = −1 2 𝑡 cos 𝑡 + 𝑐1 cos 𝑡 + 𝑐2 sin 𝑡. We observe that a particular solution is the imaginary part of a solution to 𝑦 ′′ + 𝑦 = 𝑒𝑖𝑡 . We use the incomplete partial fraction method and get 𝑝(𝑠) 1 1 𝑌 (𝑠) = (𝑠−𝑖)12 (𝑠+𝑖) . This can be written 𝑌 (𝑠) = 2𝑖 (𝑠−𝑖)2 + (𝑠−𝑖)(𝑠+𝑖) . From ( ) 1 −1 1 𝑖𝑡 this we get 𝑦𝑝 (𝑡) = Im 2𝑖 ℒ { (𝑠−𝑖) = Im −𝑖 = −1 2} 2 𝑡𝑒 2 𝑡 cos 𝑡. The general solution is 𝑦(𝑡) =

−1 2 𝑡 cos 𝑡

+ 𝑐1 cos 𝑡 + 𝑐2 sin 𝑡. { } 2. A fundamental set for 𝑦 ′′ − 4𝑦 = 0 is 𝑒2𝑡 , 𝑒−2𝑡 . Let 𝑦𝑝 (𝑡) = 𝑢1 (𝑡)𝑒2𝑡 + 𝑢2 (𝑡)𝑒−2𝑡 . Then ( ) (the )matrix ( equation ) 4𝑡 𝑒2𝑡 𝑒−2𝑡 𝑢′1 0 = implies 𝑢′1 (𝑡) = 14 and 𝑢′2 (𝑡) = −𝑒4 and 2𝑒2𝑡 −2𝑒−2𝑡 𝑢′2 𝑒2𝑡 4𝑡 𝑡 2𝑡 𝑒2𝑡 𝑒2𝑡 hence 𝑢1 (𝑡) = 4𝑡 and 𝑢2 (𝑡) = −𝑒 16 . Now 𝑦𝑝 (𝑡) = 4 𝑒 − 16 . Since 16 is a homogeneous solution we can write the general solution as 𝑦(𝑡) = 𝑡 2𝑡 2𝑡 −2𝑡 . On the other hand, the incomplete partial fraction 4 𝑒 + 𝑐1 𝑒 + 𝑐2 𝑒 1

𝑝(𝑠) 4 method gives 𝑌 (𝑠) = (𝑠−2)12 (𝑠+2 = (𝑠−2) 2 + (𝑠−2)(𝑠+2) . From this we see that a particular solution is 𝑦𝑝 (𝑡) = 14 𝑡𝑒2𝑡 . The general solution is 𝑦(𝑡) = 1 2𝑡 2𝑡 −2𝑡 . 4 𝑡𝑒 + 𝑐1 𝑒 + 𝑐2 𝑒

3. The functions 𝑒𝑡 cos 2𝑡 and 𝑒𝑡 sin 2𝑡 form a fundamental set. Let 𝑦𝑝 (𝑡) = 𝑐1 𝑒𝑡 cos 2𝑡 + 𝑐2 𝑒𝑡 sin 2𝑡. equation ) the ( matrix ) ( Then 𝑢′1 0 𝑡 𝑡 𝑊 (𝑒 cos 2𝑡, 𝑒 sin 2𝑡) ′ = implies that 𝑢′1 (𝑡) = −1 2 sin 2𝑡 and 𝑢2 𝑒𝑡 𝑢′2 (𝑡) = 12 cos 2𝑡. Hence, 𝑢1 (𝑡) = 41 𝑐𝑜𝑠2𝑡 and 𝑢2 (𝑡) = 14 sin 2𝑡. From this we get 𝑦𝑝 (𝑡) = 41 𝑒𝑡 cos2 2𝑡 + 14 𝑒𝑡 sin2 2𝑡 = 41 𝑒𝑡 . On the other hand, the method of undetermined coefficients implies that a particular solution is of the form 𝑦𝑝 (𝑡) = 𝐶𝑒𝑡 . Substitution gives 4𝐶𝑒𝑡 = 𝑒𝑡 and hence 𝐶 = 41 . It follows that 𝑦𝑝 (𝑡) = 41 𝑒𝑡 . Furthermore, the general solution is 𝑦(𝑡) = 1 𝑡 𝑡 𝑡 4 𝑒 + 𝑐1 𝑒 cos 2𝑡 + 𝑐2 𝑒 sin 2𝑡.

1 Solutions

107

{ } 4. A set is (1, 𝑒−3𝑡 . The matrix equation ( fundamental ) ( ) ) 1 𝑒−3𝑡 𝑢′1 0 = −3𝑡 implies 𝑢′1 (𝑡) = 13 𝑒−3𝑡 and 𝑢′2 (𝑡) = −1 3 . Thus 0 −3𝑒−3𝑡 𝑢′2 𝑒 −3𝑡 −3𝑡 −𝑒 𝑢1 (𝑡) = −𝑒9 , 𝑢2 (𝑡) = −𝑡 − 3𝑡 𝑒−3𝑡 . Observe though 3 , and 𝑦𝑝 (𝑡) = 9 −3𝑡 that −𝑒9 is a homogeneous solution and so the general solution can be written 𝑦(𝑡) = − 3𝑡 𝑒−3𝑡 + 𝑐1 + 𝑐2 𝑒−3𝑡 . The incomplete partial fraction −1

𝑝(𝑠) −1 −3𝑡 1 3 method gives 𝑌 (𝑠) = (𝑠+3) 2 𝑠 = (𝑠+3)2 + (𝑠+3)𝑠 which implies that 3 𝑡𝑒 is a particular solution. The general solution is as above. { } 5. A set is( 𝑒𝑡) , 𝑒2𝑡 . The matrix equation ( fundamental ) ( ) 𝑒𝑡 𝑒2𝑡 𝑢′1 0 = implies 𝑢′1 (𝑡) = −𝑒2𝑡 and 𝑢′2 (𝑡) = 𝑒𝑡 . Hence 𝑡 2𝑡 𝑒 2𝑒 𝑢′2 𝑒3𝑡 −1 2𝑡 𝑡 2𝑡 𝑡 𝑡 2𝑡 𝑢1 (𝑡) = −1 = 21 𝑒3𝑡 . The 2 𝑒 , 𝑢2 (𝑡) = 𝑒 , and 𝑦𝑝 (𝑡) = 2 𝑒 𝑒 + 𝑒 𝑒 1 3𝑡 general solution is 𝑦(𝑡) = 2 𝑒 + 𝑐1 𝑒𝑡 + 𝑐2 𝑒2𝑡 . The method of undetermined coefficients implies that a particular solution is of the form 𝑦𝑝 = 𝐶𝑒3𝑡 . Substitution gives 2𝐶𝑒3𝑡 = 3𝑒3𝑡 and hence 𝐶 = 12 . The general solution is as above.

6. sin 𝑡 and cos 𝑡 form a fundamental set for the homogeneous solutions. Let 𝑦(𝑝 (𝑡) = 𝑢1 cos)𝑡( + 𝑢) the matrix equation 2 sin 𝑡.(Then ) 𝑢′1 0 cos 𝑡 sin 𝑡 = implies that 𝑢′1 (𝑡) = cos 𝑡 − sec 𝑡 and − sin 𝑡 cos 𝑡 𝑢′2 tan 𝑡 𝑢′2 (𝑡) = sin 𝑡. From this we get 𝑢1 (𝑡) = sin 𝑡 − ln ∣sec 𝑡 + tan 𝑡∣ and 𝑢2 (𝑡) = − cos 𝑡. Therefore 𝑦𝑝 (𝑡) = − cos 𝑡 ln ∣sec 𝑡 + tan 𝑡∣. The general solution is thus 𝑦(𝑡) = − cos 𝑡 ln ∣sec 𝑡 + tan 𝑡∣ + 𝑐1 cos 𝑡 + 𝑐2 sin 𝑡. 𝑡 𝑡 7. A ( fundamental ) ( set )is {𝑒(, 𝑡𝑒)}. The matrix equation 𝑡 𝑡 ′ 0 𝑒 𝑡𝑒 𝑢1 = 𝑒𝑡 implies 𝑢′1 (𝑡) = −1 and 𝑢′2 (𝑡) = 1𝑡 . Hence, 𝑒𝑡 𝑒𝑡 + 𝑡𝑒𝑡 𝑢′2 𝑡 𝑢1 (𝑡) = −𝑡, 𝑢2 (𝑡) = ln 𝑡, and 𝑦𝑝 (𝑡) = −𝑡𝑒𝑡 + 𝑡 ln 𝑡𝑒𝑡 . Since −𝑡𝑒𝑡 is a homogeneous solution we can write the general solution as 𝑦(𝑡) = 𝑡 ln 𝑡𝑒𝑡 + 𝑐1 𝑒𝑡 + 𝑐2 𝑡𝑒𝑡 .

8. A is {cos ( fundamental ) (set′ ) (𝑡, sin)𝑡}. The matrix equation cos 𝑡 sin 𝑡 𝑢1 0 = implies 𝑢′1 (𝑡) = − tan 𝑡 and 𝑢′2 (𝑡) = 1. − sin 𝑡 cos 𝑡 𝑢′2 sec 𝑡 Hence 𝑢1 (𝑡) = ln(cos 𝑡), 𝑢2 (𝑡) = 𝑡, and 𝑦𝑝 (𝑡) = cos 𝑡 ln(cos 𝑡) + 𝑡 sin 𝑡. The general solution is 𝑦(𝑡) = cos 𝑡 ln(cos 𝑡) + 𝑡 sin 𝑡 + 𝑐1 cos 𝑡 + 𝑐2 sin 𝑡. 9. The associated homogeneous equation is Cauchy-Euler { } with indicial equation 𝑠2 −3𝑠+2 = (𝑠−2)(𝑠−1). It follows that 𝑡, 𝑡2 forms a fundamental set. We put the given equation is standard form to get 𝑦 ′′ − 2𝑡 𝑦 ′ + 𝑡22 𝑦 = 𝑡2 . 2 Thus 𝑓 (𝑡) ( 2) ( = )𝑡 . The ( )matrix equation ′ 𝑡𝑡 𝑢1 0 = 2 implies 𝑢′1 (𝑡) = −𝑡2 and 𝑢′2 (𝑡) = 𝑡. Hence 𝑢1 (𝑡) = 1 2𝑡 𝑢′2 𝑡

108

1 Solutions −𝑡 3

3

2

3

, 𝑢2 (𝑡) = 𝑡2 , and 𝑦𝑝 (𝑡) = −𝑡3 𝑡 + 4 solution is 𝑦(𝑡) = 𝑡6 + 𝑐1 𝑡 + 𝑐2 𝑡2 .

𝑡2 2 2𝑡

=

𝑡4 6.

It follows that the general

1 1 ′ ′′ 10. (In standard ( we get ) ( ′ ) form ) 𝑦 − 𝑡 𝑦 = 3𝑡 − 𝑡 . The matrix equation 2 1𝑡 𝑢1 0 1 3 1 2 ′ = implies 𝑢′1 (𝑡) = −3 2 𝑡 + 2 and 𝑢2 (𝑡) = 2 − 2𝑡2 . 0 2𝑡 𝑢′2 3𝑡 − 1𝑡 1 3 1 3 3 Hence 𝑢1 (𝑡) = −1 2 𝑡 + 2 𝑡, 𝑢2 (𝑡) = 2 𝑡 + 2𝑡 , and 𝑦𝑝 (𝑡) = 𝑡 + 𝑡. The general 3 2 solution is 𝑦(𝑡) = 𝑡 + 𝑡 + 𝑐1 + 𝑐2 𝑡 .

