• Author / Uploaded
• Abdu Abdoulaye

Citation preview

1|Page

Differential Equations with Boundary Value Problems Authors: Dennis G. Zill, Michael R. Cullen Exercise 1.1 In Problems 1–8 state the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear. 1.

1

4

5

cos ⋯

A differential equation is linear if it is in the form .

If we compare given differential equation with the standard form a linear differential equation, we see that it is linear. 0

2.

It is non-linear because of 4th power of 6

3.

It is linear. Note that

0 denotes fourth derivative and not

raised to power 4.

cos

4.

It is non-linear due to the term cos 5.

.

.

1 It is non-linear because of two things. (1) Presence of dependent variable under radical sign. (2) Square term of derivative

6. It is non-linear. It was going to be linear if

was

2|Page 7.

cos

sin

2

It is linear. 1

8.

0

It is non-linear due to square term of a derivative which is . Note that in this question denotes second derivative while denotes square of first derivative. For better understanding, we can write given differential equation as ′′

1

0

In Problems 9 and 10 determine whether the given first-order differential equation is linear in the indicated dependent variable. 1

9.



0; in ; in

Solution: 1

1

Clearly it in not linear in . Write as

or

0 or or

We see that it is linear in . 10.

0; in ; in

Solution:

1 1 1

0

3|Page 1 Clearly it is linear in . Write

as

or

.

We see that it is clearly not linear in . In Problems 11 –14 verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval I of definition for each solution. 11. 2

/

0 ;

L. H. S

2

/

/

Since L.H.S. = R.H.S., therefore 12.

1 2

2 /

/

/

is a solution of 2

/

/

0

R. H. S

0

dy 6 6  20 y  24 ; y   e 20t dt 5 5

d  6 6 20t   6 6 20t    e   20   e  dt  5 5  5 5  6   (20e 20t )  24  24e 20t 5  24e20t  24  24e 20t  24  R.H.S.

L.H.S. 

Since L.H.S. = R.H.S., therefore y 

6 6 20t dy is a solution of  e  20 y  24 . dt 5 5

13. y //  6 y /  13 y  0 ; y  e3 x cos 2 x L.H.S.  y //  6 y /  13 y d 2 3x d e cos 2 x   6  e3 x cos 2 x   13e3 x cos 2 x 2  dx dx d   3e3 x cos 2 x  2e3 x sin 2 x   6  3e3 x cos 2 x  2e3 x sin 2 x   13e3 x cos 2 x dx  9e3 x cos 2 x  6e3 x sin 2 x  6e3 x sin 2 x  4e3 x cos 2 x  18e3 x cos 2 x  12e3 x sin 2 x  13e3 x cos 2 x 

 0  R.H.S.

Since L.H.S. = R.H.S., therefore y  e3 x cos 2 x is a solution of y //  6 y /  13 y  0 .

4|Page

In Problems 33– 36 use the concept that , ∞ ∞ is a constant function if and only if ′ 0 to determine whether the given differential equation possesses constant solutions. 33. 3

5

10 0 in given differential equation, we get 3 0 5 10 or 5 10. Solving 2. Therefore, given differential equation has one constant solution 2.

Substitute for , we get

2

34.

3

Substitute 0 in given differential equation, we get 0 2 3. Now, if we solve given equation for , then we get two real solutions 3 and 1 which are the constant solutions of given differential equation. 1

1

35.

0, we get 0 If we substitute has no constant solution. 4

35.

6 0 and

Substitute 6

1, which is not true. Therefore, given differential equation

10 0 since derivatives of constant equals zero. Doing so, we get

10. Solving for , we get the constant solution

.

Exercise 1.2 In problems 1 and 2, 1/ 1 is a one parameter family of solutions of the first-order . Find a solution of the first-order IVP consisting of this differential equation and DE the given initial condition. 0

1.

when

0. Substitute in

3⟹

4 . Therefore, solution of given IVP is

i.e. 1 3 1 3

, we get

1 1 1 1

1 2.