11. The homogeneous equation is Cauchy-Euler with indicial equation 𝑠2 − 2𝑠 + 1 = (𝑠 − 1)2 . It follows that {𝑡, 𝑡 ln 𝑡} is a fundamental set. After writing in standard form we see the forcing function 𝑓 (𝑡) is 1𝑡 . The matrix equation ( )( ′) ( ) 𝑡 𝑡 ln 𝑡 𝑢1 0 𝑡 = 1 implies 𝑢′1 (𝑡) = − ln and 𝑢′2 (𝑡) = 1𝑡 . Hence 𝑡 1 ln 𝑡 + 1 𝑢′2 𝑡 2 2 2 2 𝑡 𝑡 , 𝑢2 (𝑡) = ln 𝑡, and 𝑦𝑝 (𝑡) = −𝑡 𝑢1 (𝑡) = − ln 2 2 ln 𝑡 + 𝑡 ln 𝑡 = 2 ln 𝑡. The 2 𝑡 general solution is 𝑦(𝑡) = 2 ln 𝑡 + 𝑐1 𝑡 + 𝑐2 𝑡 ln 𝑡. { } 12. A fundamental set is 𝑒2𝑡 , 𝑡𝑒2𝑡 . The matrix equation ( 2𝑡 ( )( ′) ) 0 𝑒 𝑡𝑒2𝑡 𝑢1 ′ 2𝑡 = implies 𝑢′1 (𝑡) = 𝑡2−𝑡 𝑒 +1 and 𝑢2 (𝑡) = 2𝑒2𝑡 𝑒2𝑡 (1 + 2𝑡) 𝑢′2 2 𝑡 +1 1 −1 −1 2 2𝑡 2 𝑡, and 𝑦𝑝 (𝑡) = −1 𝑡2 +1 . Hence 𝑢1 (𝑡) = 2 ln(𝑡 +1), 𝑢2 (𝑡) = tan 2 𝑒 ln(𝑡 + −1 2𝑡 2𝑡 2 1)+𝑡 tan−1 𝑒2𝑡 . The general solution is 𝑦(𝑡) = −1 𝑒 + 2 𝑒 ln(𝑡 +1)+𝑡 tan 2𝑡 2𝑡 𝑐1 𝑒 + 𝑐2 𝑡𝑒 . 13. (The matrix equation )( ′) ( ) tan 𝑡 sec 𝑡 𝑢1 0 = implies 𝑢′1 (𝑡) = 𝑡 and 𝑢′2 (𝑡) = −𝑡 sin 𝑡. sec2 𝑡 sec 𝑡 tan 𝑡 𝑢′2 𝑡 2 2 Hence 𝑢1 (𝑡) = 𝑡2 , 𝑢2 (𝑡) = 𝑡 cos 𝑡 − sin 𝑡, and 𝑦𝑝 (𝑡) = 𝑡2 tan 𝑡 + (𝑡 cos 𝑡 − 2 sin 𝑡) sec 𝑡 = 𝑡2 tan 𝑡 + 𝑡 − tan 𝑡. Since tan 𝑡 is a homogeneous solution we 2 can write the general solution as 𝑦(𝑡) = 𝑡2 tan 𝑡 + 𝑡 + 𝑐1 tan 𝑡 + 𝑐2 sec 𝑡. 14. When put in standard form one sees that 𝑓 (𝑡) = 𝑡𝑒−𝑡 . The matrix equation ( )( ′ ) ( ) 𝑡 − 1 𝑒−𝑡 𝑢1 0 = implies 𝑢′1 (𝑡) = 𝑒−𝑡 and 𝑢′2 (𝑡) = 1 − 𝑡. 1 −𝑒−𝑡 𝑢′2 𝑡𝑒−𝑡 2 2 Hence 𝑢1 (𝑡) = −𝑒−𝑡 , 𝑢2 (𝑡) = 𝑡− 𝑡2 , and 𝑦𝑝 (𝑡) = −(𝑡−1)𝑒−𝑡 +(𝑡− 𝑡2 )𝑒−𝑡 = −𝑡2 −𝑡 −𝑡 −𝑡 is a homogeneous solution we can write the general 2 𝑒 +𝑒 . Since 𝑒 −𝑡2 −𝑡 solution as 𝑦(𝑡) = 2 𝑒 + 𝑐1 (𝑡 − 1) + 𝑐2 𝑒−𝑡 . 15. After put in standard form the forcing function 𝑓 is 4𝑡4 . The matrix equation ( )( ′) ( ) cos 𝑡2 sin 𝑡2 𝑢1 0 = implies 𝑢′1 (𝑡) = −2𝑡3 sin 𝑡2 and −2𝑡 sin 𝑡2 2𝑡 cos 2𝑡 𝑢′2 4𝑡4

1 Solutions

109

𝑢′2 (𝑡) = 2𝑡3 cos 𝑡2 . Integration by parts gives 𝑢1 (𝑡) = 𝑡2 cos 𝑡2 − sin 𝑡2 and 𝑢2 (𝑡) = 𝑡2 sin 𝑡2 + cos 𝑡2 . Hence 𝑦𝑝 (𝑡) = 𝑡2 cos 𝑡2 − cos 𝑡2 sin 𝑡2 + 𝑡2 sin 𝑡2 + cos 𝑡2 sin 𝑡2 = 𝑡2 . The general solution is 𝑦(𝑡) = 𝑡2 + 𝑐1 cos 𝑡2 + 𝑐2 sin 𝑡2 . 𝑡 −𝑡 16. (A fundamental ) ( ′ )set is ( {𝑒 , 𝑒 ) }. The matrix equation 𝑡 −𝑡 0 𝑒 𝑒 𝑢1 1 −1 𝑒𝑡 ′ = implies 𝑢′1 (𝑡) = 12 1+𝑒 𝑡 and 𝑢2 (𝑡) = 2 1+𝑒𝑡 . 1 𝑒𝑡 −𝑒−𝑡 𝑢′2 −𝑡 1+𝑒 𝑡 𝑡 Hence 𝑢1 (𝑡) = 21 (𝑡 − ln(1 + 𝑒𝑡 ), 𝑢2 (𝑡) = −1 2 (𝑒 − ln(1 + 𝑒 )), and 𝑦𝑝 (𝑡) = 1 𝑡 𝑡 −𝑡 𝑡 (𝑡𝑒 − 1 − (𝑒 − 𝑒 ) ln(1 + 𝑒 )). (Note: in the integrations of 𝑢′1 and 𝑢′2 2 use the substitution 𝑢 = 𝑒𝑡 .) The general solution can now be written 𝑦(𝑡) = 12 (𝑡𝑒𝑡 − 1 − (𝑒𝑡 − 𝑒−𝑡 ) ln(1 + 𝑒𝑡 )) + 𝑐1 𝑒𝑡 + 𝑐2 𝑒−𝑡 .

18. 𝑦𝑝 (𝑡) = 𝑎1 𝑓 (𝑡) ∗ sin 𝑎𝑡 19. 𝑦𝑝 (𝑡) = 𝑎1 𝑓 (𝑡) ∗ sinh 𝑎𝑡 20. 𝑦𝑝 (𝑡) = 𝑎1 𝑓 (𝑡) ∗ 𝑡𝑒−𝑎𝑡 21. 𝑦𝑝 (𝑡) =

1 𝑏−𝑎 𝑓 (𝑡)

Section 5.1 1. Graph (c) 2. Graph (g) 3. Graph (e) 4. Graph (a) 5. Graph (f) 6. Graph (d) 7. Graph (h) 8. Graph (b)

∗ (𝑒𝑏𝑡 − 𝑒𝑎𝑡 )

110

1 Solutions

Graphs for problems 1 through 8 1

2 1

0

0

1

2

3 (a)

4

5

6

𝑡

0

0

1

2

3 (b)

4

5

6

0

1

2

3 (d)

4

5

6

0

1

2

3 (f)

4

5

6

1

2

3

4

5

6

) 2 2 3 0 (𝑡 − 4) 𝑑𝑡 + 2 0 𝑑𝑡 + 3 (−𝑡 + 3) 𝑑𝑡 = 𝑡 /3 − 4𝑡 0 + ) 5 3𝑡 3 = (8/3 − 8) + (−25/2 + 15) − (−9/2 + 9) = −22/3.

0+

𝑡

4 1 2 0

1

2

3

−1

4

5

6

𝑡 0

(c)

1

0

𝑡

1

0

1

2

3 (e)

4

5

6

𝑡

0

𝑡

1 1 0 0

9.

∫5 0

(

0

1

𝑓 (𝑡) 𝑑𝑡 = 2

−𝑡 /2 +

2

3 (g)

4

5

∫2

6

𝑡

∫3

−1 ∫5

(h) (

1 2 ∫1 ∫2 𝑓 (𝑢) 𝑑𝑢 = 0 (2 − 𝑢) 𝑑𝑢 + 1 𝑢3 𝑑𝑢 = (2𝑢 − 𝑢2 /2) 0 + 𝑢4 /4 1 = 3/2 + 4 − 1/4 = 21/4. ∫ 2𝜋 ∫𝜋 ∫ 2𝜋 11. 0 ∣sin 𝑥∣ 𝑑𝑥 = 0 sin 𝑥 𝑑𝑥 + 𝜋 − sin 𝑥 𝑑𝑥 = − cos 𝑥∣𝜋0 + cos 𝑥∣2𝜋 𝜋 = 4.

10.

12.

∫2 0

∫3 0

𝑓 (𝑤) 𝑑𝑤 =

∫1 0

𝑤 𝑑𝑤 +

∫2

1 1 𝑤

𝑑𝑤 +

∫3

1 2 2

𝑑𝑤 = 1/2 + ln 2 + 1/2 = 1 + ln 2

𝑡

1 Solutions

13. 14. 15. 16.

∫5 2

∫6 0

∫6 0

∫6 0

𝑓 (𝑡) 𝑑𝑡 = 𝑓 (𝑡) 𝑑𝑡 =

∫3 2

∫2 0

𝑓 (𝑢) 𝑑𝑢 = 𝑓 (𝑡) 𝑑𝑡 =

(3 − 𝑡) 𝑑𝑡 + (1 − 𝑡) 𝑑𝑡 +

∫1 0

∫2 0

𝑢 𝑑𝑢 +

𝑡2 𝑑𝑡 +

∫2 1

∫3 2

∫4 3

∫4 2

2(𝑡 − 3) 𝑑𝑡 + (3 − 𝑡) 𝑑𝑡 +

(2 − 𝑢) 𝑑𝑢 +

4 𝑑𝑡 +

∫6 3

∫6 2

∫6 4

∫6 4

111

2 𝑑𝑡 = 1/2 + 1 + 4 = 11/2

(5 − 𝑡) 𝑑𝑡 = 0 + 0 + 0 = 0.

1 𝑑𝑢 = 1/2 + 1/2 + 4 = 5.

(7 − 𝑡) 𝑑𝑡 = 8/3 + 4 + 15/2 = 85/6

17. A is true since 𝑦(𝑡) satisfies the differential equation on each subinterval. B is true since the left and right limits agree at 𝑡 = 2. C is not true since 𝑦(0) = 1 ∕= 2. 18. A is true since 𝑦(𝑡) satisfies the differential equation on each subinterval. B is true since lim𝑡→2− 𝑦(𝑡) = 1 + 𝑒−8 = lim𝑡→2+ 𝑦(𝑡). C is true since 𝑦(0) = 2. 19. A is true since 𝑦(𝑡) satisfies the differential equation on each subinterval. B is false since lim𝑡→2− 𝑦(𝑡) = 1 + 𝑒−8 while lim𝑡→2+ 𝑦(𝑡) = 1. C is false since B is false. 20. A is false since 2𝑒−4𝑡 does not satisfy the differential equation 𝑦 ′ + 4𝑦 = 4 on the interval [0, 2). Since 𝐴 is false, B and C are necessarily false. 21. A is true since 𝑦(𝑡) satisfies the differential equation on each subinterval. B is true since lim𝑡→1− 𝑦(𝑡) = −2𝑒 + 𝑒2 = lim𝑡→1+ 𝑦(𝑡). C is false since lim𝑡→1− 𝑦 ′ (𝑡) = −3𝑒 + 2𝑒2 while lim𝑡→1+ 𝑦 ′ (𝑡) = 3𝑒2 − 2𝑒. D is false since C is false. 22. A is true since 𝑦(𝑡) satisfies the differential equation on each subinterval. B is false since lim𝑡→1− 𝑦(𝑡) = −2𝑒 + 𝑒2 while lim𝑡→1+ 𝑦(𝑡) = (1/2)𝑒2 − 3𝑒. C and D are false since B is false. You cannot have a continuous derivative if the function is not continuous. 23. A is true since 𝑦(𝑡) satisfies the differential equation on each subinterval. B is true since lim𝑡→1− 𝑦(𝑡) = −2𝑒 + 𝑒2 = lim𝑡→1+ 𝑦(𝑡). C is true since lim𝑡→1− 𝑦 ′ (𝑡) = −3𝑒 + 2𝑒2 = lim𝑡→1+ 𝑦 ′ (𝑡). D is true since 𝑦(0) = 𝑦 ′ (0) = 0. 24. A is true since 𝑦(𝑡) satisfies the differential equation on each subinterval. B is true since lim𝑡→1− 𝑦(𝑡) = −𝑒2 = lim𝑡→1+ 𝑦(𝑡). C is true since lim𝑡→1− 𝑦 ′ (𝑡) = −𝑒 − 2𝑒2 = lim𝑡→1+ 𝑦 ′ (𝑡). D is false since 𝑦(0) = −2 ∕= 0. 25. The general solution of 𝑦 ′ − 𝑦 = 1 on the interval [0, 2) is found by using the integrating factor 𝑒−𝑡 . The general solution is 𝑦(𝑡) = −1 + 𝑐𝑒𝑡 and the initial condition 𝑦(0) = 0 gives 𝑐 = 1, so that 𝑦(𝑡) = −1 + 𝑒𝑡 for 𝑡 ∈ [0, 2). Continuity of 𝑦(𝑡) at 𝑡 = 2 will then give 𝑦(2) = lim𝑡→2− 𝑦(𝑡) = −1 + 𝑒2 , which will provide the initial condition for the next interval

112

1 Solutions

[2, 4). The general solution of 𝑦 ′ − 𝑦 = −1 on [2, 4) is 𝑦(𝑡) = 1 + 𝑘𝑒𝑡 . Thus −1 + 𝑒2 = 𝑦(2) = 1 + 𝑘𝑒2 and solve for 𝑘 to get 𝑘 = −2𝑒−2 + 1, so that 𝑦(𝑡) = 1 + (−2𝑒−2 + 1)𝑒𝑡 for 𝑡 ∈ [2, 4). Continuity will then give 𝑦(4) = 1 + (−2𝑒−2 + 1)𝑒4 , which will provide the initial condition for the next interval [4, ∞). The general solution to 𝑦 ′ − 𝑦 = 0 on [4, ∞) is 𝑦(𝑡) = 𝑏𝑒𝑡 and the constant 𝑏 is obtained from the initial condition 𝑏𝑒4 = 𝑦(4) = 1 + (−2𝑒−2 + 1)𝑒4 , which gives 𝑏 = 𝑒−4 − 2𝑒−2 + 1, so that 𝑦(𝑡) = (𝑒−4 − 2𝑒−2 + 1)𝑒𝑡 for 𝑡 ∈ [4, ∞). Putting these three pieces together, we find that the solution is ⎧ 𝑡  if 0 ≤ 𝑡 < 2, ⎨−1 + 𝑒 𝑡−2 𝑦(𝑡) = 1 − 2𝑒 + 𝑒𝑡 if 2 ≤ 𝑡 < 4  ⎩ 𝑡−4 𝑒 − 2𝑒𝑡−2 + 𝑒𝑡 if 4 ≤ 𝑡 < ∞. 26. The general solution of 𝑦 ′ + 3𝑦 = 𝑡 on the interval [0, 1) is found by using the integrating factor 𝑒3𝑡 . The general solution is 𝑦(𝑡) = 31 𝑡 − 19 + 𝑐𝑒−3𝑡 and the initial condition gives 𝑐 = 19 . By continuity we must have 𝑦(1) = 1 1 1 −3 = 92 + 19 𝑒−3 , which will provide the initial condition for 3 − 9 + 9𝑒 the next interval [1, ∞). The general solution of 𝑦 ′ + 3𝑦 = 1 on (1, ∞) is 𝑦(𝑡) = 31 + 𝑘𝑒−3𝑡 . Setting 𝑦(1) = 31 − 19 + 19 𝑒−3 = 29 + 19 𝑒−3 found from the interval [0, 1) equal to 𝑦(1) = 31 + 𝑘𝑒−3 found from the interval [1, ∞) and solving for 𝑘 gives 𝑘 = 91 − 19 𝑒3 . Thus the complete solution is { 1 𝑡 − 19 + 19 𝑒−3𝑡 if 0 ≤ 𝑡 < 1, 𝑦(𝑡) = 13 1 1 3 −3𝑡 + ( − 𝑒 )𝑒 if 1 ≤ 𝑡 < ∞. 3 9 9 27. The general solution of 𝑦 ′ − 𝑦 = 𝑓 (𝑡) on any interval is found by using the integrating factor 𝑒−𝑡 . The general solution on the interval [0, 1) is 𝑦(𝑡) = 𝑎𝑒𝑡 and since the initial condition is 𝑦(0) = 0, the solution on [0, 1) is 𝑦(𝑡) = 0. Continuity then given 𝑦(1) = 0, which will be the initial condition for the interval [1, 2). The general solution of 𝑦 ′ − 𝑦 = 𝑡 − 1 on the interval [1, 2) is 𝑦(𝑡) = −𝑡 + 𝑏𝑒𝑡 and the initial condition 𝑦(1) = 0 gives 0 = −1+𝑏𝑒1 so that 𝑏 = 𝑒−1 . Thus 𝑦(𝑡) = −𝑡+𝑒−1 𝑒𝑡 = −𝑡+𝑒𝑡−1 for 𝑡 ∈ [0, 2). Continuity of 𝑦(𝑡) at 𝑡 = 2 will then give 𝑦(2) = lim𝑡→2− 𝑦(𝑡) = −2+𝑒1 , which will provide the initial condition for the next interval [2, 3). The general solution of 𝑦 ′ − 𝑦 = 3 − 𝑡 on [2, 3) is 𝑦(𝑡) = 𝑡 − 2 + 𝑐𝑒𝑡 . Thus −2 + 𝑒1 = 𝑦(2) = 𝑐𝑒2 and solve for 𝑐 to get 𝑐 = −2𝑒−2 + 𝑒−1 , so that 𝑦(𝑡) = 𝑡 − 2 + (−2𝑒−2 + 𝑒−1 )𝑒𝑡 = 𝑡 − 2 − 2𝑒𝑡−2 + 𝑒𝑡−1 for 𝑡 ∈ [2, 3). Continuity will then give 𝑦(3) = 1 − 2𝑒1 + 𝑒2 , which will provide the initial condition for the next interval [3, ∞). The general solution to 𝑦 ′ − 𝑦 = 0 on [4, ∞) is 𝑦(𝑡) = 𝑘𝑒𝑡 and the constant 𝑘 is obtained from the initial condition 𝑘𝑒3 = 𝑦(3) = 1 − 2𝑒1 + 𝑒2 , which gives 𝑐 = 𝑒−3 − 2𝑒−2 + 𝑒−1 ,

1 Solutions

113

so that 𝑦(𝑡) = (𝑒−3 − 2𝑒−2 + 𝑒−1 )𝑒𝑡 = 𝑒𝑡−3 − 2𝑒𝑡−2 + 𝑒𝑡−1 for 𝑡 ∈ [3, ∞). Putting these three pieces together, we find that the solution is ⎧ if 0 ≤ 𝑡 < 1,  0  ⎨−𝑡 + 𝑒𝑡−1 if 1 ≤ 𝑡 < 2, 𝑦(𝑡) = 𝑡−2 𝑡−1 𝑡 − 2 − 2𝑒 +𝑒 if 2 ≤ 𝑡 < 3   ⎩ 𝑡−3 𝑒 − 2𝑒𝑡−2 + 𝑒𝑡−1 if 3 ≤ 𝑡 < ∞.

28. The general solution of 𝑦 ′ + 𝑦 = 𝑓 (𝑡) on any interval is found by using the integrating factor 𝑒𝑡 . Using this integrating factor the general solution of 𝑦 ′ + 𝑦 = sin 𝑡 on the interval [0, 𝜋) is found to be 𝑦(𝑡) = (sin 𝑡 − cos 𝑡)/2 + 𝑐𝑒−𝑡 . The initial condition actually occurs at the end of this interval, but by continuity we can substitute 𝑡 = 𝜋 in this formula to get −1 = 𝑦(𝜋) = 1/2 + 𝑐𝑒−𝜋 so 𝑐 = −(3/2)𝑒𝜋 . Hence on the interval [0, 𝜋] the solution is 𝑦(𝑡) = (sin 𝑡 − cos 𝑡)/2 − (3/2)𝑒𝜋 𝑒−𝑡 = (sin 𝑡 − cos 𝑡)/2 − (3/2)𝑒−(𝑡−𝜋). The general solution of 𝑦 ′ + 𝑦 = 0 on the interval [𝜋, ∞) is 𝑦(𝑡) = 𝑘𝑒−𝑡 and 𝑦(𝜋) = −1 this gives −1 = 𝑘𝑒−𝜋 so 𝑘 = −𝑒𝜋 and 𝑦(𝑡) = −𝑒−(𝑡−𝜋) on [𝜋, ∞). Putting these two pieces together, we find that the solution is { (sin 𝑡 − cos 𝑡)/2 − (3/2)𝑒−(𝑡−𝜋) if 0 ≤ 𝑡 < 𝜋, 𝑦(𝑡) = −𝑒−(𝑡−𝜋) if 𝜋 ≤ 𝑡 < ∞. 29. The characteristic polynomial of the equation 𝑦 ′′ −𝑦 = 𝑓 (𝑡) is 𝑠2 −1 = (𝑠− 1)(𝑠 + 1) so the homogeneous equation has the solution 𝑦ℎ (𝑡) = 𝑎𝑒𝑡 + 𝑏𝑒−𝑡 for constants 𝑎 and 𝑏. On the interval [0, 1] the equation 𝑦 ′′ − 𝑦 = 𝑡 has a particular solution 𝑦𝑝 (𝑡) = −𝑡 so the general solution has the form 𝑦(𝑡) = −𝑡 + 𝑎𝑒𝑡 + 𝑏𝑒−𝑡 . The initial conditions give 0 = 𝑦(0) = 𝑎 + 𝑏 and 1 = 𝑦 ′ (0) = −1 + 𝑎 − 𝑏. Solving gives 𝑎 = 1, 𝑏 = −1 so 𝑦(𝑡) = −𝑡 + 𝑒𝑡 − 𝑒−𝑡 on [0, 1). By continuity it follows that 𝑦(1) = −1 + 𝑒1 − 𝑒−1 and 𝑦 ′ (1) = −1 + 𝑒1 + 𝑒−1 and these constitute the initial values for the equation 𝑦 ′′ − 𝑦 = 0 on the interval [1, ∞). The general solution on this interval is 𝑦(𝑡) = 𝑎𝑒𝑡 +𝑏𝑒−𝑡 and at 𝑡 = 1 we get 𝑦(1) = 𝑎𝑒1 +𝑏𝑒−1 = −1+𝑒1 −𝑒−1 and 𝑦 ′ (1) = 𝑎𝑒1 − 𝑏𝑒−1 = −1 + 𝑒1 + 𝑒−1 . Solving for 𝑎 and 𝑏 gives 𝑎 = 1 − 𝑒−1 and 𝑏 = −1 so that 𝑦(𝑡) = (1 − 𝑒−1 )𝑒𝑡 − 𝑒−𝑡 = 𝑒𝑡 − 𝑒𝑡−1 − 𝑒−1 . Putting the two pieces together gives { −𝑡 + 𝑒𝑡 − 𝑒−𝑡 if 0 ≤ 𝑡 < 1, 𝑦(𝑡) = 𝑡 𝑡−1 −1 𝑒 −𝑒 −𝑒 1 ≤ 𝑡 < ∞. 30. The characteristic polynomial of the equation 𝑦 ′′ − 4𝑦 ′ + 4𝑦 = 𝑓 (𝑡) is 𝑠2 − 4𝑠 + 4 = (𝑠 − 2)2 so the homogeneous equation has the solution 𝑦ℎ (𝑡) = 𝑎𝑒2𝑡 + 𝑏𝑡𝑒2𝑡 for constants 𝑎 and 𝑏, which is valid on the interval