1/ 1

1

2

1/ 1

4

5|Page 2 at

i.e. 2 2 2 2

1. Substitute in

1/ 1

, we get

1/ 1 1 1 1 1 2

1

2

1 1/2

Therefore, solution of given IVP is

1/ 1

Exercise 2.1 In Problems 1-22 solve the given differential equation by separation of variables. 1.

dy  sin 5 x dx

dy  sin 5 xdx

 dy   sin 5 xdx y

2.

cos 5 x C 5

dy 2   x  1 dx

dy   x  1 dx 2

 dy    x  1 y

 x  1 3

2

dx

3

C

3. dx  e3 x dy  0

6|Page

e3 x dy  dx 1 dy   3 x dx e dy  e3 x dx

 dy   e

3x

dx

e3 x y C 3 4. dy  ( y  1) 2 dx  0 dy  ( y  1) 2 dx 1 dy  dx ( y  1) 2 1  ( y  1)2 dy   dx 1   xC y 1 1   y 1 xC 1 1 y xC 1 y 1 xC 5. x

dy  4y dx

1 4 dy  dx y x 1 4  y dy   x dx ln y  4 ln x  C1 y  e4ln xC1 y  e4ln x eC1 y  eln x eC1 4

y  Cx 4

 eC1  C

7|Page

6.

dy  2 xy 2  0 dx

dy  2 xy 2 dx 1  2 dy  2 xdx y 1   2 dy   2 xdx y 1  x2  C y 1 y 2 x C

7.

dy  e3 x  2 y dx

dy  e3 x e 2 y dx 1 dy  e3 x dx 2y e e 2 y dy  e3 x dx

e

2 y

dy   e3 x dx

e 2 y e3 x    C1 2 3  e3 x   C1  e 2 y  2   3  2 e 2 y   e3 x  2C1 3 2  2C1  C e 2 y   e3 x  C 3  2  2 y  ln   e3 x  C   3  1  2  y   ln   e3 x  C  2  3  8. e x y

dy  e  y  e 2 x y dx

8|Page dy  e y  e 2 x e  y dx dy  e y 1  e 2 x  ex y dx 1  e 2 x y  dy dx e y ex ye y dy   e x  e3 x  dx ex y

 ye dy    e y

x

 e 3 x  dx

1 ye y  e y  e  x  e 3 x  C 3

dx  y  1   9. y ln x  dy  x  y ln x

2

dx ( y  1) 2  dy x2

x 2 ln xdx 

( y  1) 2 dy y

x 2 ln xdx 

y2  2 y  1 dy y

 1 x 2 ln xdx   y  2   dy y   1 ln xdx    y  2   dy y  y2 1 3 1  2 y  ln y  C x ln x  x3  3 9 2

x

10.

2

dy  2 y  3    dx  4 x  5 

2

dy (2 y  3) 2  dx (4 x  5) 2 1 1 dy  dx 2 (2 y  3) (4 x  5) 2

9|Page

1

1 dx (4 x  5) 2 1 1    C1 2(2 y  3) 4(4 x  5)

 (2 y  3)

2

dy  

  1 1  2    C1  2y  3  4(4 x  5)  1 1   2C1 2 y  3 2(4 x  5) 1 1  C 2 y  3 2(4 x  5) 1  2y  3 1 2(4 x 5)  C 2y 

1 1 2(4 x 5)  C

1 y  2 

1 1 2(4 x 5)  C

 2C1  C

3   3  

11. csc y dx  sec2 x dy  0

sec2 x dy   csc y dx 1 1 dy   dx 2 cos x sin y  sin y dy  cos 2 xdx 1 1   sin y dy    cos 2 x  dx 2 2  1 1    sin y dy     cos 2 x  dx 2 2  1 1 cos y  x  sin 4 x  C 2 4 1 1  y  cos 1  x  sin 4 x  C  4 2  12.