114

1 Solutions

[0, 2). The initial conditions 𝑦(0) = 1 and 𝑦 ′ (0) = 0 imply that 𝑐1 = 1 and 𝑐2 = −2. So 𝑦(𝑡) = 𝑒2𝑡 − 2𝑡𝑒2𝑡 on [0, 2). By continuity it follows that 𝑦(2) = 𝑒4 − 4𝑒4 = −3𝑒4 and 𝑦 ′ (2) = −8𝑒4 and these constitute the initial values for the equation 𝑦 ′′ − 4𝑦 ′ + 4𝑦 = 4 on the interval [2, ∞). The general solution on this interval is 𝑦(𝑡) = 1 + 𝑎𝑒2𝑡 + 𝑏𝑡𝑒2𝑡 and at 𝑡 = 2 we get 𝑦(2) = 1 + 𝑎𝑒4 + 2𝑏𝑒4 = −3𝑒4 and 𝑦 ′ (1) = (2𝑎 + 𝑏)𝑒4 + 4𝑏𝑒4 = −8𝑒4 . Solving for 𝑎 and 𝑏 gives 𝑎 = 1 − 5𝑒−4 and 𝑏 = −2 + 2𝑒−4 so that 𝑦(𝑡) = 1 + (1 − 5𝑒−4)𝑒2𝑡 + (−2 + 2𝑒−4)𝑡𝑒2𝑡 = 1 + 𝑒2𝑡 − 5𝑒2(𝑡−2) − 2𝑡𝑒2𝑡 + 2𝑡𝑒2(𝑡−2) . Putting the two pieces together gives { 𝑒2𝑡 − 2𝑡𝑒2𝑡 if 0 ≤ 𝑡 < 2, 𝑦(𝑡) = 1 + 𝑒2𝑡 − 5𝑒2(𝑡−2) − 2𝑡𝑒2𝑡 + 2𝑡𝑒2(𝑡−2) 2 ≤ 𝑡 < ∞. 33. 1. ∣𝑓 (𝑡)∣ = ∣sin(1/𝑡)∣ ≤ 1 for all 𝑡 ∕= 0, while ∣𝑓 (0)∣ = ∣0∣ = 0 ≤ 1. 2. It is enough to observe that lim𝑡→0+ does not exist. But letting 𝑡𝑛 = 1 𝑛𝜋 gives 𝑓 (𝑡𝑛 ) = sin 𝑛𝜋 = 0 for all positive integers 𝑛, while letting 2 𝑡𝑛 = (4𝑛+1)𝜋 gives 𝑓 (𝑡𝑛 ) = sin(1/𝑡𝑛 ) = sin((4𝑛 + 1)𝜋/2) = sin(2𝑛𝜋 + 𝜋 2 ) = 1 so there is one sequence 𝑡𝑛 → 0 with 𝑓 (𝑡𝑛 ) → 0 while another sequence 𝑡𝑛 → 0 with 𝑓 (𝑡𝑛 ) → 1 so 𝑓 (𝑡) cannot be continuous at 0. 3. To be piecewise continuous, 𝑓 (𝑡) would have to have a limit at 𝑡 approaches 0 from above, and this is not true as shown in part 2.

Section 5.2 ⎧  ⎨0 1. 𝑓 (𝑡) = 3ℎ(𝑡 − 2) − ℎ(𝑡 − 5) = 3  ⎩ 2 𝑦 3 2 1 0 0 1 2 3

if 𝑡 < 2, if 2 ≤ 𝑡 < 5, Thus, the graph is if 𝑡 ≥ 5.

𝑡 4 5 6 7 8 ⎧ 0 if    ⎨2 if 2. 𝑓 (𝑡) = 2ℎ(𝑡 − 2) − 3ℎ(𝑡 − 3) + 4ℎ(𝑡 − 4) =   −1 if ⎩ 3 if graph is

𝑡 < 2, 2 ≤ 𝑡 < 3, Thus, the 2 ≤ 𝑡 < 4, 𝑡 ≥ 4.

1 Solutions

115

𝑦 3 2 1 1 2 3 4 5 6 7

−1

𝑡

3. This function is 𝑔(𝑡 − 1)ℎ(𝑡 − 1) where 𝑔(𝑡) = 𝑡, so the graph of 𝑓 (𝑡) is the graph of 𝑔(𝑡) = 𝑡 translated 1 unit to the right and then truncated at 𝑡 = 1, with the graph before 𝑡 = 1 replaced by the line 𝑦 = 0. Thus the graph is 𝑦 2 1 0

0

1

2

3

𝑡

4. This function is 𝑔(𝑡 − 2)ℎ(𝑡 − 2) where 𝑔(𝑡) = 𝑡2 , so the graph of 𝑓 (𝑡) is the graph of 𝑔(𝑡) = 𝑡2 translated 1 unit to the right and then truncated at 𝑡 = 2, with the graph before 𝑡 = 2 replaced by the line 𝑦 = 0. Thus the graph is 𝑦 2 1 0

0

1

2

3

𝑡

5. This function is just 𝑡2 truncated at 𝑡 = 2, with the graph before 𝑡 = 2 replaced by the line 𝑦 = 0. Thus the graph is 𝑦 8 6 4 2 𝑡 0 1 2 3 where the dashed line is the part of the 𝑡2 graph that has been truncated. It is only shown for emphasis and it is not part of the graph. 0

116

1 Solutions

6. This is the sin 𝑡 function, truncated at 𝑡 = 𝜋. The graph is 𝑦 1 0

𝜋

2𝜋

3𝜋

𝑡

−1 7. This is the function cos 2𝑡 shifted 𝜋 units to the right and then truncated at 𝑡 = 𝜋. The graph is 𝑦 1 0

𝜋

2𝜋

3𝜋

𝑡

−1 8. 𝑓 (𝑡) = 𝑡2 𝜒[0, 1) (𝑡) + (2 − 𝑡)𝜒[1, 3) (𝑡) + 3𝜒[3, ∞) (𝑡) ⎧ 𝑡2 if 0 ≤ 𝑡 < 1, ⎨ = 2 − 𝑡 if 1 ≤ 𝑡 < 3, Thus, the graph is  ⎩ 3 if 𝑡 ≥ 3. 𝑦 3 2 1 1

2

3

4

5

𝑡

−1 9. (a) (𝑡 − 2)𝜒[2, ∞) (𝑡); (b) (𝑡 − 2)ℎ(𝑡 − 2); (c) ℒ {(𝑡 − 2)ℎ(𝑡 − 2)} = 𝑒−2𝑠 ℒ {𝑡} = 𝑒−2𝑠 /𝑠2 . 10. (a) 𝑡𝜒[2, ∞) (𝑡); (b) 𝑡ℎ(𝑡 − 2); (c) ℒ {𝑡ℎ(𝑡 − 2)} = 𝑒−2𝑠 ℒ {𝑡 + 2} = 𝑒−2𝑠

(

) 2 1 + . 𝑠2 𝑠

11. (a) (𝑡 + 2)𝜒[2, ∞) (𝑡); (b) (𝑡 + 2)ℎ(𝑡 − 2); (c) ℒ {(𝑡 + 2)ℎ(𝑡 − 2)} = 𝑒

−2𝑠

ℒ {(𝑡 + 2) + 2} = 𝑒

−2𝑠

(

) 1 4 + . 𝑠2 𝑠

1 Solutions

117

12. (a) (𝑡 − 4)2 𝜒[4, ∞) (𝑡); (b) (𝑡 − 4)2 ℎ(𝑡 − 4); { } { } 2 (c) ℒ (𝑡 − 4)2 ℎ(𝑡 − 4) = 𝑒−2𝑠 ℒ 𝑡2 = 𝑒−4𝑠 3 . 𝑠

2 13. (a) 𝑡2{𝜒[4, ∞) (𝑡); (b) ℎ(𝑡 −{4); } 𝑡 −4𝑠 } { } 2 (c) ℒ 𝑡(ℎ(𝑡 − 4) = 𝑒 ) ℒ (𝑡 + 4)2 = 𝑒−4𝑠 ℒ 𝑡2 + 8𝑡 + 16 2 8 16 = 𝑒−4𝑠 + 2+ . 𝑠3 𝑠 𝑠

14. (a) (𝑡2 − 4)𝜒[4, ∞) (𝑡); (b) (𝑡2 − 4)ℎ(𝑡 − 4); { 2 } { } {2 } −4𝑠 2 −4𝑠 (c) ℒ (𝑡 ( − 4)ℎ(𝑡 − 4) )= 𝑒 ℒ (𝑡 + 4) − 4 = 𝑒 ℒ 𝑡 + 8𝑡 + 12 2 8 12 + 2+ . = 𝑒−4𝑠 𝑠3 𝑠 𝑠 15. (a) (𝑡{− 4)2 𝜒[2, ∞) (𝑡); (b) − 4)2{ℎ(𝑡 − 2); } (𝑡−2𝑠 } {2 } 2 2 −2𝑠 (c) ℒ (𝑡 − 4) ℎ(𝑡 − 2) = 𝑒 ℒ ((𝑡 + 2) − 4) = 𝑒 ℒ 𝑡 − 4𝑡 + 4 ( ) 2 4 4 = 𝑒−2𝑠 − + . 𝑠3 𝑠2 𝑠 16. (a) 𝑒𝑡−4 𝜒[4, ∞) (𝑡); (b) 𝑒𝑡−4 ℎ(𝑡 − 4); { } (c) ℒ 𝑒𝑡−4 ℎ(𝑡 − 4) = 𝑒−4𝑠 ℒ {𝑒𝑡 } = 𝑒−4𝑠

1 . 𝑠−1

17. (a) 𝑒𝑡 𝜒[4, ∞) (𝑡); (b) 𝑒𝑡 ℎ(𝑡 −{4); } (c) ℒ {𝑒𝑡 ℎ(𝑡 − 4)} = 𝑒−4𝑠 ℒ 𝑒𝑡+4 = 𝑒−4𝑠 𝑒4 ℒ {𝑒𝑡 } 1 = 𝑒−4(𝑠−1) . 𝑠−1 18. (a) 𝑒𝑡−4 𝜒[6, ∞) (𝑡); (b) 𝑒𝑡−4 ℎ(𝑡 − 6); { } { } (c) ℒ 𝑒𝑡−4 ℎ(𝑡 − 6) = 𝑒−6𝑠 ℒ 𝑒(𝑡+6)−4 = 𝑒−6𝑠 𝑒2 ℒ {𝑒𝑡 } 1 = 𝑒−6𝑠+2 . 𝑠−1

19. (a) 𝑡𝑒𝑡 𝜒[4, ∞) (𝑡); (b) 𝑡𝑒𝑡 ℎ(𝑡 − { 4); } −4𝑠 𝑡+4 (c) ℒ {𝑡𝑒𝑡 ℎ(𝑡 − 4)} = 𝑒 ℒ (𝑡 = 𝑒−4𝑠 𝑒4 ℒ {𝑡𝑒𝑡 + 4𝑒𝑡 } ( ) + 4)𝑒 1 4 = 𝑒−4(𝑠−1) + . (𝑠 − 1)2 𝑠−1 20. (a) 𝜒[0,4) (𝑡) − 𝜒[4,5) (𝑡); (b) 1 − 2ℎ(𝑡 − 4) + ℎ(𝑡 − 5); 1 𝑒−4𝑠 𝑒−5𝑠 (c) ℒ {1 − 2ℎ(𝑡 − 4) + ℎ(𝑡 − 5)} = − 2 + . 𝑠 𝑠 𝑠 21. (a) 𝑡𝜒[0,1) (𝑡) + (2 − 𝑡)𝜒[1,∞) (𝑡); (b) 𝑡 + (2 − 2𝑡)ℎ(𝑡 − 1); (c) ℒ {𝑡 + (2 − 2𝑡)ℎ(𝑡 − 1)} = ℒ {𝑡} + 𝑒−𝑠 ℒ {(2 − 2(𝑡 + 1))} 1 2𝑒−𝑠 = ℒ {𝑡} + 𝑒−𝑠 ℒ {−2𝑡} = 2 − 2 . 𝑠 𝑠

118

1 Solutions

22. (a) 𝑡𝜒[0,1) (𝑡) + (2 − 𝑡)𝜒[1,2) (𝑡) + 𝜒[2,∞) (𝑡); (b) 𝑡 − (2 − 2𝑡)ℎ(𝑡 − 1) + (𝑡 − 1)ℎ(𝑡 − 2); (c) ℒ {𝑡 + (2 − 2𝑡)ℎ(𝑡 − 1) + (𝑡 − 1)ℎ(𝑡 − 2)} = ℒ {𝑡} + 𝑒−𝑠 ℒ {(2 − 2(𝑡 + 1))} + 𝑒−2𝑠 ℒ {(𝑡 + 2) − 1} −2𝑠 = ℒ {𝑡} + 𝑒−𝑠 ℒ {−2𝑡} (+ 𝑒 ℒ){𝑡 + 1} −𝑠 1 2𝑒 1 1 = 2 − 2 + 𝑒−2𝑠 + . 𝑠 𝑠 𝑠2 𝑠 23. (a) 𝑡2 𝜒[0, 2) (𝑡) + 4𝜒[2, 3) (𝑡) + (7 − 𝑡)𝜒[3, ∞) (𝑡); (b) 𝑡2{+ (4 − 𝑡2 )ℎ(𝑡 − 2) + (3 − 𝑡)ℎ(𝑡 − 3); } 2 (c) ℒ{ 𝑡} + (4 − 𝑡2{ )ℎ(𝑡 − 2) + (3}− 𝑡)ℎ(𝑡 − 3) = ℒ 𝑡2 + 𝑒−2𝑠 ℒ 4 − (𝑡 + 2)2 + 𝑒−3𝑠 ℒ {3 − (𝑡 + 3)} ( ) 2 2 4 𝑒−3𝑠 = 3 − 𝑒−2𝑠 + − 2 . 3 2 𝑠 𝑠 𝑠 𝑠 24. (a) 𝜒[0,2) (𝑡) + (3 − 𝑡)𝜒[2,3) (𝑡) + 2(𝑡 − 3)𝜒[3,4) (𝑡) + 2𝜒[4,∞) (𝑡); (b) 1 + (2 − 𝑡)ℎ(𝑡 − 2) + (3𝑡 − 9)ℎ(𝑡 − 3) − (2𝑡 − 4)ℎ(𝑡 − 4); (c) ℒ {1 + (2 − 𝑡)ℎ(𝑡 − 2) + (3𝑡 − 9)ℎ(𝑡 − 3) − (2𝑡 − 4)ℎ(𝑡 − 4)} −3𝑠 −4𝑠 = ℒ {1}+𝑒−2𝑠 ℒ {2 − (𝑡 + 2)}+𝑒 ℒ {3(𝑡 ( ) + 3) − 9}−𝑒 ℒ {2(𝑡 + 4) − 4} −2𝑠 −3𝑠 3𝑒 2 4 1 𝑒 = − 2 + − 𝑒−4𝑠 + . 𝑠 𝑠 𝑠2 𝑠2 𝑠 ∑∞ 25. (a) 𝑛=0 ∑ (𝑡 − 𝑛)𝜒[𝑛,𝑛+1) (𝑡); (b) 𝑡 − ∞ 𝑛=1 ∑∞ℎ(𝑡 − 𝑛); ∑∞ (c) ℒ {𝑡 − 𝑛=1 ℎ(𝑡 − 𝑛)} = ℒ {𝑡} − 𝑛=1 ℒ {ℎ(𝑡 − 𝑛)} ∑ 𝑒−𝑛𝑠 1 1 ∑∞ 1 𝑛 = 2− ∞ = 2− (𝑒−𝑠 ) 𝑛=1 𝑠 𝑠 𝑠 𝑠 𝑛=1 1 𝑒−𝑠 = 2− . 𝑠 𝑠(1 − 𝑒−𝑠 ) ∑∞ ∑∞ 26. (a) 𝑛=0 𝜒[2𝑛,2𝑛+1) (𝑡); (b) 𝑛=0 (−1)𝑛 ℎ(𝑡 − 𝑛); 1 (c) . 𝑠(1 + 𝑒−𝑠 ) ∑∞ ∑∞ 27. (a) 𝑛=0 (2𝑛 + 1 − 𝑡)𝜒[2𝑛,2𝑛+2) (𝑡); (b) −(𝑡 + 1) + 2 𝑛=0 ℎ(𝑡 − 2𝑛); 1 1 2 (c) − 2 − + . 𝑠 𝑠 𝑠(1 − 𝑒−2𝑠 ) { −3𝑠 } { } 𝑒 1 28. ℒ−1 = ℎ(𝑡 − 3) ℒ−1 𝑠−1 𝑠 − 1 𝑡→𝑡−3 { 0 if 0 ≤ 𝑡 < 3, = ℎ(𝑡 − 3) (𝑒𝑡 )∣𝑡→𝑡−3 = 𝑒𝑡−3 ℎ(𝑡 − 3) = 𝑒𝑡−3 if 𝑡 ≥ 3.

1 Solutions

29. ℒ−1

{

𝑒

} −3𝑠 𝑠2

= ℎ(𝑡 − 3) ℒ−1

{

} 1 𝑠2

𝑡→𝑡−3 {

= ℎ(𝑡 − 3) (𝑡)∣𝑡→𝑡−3 = (𝑡 − 3)ℎ(𝑡 − 3) =

0 if 0 ≤ 𝑡 < 3, 𝑡 − 3 if 𝑡 ≥ 3.

} { } 1 𝑒−3𝑠 −1 = ℎ(𝑡 − 3) ℒ 30. ℒ (𝑠 − 1)3 (𝑠 − 1)3 𝑡→(𝑡−3) ( ) = ℎ(𝑡 − 3) 12 𝑡2 𝑒𝑡 𝑡→𝑡−3 = 21 (𝑡 − 3)2 𝑒𝑡−3 ℎ(𝑡 − 3) { 0 if 0 ≤ 𝑡 < 3, = 1 2 𝑡−3 if 𝑡 ≥ 3. 2 (𝑡 − 3) 𝑒 −1

{

} { } 𝑒−𝜋𝑠 1 −1 31. ℒ = ℎ(𝑡 − 𝜋) ℒ 2 2 𝑠 +1 𝑠 + 1 𝑡→𝑡−𝜋 = ℎ(𝑡 − 𝜋) (sin 𝑡)∣𝑡→𝑡−𝜋 = ℎ(𝑡 − 𝜋) sin(𝑡 − 𝜋) { { 0 if 0 ≤ 𝑡 < 𝜋, 0 if 0 ≤ 𝑡 < 𝜋, = = sin(𝑡 − 𝜋) if 𝑡 ≥ 𝜋 − sin 𝑡 if 𝑡 ≥ 𝜋. −1

{

} { } 𝑠 𝑠𝑒−3𝜋𝑠 −1 32. ℒ = ℎ(𝑡 − 3𝜋) ℒ 𝑠2 + 1 𝑠2 + 1 𝑡→𝑡−3𝜋 = ℎ(𝑡 − 3𝜋) (cos 𝑡)∣𝑡→𝑡−3𝜋 = ℎ(𝑡 − 3𝜋) cos(𝑡 − 3𝜋) { { 0 if 0 ≤ 𝑡 < 3𝜋, 0 if 0 ≤ 𝑡 < 3𝜋, = = cos(𝑡 − 3𝜋) if 𝑡 ≥ 3𝜋 − cos 𝑡 if 𝑡 ≥ 3𝜋. −1

{

} { } 𝑒−𝜋𝑠 1 −1 = ℎ(𝑡 − 𝜋) ℒ 2 2 𝑠 + 2𝑠 + 5 𝑠 + 2𝑠 + 5 𝑡→𝑡−𝜋 { } 1 = ℎ(𝑡 − 𝜋) ℒ−1 = ℎ(𝑡 − 𝜋) ( 21 𝑒−𝑡 sin 2𝑡) 𝑡→𝑡−𝜋 (𝑠 + 1)2 + 22 𝑡→𝑡−𝜋 { 0 if 0 ≤ 𝑡 < 𝜋, 1 −(𝑡−𝜋) = 2𝑒 sin 2(𝑡 − 𝜋)ℎ(𝑡 − 𝜋) = 1 −(𝑡−𝜋) sin 2𝑡 if 𝑡 ≥ 𝜋. 2𝑒

33. ℒ−1

{

} 𝑒−𝑠 𝑒−2𝑠 + 3 𝑠2 (𝑠{− 1)} { } 1 1 −1 + ℎ(𝑡 − 2) ℒ = ℎ(𝑡 − 1) ℒ−1 2 3 𝑠 (𝑠 − 1) 𝑡→𝑡−2 𝑡→𝑡−1 = ℎ(𝑡 − 1) (𝑡)∣𝑡→𝑡−1 + ℎ(𝑡 − 2) ( 21 𝑡2 𝑒𝑡 ) 𝑡→𝑡−2 = (𝑡 − 1)ℎ(𝑡 − 1) + 21 (𝑡 − 2)2 𝑒𝑡−2 ℎ(𝑡 − 2) ⎧ 0 if 0 ≤ 𝑡 < 1, ⎨ = 𝑡−1 if 1 ≤ 𝑡 < 2,  ⎩ 𝑡 − 1 + 12 (𝑡 − 2)2 𝑒𝑡−2 if 𝑡 ≥ 2.

34. ℒ−1

{

119

120

1 Solutions

} { } 𝑒−2𝑠 1 −1 = ℎ(𝑡 − 2) ℒ 2 2 𝑠 +4 𝑠 + 4 𝑡→𝑡−2 = ℎ(𝑡 − 2) ( 21 sin 2𝑡) 𝑡→𝑡−2 = 12 ℎ(𝑡 − 2) sin 2(𝑡 − 2) { 0 if 0 ≤ 𝑡 < 2, = 1 2 sin 2(𝑡 − 2) if 𝑡 ≥ 2.

35. ℒ−1

{

{

𝑒−2𝑠 𝑠2 − 4

{ } 1 = ℎ(𝑡 − 2) ℒ−1 2 𝑠 − 4 𝑡→𝑡−2 { ( )} 1 1 1 = ℎ(𝑡 − 2) ℒ−1 − 4 𝑠−2 𝑠+2 𝑡→𝑡−2( ) 1 1 2𝑡 −2𝑡 = ℎ(𝑡 − 2) ( 4 (𝑒 − 𝑒 ) 𝑡→𝑡−2 = 4 ℎ(𝑡 − 2) 𝑒2(𝑡−2) − 𝑒−2(𝑡−2) { 0 if 0 ≤ 𝑡 < 2, ) = 1 ( 2(𝑡−2) −2(𝑡−2) −𝑒 if 𝑡 ≥ 2. 4 𝑒

36. ℒ−1

}

} { } 𝑠𝑒−4𝑠 𝑠 −1 = ℎ(𝑡 − 4) ℒ 2 2 𝑠 + 3𝑠 + 2 𝑠 + 3𝑠 + 2 𝑡→𝑡−4 { } 1 2 −1 = ℎ(𝑡 − 4) ℒ − 𝑠 + 2 𝑠 + 1 𝑡→𝑡−4 ( ) = ℎ(𝑡 − 4) (2𝑒−2𝑡 − 𝑒−𝑡 ) 𝑡→𝑡−4 = ℎ(𝑡 − 4) 2𝑒−2(𝑡−4) − 𝑒−(𝑡−4) { 0 if 0 ≤ 𝑡 < 4, = 2𝑒−2(𝑡−4) − 𝑒−(𝑡−4) if 𝑡 ≥ 4.

37. ℒ−1

{

} { } { } 𝑒−2𝑠 + 𝑒−3𝑠 1 1 −1 −1 38. ℒ = ℎ(𝑡−2) ℒ +ℎ(𝑡−3) ℒ 𝑠2 − 3𝑠 + 2 𝑠2 − 3𝑠 + 2 𝑡→𝑡−2 𝑠2 − 3𝑠 + 2 𝑡→𝑡−3 { } 1 1 = ℎ(𝑡 − 2) ℒ−1 − 𝑠 − 2 𝑠 − 1 𝑡→𝑡−2 { } 1 1 + ℎ(𝑡 − 3) ℒ−1 − 𝑠 − 2 𝑠 − 1 𝑡→𝑡−3 = ℎ(𝑡 − 2) (𝑒2𝑡 − 𝑒𝑡 ) 𝑡→𝑡−2 + ℎ(𝑡 − 3) (𝑒2𝑡 − 𝑒𝑡 ) 𝑡→𝑡−3 ( ) ( ) = ℎ(𝑡 − 2) 𝑒2(𝑡−2) − 𝑒𝑡−2 + ℎ(𝑡 − 3) 𝑒2(𝑡−3) − 𝑒𝑡−3 { } { } { } 1 − 𝑒−5𝑠 1 1 −1 −1 −1 39. ℒ = ℒ − ℎ(𝑡 − 5) ℒ 𝑠2 𝑠2 𝑠2 𝑡→𝑡−5 { 𝑡 if 0 ≤ 𝑡 < 5, = 𝑡 − ℎ(𝑡 − 5) (𝑡)∣𝑡→𝑡−5 = 𝑡 − (𝑡 − 5)ℎ(𝑡 − 5) = 5 if 𝑡 ≥ 5. −1

{

1 Solutions

{

} −3𝑠

{

}

{

121

} 1 𝑠4

1+𝑒 1 = ℒ−1 + ℎ(𝑡 − 3) ℒ−1 𝑠4 𝑠4 𝑡→𝑡−3 ( ) = 61 𝑡3 + ℎ(𝑡 − 3) 61 𝑡3 𝑡→𝑡−3 = 16 𝑡3 + 16 (𝑡 − 3)3 ℎ(𝑡 − 3) { 1 3 𝑡 if 0 ≤ 𝑡 < 3, = 16 3 1 3 if 𝑡 ≥ 3. 6 𝑡 + 6 (𝑡 − 3) { } { } 2𝑠 + 1 2𝑠 + 1 = ℎ(𝑡 − 𝜋)ℒ−1 41. ℒ−1 𝑒−𝜋𝑠 2 2 𝑠 + 6𝑠 + 13 𝑠 + 6𝑠 + 13 𝑡→𝑡−𝜋 { } 2(𝑠 + 3) − 5 = ℎ(𝑡 − 𝜋) ℒ−1 (𝑠 + 3)2 + 22 𝑡→𝑡−𝜋 { } 2(𝑠 + 3) = ℎ(𝑡 − 𝜋) ℒ−1 2 2 (𝑠 + 3) + 2 𝑡→𝑡−𝜋 { } −5 + ℎ(𝑡 − 𝜋) ℒ−1 2 2 (𝑠 + 3) + 2 𝑡→𝑡−𝜋 = ℎ(𝑡 − 𝜋) (2𝑒−3𝑡 cos 2𝑡 − 25 𝑒−3𝑡 sin 2𝑡) 𝑡→𝑡−𝜋 ( ) −3(𝑡−𝜋) = ℎ(𝑡 2 cos 2(𝑡 − 𝜋) − 25 sin 2(𝑡 − 𝜋) { − 𝜋)𝑒 40. ℒ−1

=

0 ( 𝑒−3(𝑡−𝜋) 2 cos 2𝑡 −

5 2

sin 2𝑡

)

if 0 ≤ 𝑡 < 𝜋, if 𝑡 ≥ 𝜋.

42. This is the same calculation as the previous exercise with the additional term added: } { 5 2𝑠 + 1 −1 ℒ = 2𝑒−3𝑡 cos 2𝑡 − 𝑒−3𝑡 sin 2𝑡. 2 𝑠 + 6𝑠 + 13 2 { } 2𝑠 + 1 −1 −𝜋𝑠 Combining this with the previous calculation gives ℒ (1 − 𝑒 ) 2 = 𝑠 + 6𝑠 + 13 { } 2𝑠 + 1 ℒ−1 2 𝑠{ + 6𝑠 + 13 } 2𝑠 + 1 −1 −ℒ 𝑒−𝜋𝑠 2 𝑠 + 6𝑠 + 13) ( = 𝑒−3𝑡 2 cos 2𝑡 − 25 sin 2𝑡 ( ) − ℎ(𝑡 − 𝜋)𝑒−3(𝑡−𝜋) 2 cos 2(𝑡 − 𝜋) − 25 sin 2(𝑡 − 𝜋)

Section 5.3

( ) 1. 𝑦 = − 23 ℎ(𝑡 − 1) 1 − 𝑒−2(𝑡−1) ( ) 2. 𝑦 = 𝑒−2𝑡 − 1 + 2ℎ(𝑡 − 1) 1 − 𝑒−2(𝑡−1) ( ) ( ) 3. 𝑦 = ℎ(𝑡 − 1) 1 − 𝑒−2(𝑡−1) − ℎ(𝑡 − 3) 1 − 𝑒−2(𝑡−3) ( ) 4. 𝑦 = 41 𝑒−2𝑡 − 14 + 12 𝑡 + ℎ(𝑡 − 1) 41 𝑒−2(𝑡−1) − 14 + 12 (𝑡 − 1) − ( ) 1) 1 − 𝑒−2(𝑡−1)

1 2 ℎ(𝑡

−

122

1 Solutions

5. − 91 ℎ(𝑡 − 3) (−1 + cos 3(𝑡 − 3)) 6. 𝑦 = − 32 𝑒𝑡 +

5 4𝑡 12 𝑒

8. 𝑦 = cos 3𝑡 +

1 24 ℎ(𝑡

1 4

1 12 ℎ(𝑡

( ) − 5) −3 + 4𝑒𝑡−5 − 𝑒4(𝑡−5) ( ) 7. 𝑦 = 31 ℎ(𝑡 − 1) 1 − 3𝑒−2(𝑡−1) + 2𝑒−3(𝑡−1) ( ) + 31 ℎ(𝑡 − 3) −1 + 3𝑒−3(𝑡−3) − 2𝑒−3(𝑡−3) +

+

− 2𝜋) (3 sin 𝑡 − sin 3𝑡) ( ) 9. 𝑦 = 𝑡𝑒−𝑡 + ℎ(𝑡 − 3) 1 − (𝑡 − 2)𝑒−(𝑡−3) ( ) 10. 𝑦 = 𝑡𝑒−𝑡 − 14 ℎ(𝑡 − 3) −𝑒𝑡 − 5𝑒−𝑡+6 + 2𝑡𝑒−𝑡+6 ( ) 1 −5𝑡 1 1 11. 𝑦 = 20 𝑒 − 14 𝑒−𝑡 + 51 + 20 ℎ(𝑡 − 2) 4 + 𝑒−5(𝑡−2) − 5𝑒−(𝑡−2) + 20 ℎ(𝑡 − ( ) ( ) 1 −5(𝑡−4) −(𝑡−4) −5(𝑡−6) −(𝑡−6) 4) 4 + 𝑒 − 5𝑒 + 20 ℎ(𝑡 − 6) 4 + 𝑒 − 5𝑒

Section 5.4 1. 𝑦 = ℎ(𝑡 − 1)𝑒−2(𝑡−1) 2. 𝑦 = (1 + ℎ(𝑡 − 1))𝑒−2(𝑡−1) 3. 𝑦 = ℎ(𝑡 − 1)𝑒−2(𝑡−1) − ℎ(𝑡 − 3)𝑒−2(𝑡−3) { 1 sin 2𝑡 if 0 ≤ 𝑡 < 𝜋, 1 4. 𝑦 = 2 (1 + ℎ(𝑡 − 𝜋)) sin 2𝑡 = 2 sin 2𝑡 if 𝑡 ≥ 𝜋. { 1 sin 2𝑡 if 𝜋 ≤ 𝑡 < 2𝜋, 5. 𝑦 = 21 𝜒[𝜋, 2𝜋) sin 2𝑡 = 2 0 otherwise. 6. 𝑦 = cos 2𝑡 + 21 𝜒[𝜋, 2𝜋) sin 2𝑡 7. 𝑦 = (𝑡 − 1)𝑒−2(𝑡−1) ℎ(𝑡 − 1) ( ) 8. 𝑦 = (𝑡 − 1) 𝑒−2𝑡 + 𝑒−2(𝑡−1) ℎ(𝑡 − 1) 9. 𝑦 = 3ℎ(𝑡 − 1)𝑒−2(𝑡−1) sin(𝑡 − 1)

10. 𝑦 = 𝑒−2𝑡 (sin 𝑡 − cos 𝑡) + 3ℎ(𝑡 − 1)𝑒−2(𝑡−1) sin(𝑡 − 1) ( ) 11. 𝑦 = 𝑒−2𝑡 cos 4𝑡 + 21 sin 4𝑡 ( ) + 14 sin 4𝑡 ℎ(𝑡 − 𝜋)𝑒−2(𝑡−𝜋) − ℎ(𝑡 − 2𝜋)𝑒−2(𝑡−2𝜋) ( 5𝑡 ) ( ) 1 12. 𝑦 = 18 𝑒 − 𝑒−𝑡 − 6𝑡𝑒−𝑡 + 16 ℎ(𝑡 − 3) 𝑒5(𝑡−3) − 𝑒−(𝑡−3)

1 Solutions

123

Section 6.1 ⎤ ⎡ ⎤ 1 −1 −1 2 0 8 𝐵𝐴 = ⎣5 −2 18 ⎦, 𝐶𝐴 = ⎣−2 3 7⎦ 0 1 5 3 −1 7 ⎡ ⎤ [ ] 3 −1 7 3 4 3. 𝐴(𝐵 + 𝐶) = 𝐴𝐵 + 𝐴𝐶 = , (𝐵 + 𝐶)𝐴 = ⎣3 1 25⎦ 1 13 5 0 12 ⎡ ⎤ −2 5 4. 𝐶 = ⎣−13 −8⎦ 7 0 ⎡ ⎤ 6 4 −1 −8 5. 𝐴𝐵 = ⎣0 2 −8 2 ⎦ 2 −1 9 −5 [ ] 2 3 −8 6. 𝐵𝐶 = −2 0 24 ⎡ ⎤ 8 0 ⎢ 4 −5⎥ ⎥ 7. 𝐶𝐴 = ⎢ ⎣ 8 14 ⎦ 10 11 ⎡ ⎤ 6 0 2 ⎢ 4 2 −1⎥ ⎥ 8. 𝐵 𝑡 𝐴𝑡 = ⎢ ⎣−1 −8 9 ⎦ −8 2 −5 ⎡ ⎤ 8 9 −48 9. 𝐴𝐵𝐶 = ⎣ 4 0 −48⎦ . −2 3 40 ⎡ ⎤ 1 4 3 1 ⎢0 0 0 0⎥ ⎥ 10. 𝐴𝐵 = −4 and 𝐵𝐴 = ⎢ ⎣−1 −4 −3 −1⎦ −2 −8 −6 −2 [ ] 1 0 14. 1 −1 ⎡ ⎤ 0 0 1 15. ⎣3 −5 −1⎦ 0 0 5 [

] [ ] −3 1 63 2. 𝐴𝐵 = , 𝐴𝐶 = , −3 5 46

⎡

124

1 Solutions

16. 𝐴𝐵 − 𝐵𝐴 = since 1 ∕= −1.

[

] 𝑎𝑏 0 . It is not possible to have 𝑎𝑏 = 1 and −𝑎𝑏 = 1 0 −𝑎𝑏 [

] [ ] 01 00 17. (a) Choose, for example, 𝐴 = and 𝐵 = . 00 10 (b) (𝐴 + 𝐵)2 = 𝐴2 + 2𝐴𝐵 + 𝐵 2 precisely when 𝐴𝐵 = 𝐵𝐴. [ ] [ ] 11 12 18. 𝐴2 = , 𝐴3 = 12 23 [ ] 1𝑛 19. 𝐵 𝑛 = 01 [ 𝑛 ] 𝑎 0 20. 𝐴𝑛 = 0 𝑏𝑛 [ ] [ ] [ ] 01 𝑣2 1𝑐 21. (a) 𝐴 = ; the two rows of 𝐴 are switched. (b) 𝐴 = 1 0 𝑣 01 1 [ ] 𝑣1 + 𝑐𝑣2 ; to the first row is added 𝑐 times the second row while the 𝑣2 second row is unchanged, (c) to the second row is added 𝑐 times the first row while the first row is unchanged. (d) the first row is multiplied by 𝑎 while the second row is unchanged, (e) the second row is multiplied by 𝑎 while the first row is unchanged.

Section 6.2 ⎡

⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 4 3 2 14 3 2 𝑥 ⎢ ⎥ ⎢ ⎥ 1 −1⎥ ⎥, x = ⎣ 𝑦 ⎦, b = ⎢4⎥, and [𝐴∣b] = ⎢ 1 1 −1 4 ⎥. ⎣1 ⎦ ⎣2 0 1 1⎦ 0 1⎦ 𝑧 1 −1 6 0 1 −1 6 ⎡ ⎤ 𝑥1 [ ] [ ] [ ] ⎢ 𝑥2 ⎥ 2 −3 4 1 0 2 −3 4 1 0 ⎢ ⎥ (b) 𝐴 = , x = ⎣ ⎦, b = , and [𝐴∣b] = . 3 8 −3 −6 𝑥3 1 3 8 −3 −6 1 𝑥4

1 ⎢1 1. (a) 𝐴 = ⎢ ⎣2 0

𝑥1 5𝑥1 2. 3𝑥1 −8𝑥1

− 𝑥3 + 3𝑥2 − 3𝑥3 − 2𝑥2 + 8𝑥3 + 2𝑥2 ⎡ ⎤ 10 1 3. 𝑝2,3 (𝐴) = ⎣0 1 4 ⎦ 000

+ − + +

4𝑥4 𝑥4 4𝑥4 2𝑥4

+ − − +

3𝑥5 3𝑥5 3𝑥5 𝑥5

= 2 = 1 = 3 = −4

1 Solutions

4. RREF [

] 1 0 −5 −2 −1 5. 𝑡2,1 (−2)(𝐴) = 01 3 1 1 ⎡ ⎤ 010 3 6. 𝑚2 (1/2)(𝐴) = ⎣0 0 1 3 ⎦ 0000 7. RREF

⎤ 1010 3 8. 𝑡1,3 (−3)(𝐴) = ⎣0 1 3 4 1 ⎦ 00000 ⎡ ⎤ 100 2 9. ⎣0 1 0 1 ⎦ 0 0 1 −1 ⎡ ⎤ 1 0 0 −11 −8 10. ⎣0 1 0 −4 −2⎦ 001 9 6 ⎡ ⎤ 0 1 0 72 14 ⎢0 0 1 3 1 ⎥ 2⎥ 11. ⎢ ⎣0 0 0 0 0 ⎦ 00000 ⎡ ⎤ 1200 3 ⎢0 0 1 0 2 ⎥ ⎥ 12. ⎢ ⎣0 0 0 1 0 ⎦ 00000 ⎡ ⎤ 100 1 1 1 13. ⎣0 1 0 −1 3 1 ⎦ 001 2 11 ⎡ ⎤ 10 2 ⎢0 1 1 ⎥ ⎢ ⎥ ⎥ 14. ⎢ ⎢0 0 0 ⎥ ⎣0 0 0 ⎦ 000 ⎡ ⎤ 1010 ⎢0 1 3 0 ⎥ ⎥ 15. ⎢ ⎣0 0 0 1 ⎦ 0000 ⎡

125

126

16.

17.

18.

19.

20.

21.

22.

1 Solutions

⎡

⎤ 14003 ⎣0 0 1 0 1⎦ 00013 ⎡ ⎤ 1 0 0 0 0 ⎢0 ‘1 −1 0 0 ⎥ ⎢ ⎥ ⎣0 0 0 1 −1⎦ 0 0 0 0 0 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 𝑥 −1 −3 ⎣𝑦⎦ = ⎣ 1 ⎦ + 𝛼⎣ 1 ⎦ 𝑧 0 5 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 𝑥1 4 −1 −2 ⎢ 𝑥2 ⎥ ⎢−1⎥ ⎢−3⎥ ⎢−1⎥ ⎢ ⎥ = ⎢ ⎥+ 𝛼⎢ ⎥ +𝛽⎢ ⎥ ⎣ 𝑥3 ⎦ ⎣ 0 ⎦ ⎣1⎦ ⎣0⎦ 𝑥4 0 0 1 [ ] [ ] 𝑥 −2 =𝛼 𝑦 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 𝑥1 3 −4 ⎢ 𝑥2 ⎥ ⎢ 0 ⎥ ⎢1⎥ ⎢ ⎥ = ⎢ ⎥+ 𝛼⎢ ⎥ ⎣ 𝑥3 ⎦ ⎣−2⎦ ⎣0⎦ 𝑥4 5 0 ⎡ ⎤ ⎡ ⎤ 𝑥 14/3 ⎣ 𝑦 ⎦ = ⎣ 1/3 ⎦ 𝑧 −2/3

23. no solution ⎡ ⎤ ⎡ ⎤ 0 1 24. ⎣ 3 ⎦ + 𝛼 ⎣ 0 ⎦ 4 0 ⎡

⎤ ⎡ ⎤ ⎡ ⎤ 5 1 1 25. The equation ⎣−1⎦ = 𝑎 ⎣ 1 ⎦ + 𝑏 ⎣−1⎦ has solution 𝑎 = 2 and 𝑏 = 3. By 2 0 ⎡4 ⎤ 5 Proposition 6 ⎣−1⎦ is a solution. 4

26. 𝑘 = 2

27. 𝑘 = 2

1 Solutions

⎡

⎤

127

−7/2 28. (a) If x𝑖 is the solution set for 𝐴x = b𝑖 then x1 = ⎣ 7/2 ⎦, x2 = −3/2 ⎡ ⎤ ⎡ ⎤ −3/2 7 ⎣ 3/2 ⎦, and x3 = ⎣−6⎦. −1/2 3 (b) The augmented matrix [𝐴∣𝑏1 ∣𝑏2 ∣𝑏3 ] reduces to ⎡ ⎤ 1 0 0 −7/2 −3/2 7 ⎣0 1 0 7/2 3/2 −6 ⎦ . 0 0 1 −3/2 −1/2 3 The last three columns correspond in order to the solutions.

Section 6.3 [

] 4 −1 1. −3 1 ] [ 3 −2 2. −4 3 3. not invertible [ ] −2 1 4. −3/2 1/2 5. not invertible ⎡ ⎤ 1 −1 1 6. ⎣0 1 −2⎦ 0 0 1 ⎡ ⎤ −6 5 13 7. ⎣ 5 −4 −11⎦ −1 1 3 ⎡ ⎤ −1/5 2/5 2/5 8. ⎣−1/5 −1/10 2/5 ⎦ −3/5 1/5 1/5 ⎡ ⎤ −29 39/2 −22 13 ⎢ 7 −9/2 5 −3 ⎥ ⎥ 9. ⎢ ⎣−22 29/2 −17 10 ⎦ 9 −6 7 −4

128

1 Solutions

⎡

⎤ −1 0 0 −1 ⎢ 0 −1 0 −1 ⎥ ⎥ 10. 21 ⎢ ⎣ 0 0 −1 −1 ⎦ −1 −1 −1 −1 ⎡ ⎤ 0 0 −1 1 ⎢1 0 0 0⎥ ⎥ 11. ⎢ ⎣ 0 1 1 −1⎦ −1 −1 0 1

12. not invertible [ ] 5 13. b = −3 ⎡ ⎤ −2 14. b = ⎣ 6 ⎦ −3 ⎡ ⎤ 16 1 ⎣ 11 ⎦ 15. b = 10 18 ⎡ ⎤ 1 16. b = ⎣ 1 ⎦ 1 ⎤ ⎡ 19 ⎢ −4 ⎥ ⎥ 17. b = ⎢ ⎣ 15 ⎦ −6 ⎡ ⎤ 3 ⎢1⎥ ⎥ 18. b = ⎢ ⎣−4⎦ . 1

19. (𝐴𝑡 )−1 = (𝐴−1 )𝑡 20. (𝐸(𝜃))−1 = 𝐸(−𝜃) 21. 𝐹 (𝜃)−1 = 𝐹 (−𝜃) [ ] [ ] [ ] 10 12 12 22. Example. 𝐴 = ,𝐵= ,𝐶= . 00 34 56

1 Solutions

Section 6.4 1. 1 2. 0 3. 10 4. 8 5. −21 6. 6 7. 2 8. 15 9. 0 10. 11. 12.

13.

14.

15.

16.

] −2 + 𝑠 2 𝑠 = 0, 3 1 −1 + 𝑠 [ ] 𝑠−3 1 1 𝑠 = 2, 4 𝑠2 −6𝑠+8 1 𝑠−3 [ ] 𝑠−1 1 1 𝑠=1±𝑖 𝑠2 −2𝑠+𝑠 −1 𝑠 − 1 ⎡ ⎤ (𝑠 − 1)2 3 𝑠−1 1 ⎣ 0 ⎦ 𝑠=1 (𝑠 − 1)2 0 (𝑠−1)3 0 3(𝑠 − 1) (𝑠 − 1)2 ⎡ 2 ⎤ 𝑠 − 2𝑠 + 10 −3𝑠 − 6 3𝑠 − 12 1 ⎣ −3𝑠 + 12 𝑠2 − 2𝑠 − 8 3𝑠 − 12 ⎦ 𝑠 = −2, 1, 4 𝑠3 −3𝑠2 −6𝑠+8 3𝑠 + 6 −3𝑠 − 6 𝑠2 − 2𝑠 − 8 ⎡ 2 ⎤ 𝑠 + 𝑠 4𝑠 + 4 0 1 ⎣−𝑠 − 1 𝑠2 + 𝑠 0 ⎦ 𝑠 = −1, ±2𝑖 𝑠3 +𝑠2 +4𝑠+4 𝑠 − 4 4𝑠 + 4 𝑠2 + 4 [ ] 9 −4 −2 1 1 𝑠2 −3𝑠

[

17. no inverse [ ] 6 −4 1 18. 10 −2 3

129

130

19.

20.

21.

22.

23.

1 Solutions

⎡

⎤ 4 −4 4 1⎣ ⎦ 8 −1 3 −1 −5 −1 3 ⎡ ⎤ 27 −12 3 1 ⎣ −13 5 4 ⎦ 21 −29 16 −4 ⎡ ⎤ 2 −98 9502 1⎣ ⎦ 6 0 3 −297 0 0 6 ⎡ ⎤ −13 76 −80 35 ⎢ ⎥ 1 ⎢−14 76 −80 36 ⎥ 2 ⎣ 6 −34 36 −16 ⎦ 7 −36 38 −17 ⎡ ⎤ 55 −95 44 −171 ⎢ ⎥ 1 ⎢50 −85 40 −150 ⎥ 15 ⎣70 −125 59 −216 ⎦ 65 −115 52 −198

24. no inverse

Section 7.1 1. nonlinear 2. linear,( constant non homogeneous ) coefficient, ( 2) 1 1 𝑡 y′ = y+ −1 1 1 3. linear,( homogeneous, but not constant coefficient ) sin 𝑡 −1 ′ y = y 1 cos 𝑡 4. nonlinear, because of the presence of sin 𝑦1 or cos 𝑦2 . 5. linear,⎛constant ⎞ coefficient, homogeneous 1000 ⎜2 0 0 1⎟ ⎟ y′ = ⎜ ⎝0 0 0 1⎠ y 0120

6. linear,( constant ) coefficient, ( ) nonhomogeneous 1 −1 5 ′ y = 2 1 y+ −1 2 −5

1 Solutions

131

7. First note that 𝑦1 (0) = [0 and] 𝑦2 (0)[ = 1, so ]the initial is [ condition ] ′ 𝑡 3𝑡 𝑦 (𝑡) 𝑒 − 3𝑒 5 −2 1 satisfied. Then 𝒚 ′ (𝑡) = = while 𝒚(𝑡) = 𝑦2′ (𝑡)[ 2𝑒𝑡]− 3𝑒3𝑡 4[−1 ] [ 𝑡 ] 5(𝑒 − 𝑒3𝑡 ) − 2(2𝑒𝑡 − 𝑒3𝑡 ) 𝑒𝑡 − 3𝑒3𝑡 5 −2 = . Thus 𝒚 ′ (𝑡) = 𝒚, as 𝑡 3𝑡 𝑡 3𝑡 4(𝑒 − 𝑒 ) − (2𝑒 − 𝑒 ) 2𝑒𝑡 − 3𝑒3𝑡 4 −1 required. [ ] [ ] 𝑦 𝑦 In solutions, 11–16, 𝒚 = 1 = ′ . 𝑦2 𝑦 11. Let 𝑦1 = 𝑦 and 𝑦2 = 𝑦 ′ . Then 𝑦1′[ =]𝑦 ′ = 𝑦2 and 𝑦2′ = 𝑦 ′′ = −5𝑦 ′ −6𝑦+𝑒2𝑡 = 𝑦 −6𝑦1 − 5𝑦2 + 𝑒2𝑡 . Letting 𝒚 = 1 , this can be expressed in vector form 𝑦2 as [ ] [ ] 0 1 0 𝒚′ = 𝒚 + 2𝑡 −6 −5 𝑒 [ ] 1 𝒚(0) = . −2 ( ) 0 1 ′ 12. y = y 2 −𝑘 0 ( ) −1 y(0) = 0 ( ) 0 1 13. y′ = y 2 𝑘 ( 0) −1 y(0) = 0 ( ) ( ) 0 1 0 ′ 14. y = y+ −𝑘 2 0 cos 𝜔𝑡 ( ) 0 y(0) = 0 [ ] [ ] 0 1 𝛼 ′ 15. 𝒚 = 𝑐 𝑏 𝒚, 𝒚(0) = 𝛽 − − 𝑎 𝑎 [ ] [ ] 0 1 −2 ′ 16. 𝒚 = 1 2 𝒚, 𝒚(1) = 3 − 2 − 𝑡 𝑡 [ ] [ ] [ ] 0 1 0 𝛼 17. 𝒚 ′ = , 𝒚(0) = 𝑐 𝑏 𝒚+ 𝐴 sin 𝜔𝑡 𝛽 − − 𝑎 𝑎 [ ] −2 sin 2𝑡 2 cos 2𝑡 ′ 18. 𝐴 (𝑡) = −2 cos 2𝑡 −2 sin 2𝑡

132

1 Solutions

] −3𝑒−3𝑡 1 2𝑡 2𝑒2𝑡 ⎡ −𝑡 ⎤ −𝑒 (1 − 𝑡)𝑒−𝑡 (2𝑡 − 𝑡2 )𝑒−𝑡 𝐴′ (𝑡) = ⎣ 0 −𝑒−𝑡 (1 − 𝑡)𝑒−𝑡 ⎦ 0 0 −𝑒−𝑡 ⎡ ⎤ 1 𝒚 ′ (𝑡) = ⎣ 2𝑡 ⎦ 𝑡−1 [ ] 00 𝐴′ (𝑡) = 00 [ ] 𝒗 ′ (𝑡) = −2𝑒−2𝑡 𝑡22𝑡 +1 −3 sin 3𝑡 [ ] 0 1 −1 0 [ 2 ] 𝑒 − 𝑒−2 𝑒2 + 𝑒−2 − 1 1 4 1 − 𝑒2 − 𝑒−2 𝑒2 − 𝑒−2 ⎡ ⎤ 3/2 ⎣ 7/3 ⎦ ln 4 − 1 [ ] 4 8 12 16

19. 𝐴′ (𝑡) =

20.

21.

22. 23. 24. 25.

26.

27.

[

28. Continuous on 𝐼1 , 𝐼4 , and 𝐼5 [1 1 ] 𝑠 𝑠2 2 1 𝑠3 𝑠−2

29.

30.

[

𝑠 1 𝑠2 +1 𝑠2 +1 −1 𝑠 𝑠2 +1 𝑠2 +1

31.

[

3! 2𝑠 1 𝑠4 (𝑠2 +1)2 (𝑠+1)2 2−𝑠 𝑠−3 3 𝑠3 𝑠2 −6𝑠+13 𝑠

32.

33.

1 ⎤ 𝑠2 ⎢2⎥ ⎣ 𝑠3 ⎦ 6 𝑠4

]

⎡

2 𝑠2 −1

[

] 1 −1 −1 1

]

1 Solutions

133

1 1 𝑠 𝑠2+1 𝑠(𝑠2 +1) ⎥ 𝑠 1 𝑠2 +1 𝑠2 +1 ⎦ 𝑠 0 𝑠−1 +1 𝑠2 +1

⎡1

⎤

⎢ 34. ⎣ 0

[ ] 35. 1 2𝑡 3𝑡2 [ ] 1 𝑡 36. 𝑒𝑡 +𝑒−𝑡 cos 𝑡 2 [ ] 𝑒𝑡 + 𝑒−𝑡 𝑒𝑡 − 𝑒−𝑡 37. 𝑒𝑡 − 𝑒−𝑡 𝑒𝑡 + 𝑒−𝑡 ⎡

𝑒𝑡

⎢ 38. ⎣ −4 3 +

𝑒−3𝑡 3

𝑡𝑒𝑡

⎤

⎥ + 𝑒𝑡 sin 𝑡⎦ 𝑒3𝑡

3 cos 3𝑡

Section 7.2 =

[

1 𝑠−1

0

0

1 𝑠−2

2. (𝑠𝐼 − 𝐴)−1 =

[

−1 𝑠−2 𝑠(𝑠−3) 𝑠(𝑠−3) −2 𝑠−1 𝑠(𝑠−3) 𝑠(𝑠−3)

3. (𝑠𝐼 − 𝐴)−1 =

⎡1

1. (𝑠𝐼 − 𝐴)

4. (𝑠𝐼 − 𝐴) 5. 6. 7. 8.

−1

−1

=

]

and ℒ

1 𝑠+1 ⎤ 𝑠 𝑠2 𝑠3 ⎢ 1 1 ⎥ ⎣ 0 𝑠 𝑠2 ⎦ 0 0 1𝑠

[

𝑠 1 𝑠2 +1 𝑠2 +1 −1 𝑠 𝑠2 +1 𝑠2 +1

]

]

−1

(𝑠𝐼 − 𝐴)

−1

and ℒ−1 (𝑠𝐼 − 𝐴)−1 = ⎡

and ℒ

−1

(𝑠𝐼 − 𝐴)

−1

]

[2

3 2 3

+ 31 𝑒3𝑡 −

1 𝑡 𝑡+

⎢ and ℒ−1 (𝑠𝐼 − 𝐴)−1 = ⎣0 1 00

] [ −𝑡 ] 𝑒−𝑡 0 𝑒 ; 𝒚(𝑡) = 0 𝑒3𝑡 −2𝑒3𝑡 [ ] [ ] cos 2𝑡 sin 2𝑡 cos 2𝑡 − sin 2𝑡 ; − sin 2𝑡 cos 2𝑡 1 sin 2𝑡 + cos 2𝑡 [ 2𝑡 2𝑡 ] [ 2𝑡 ] 𝑒 𝑡𝑒 −𝑒 + 2𝑡𝑒2𝑡 ; y(𝑡) = 0 𝑒2𝑡 2𝑒2𝑡 [ −𝑡 ] [ −𝑡 ] 𝑒 cos 2𝑡 𝑒−𝑡 sin 2𝑡 𝑒 cos 2𝑡 ; y(𝑡) = −𝑒−𝑡 sin 2𝑡 𝑒−𝑡 cos 2𝑡 −𝑒−𝑡 sin 2𝑡 [

𝑒𝑡 0 = 0 𝑒2 𝑡 [

[

𝑡 1

1 3 2 3𝑡 1 3𝑒 3

𝑡2 2

⎤ ⎥ ⎦

cos 𝑡 sin 𝑡 = − sin 𝑡 cos 𝑡

]

− 13 𝑒3𝑡 + 23 𝑒3𝑡

]

134

9. 10. 11.

12.

13. 14.

15.

16.

17.

18.

19.

20. 21. 22.

1 Solutions

] [ −𝑡 ] 3𝑒𝑡 − 𝑒−𝑡 −𝑒𝑡 + 𝑒−𝑡 𝑒 ; y(𝑡) = 3𝑒𝑡 − 3𝑒−𝑡 −𝑒𝑡 + 3𝑒−𝑡 3𝑒−𝑡 [ 𝑡 ] [ 𝑡 ] 𝑒 + 2𝑡𝑒𝑡 −4𝑡𝑒𝑡 𝑒 − 2𝑡𝑒𝑡 ; y(𝑡) = 𝑡𝑒𝑡 𝑒𝑡 − 2𝑡𝑒𝑡 𝑒𝑡 − 𝑡𝑒𝑡 [ ] [ ] cos 𝑡 + 2 sin 𝑡 −5 sin 𝑡 cos 𝑡 + 7 sin 𝑡 ; y(𝑡) = sin 𝑡 cos 𝑡 − 2 sin 𝑡 3 sin 𝑡 − cos 𝑡 [ ] [ ] 𝑒−𝑡 cos 2𝑡 −2𝑒−𝑡 sin 2𝑡 2𝑒−𝑡 (cos 2𝑡 + sin 2𝑡) ; y(t) = −𝑡 1 −𝑡 𝑒 (sin 2𝑡 − 2 cos 2𝑡) sin 2𝑡 𝑒−𝑡 cos 2𝑡 2𝑒 [ 𝑡 ] [ 3𝑡 ] 3𝑡 3𝑡 𝑡 𝑒 1 𝑒 +𝑒 𝑒 −𝑒 ; y(𝑡) = 2 𝑒3𝑡 − 𝑒𝑡 𝑒𝑡 + 𝑒3𝑡 𝑒3𝑡 [ 𝑡 ] [ 𝑡 ] 𝑒 + 4𝑡𝑒𝑡 2𝑡𝑒𝑡 3𝑒 + 10𝑡𝑒𝑡 ; y(𝑡) = −8𝑡𝑒𝑡 𝑒𝑡 − 4𝑡𝑒𝑡 −𝑒𝑡 − 20𝑡𝑒𝑡 ⎤ ⎡ ⎤ ⎡ −𝑡 −2𝑒−𝑡 + 3𝑒𝑡 𝑒 0 32 (𝑒𝑡 − 𝑒−𝑡 ) ⎦; y(𝑡) = ⎣ ⎦ ⎣ 0 𝑒2𝑡 0 𝑒2𝑡 𝑡 2𝑒𝑡 0 0 𝑒 ⎡ ⎤ ⎡ ⎤ cos 2𝑡 2 sin 2𝑡 0 2 cos 2𝑡 + 2 sin 2𝑡 ⎣ − 21 sin 2𝑡 cos 2𝑡 0 ⎦; y(𝑡) = ⎣ cos 2𝑡 − sin 2𝑡 ⎦ −𝑡 2 cos 2𝑡 + 2 sin 2𝑡 −𝑒 + cos 2𝑡 2 sin 2𝑡 𝑒−𝑡 ⎡ −3𝑡 ⎤ ⎡ ⎤ 2𝑒 + 1 3𝑒−𝑡 − 3𝑒−3𝑡 1 − 𝑒−3𝑡 2 + 𝑒−𝑡 1⎣ ⎦; y(𝑡) = ⎣ 𝑒−𝑡 ⎦ 0 3𝑒−𝑡 0 3 −3𝑡 −𝑡 −3𝑡 −3𝑡 2 − 2𝑒 −3𝑒 + 3𝑒 𝑒 +2 4 − 𝑒−𝑡 ⎡ ⎤ ⎡ 𝑡 ⎤ −𝑡𝑒𝑡 + 𝑒𝑡 𝑡𝑒𝑡 𝑡𝑒𝑡 3𝑒 + 𝑡𝑒𝑡 ⎣ 𝑒2𝑡 − 𝑒𝑡 𝑒𝑡 −𝑒2𝑡 + 𝑒𝑡 ⎦; y(𝑡) = ⎣ 𝑒𝑡 ⎦ 𝑡 2𝑡 𝑡 −𝑡𝑒 − 𝑒 + 𝑒 𝑡𝑒𝑡 𝑒2𝑡 + 𝑡𝑒𝑡 3𝑒𝑡 + 𝑡𝑒𝑡 ⎡ 3𝑡 3𝑡 ⎤ ⎡ 3𝑡 ⎤ 𝑒 − 𝑡𝑒3𝑡 𝑒 𝑡𝑒 − 21 𝑡2 𝑒3𝑡 − 𝑡𝑒3𝑡 ⎣ 0 𝑒3𝑡 ⎦; y(𝑡) = ⎣ −𝑒3𝑡 ⎦ −𝑡𝑒3𝑡 3𝑡 0 0 𝑒 0 [ −𝑡 ] 𝑒 + 𝑡𝑒𝑡 y(𝑡) = 3𝑒−𝑡 + 𝑡𝑒𝑡 [ ] 1 + 2𝑒−𝑡 sin 2𝑡 y(𝑡) = −2 + 2𝑒−𝑡 cos 2𝑡 [ ] 1 2 cos 𝑡 + 16 sin 𝑡 + 2𝑡 cos 𝑡 − 𝑡 sin 𝑡 y(𝑡) = 2 −2 cos 𝑡 + 7 sin 𝑡 + 𝑡 cos 𝑡

1 2

[

1 Solutions

135