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DIFFERENTIAL EQUATIONS

have any arbitrary constant, then it is classified as particular solution.

A differential equation is an equation that contains derivatives or differentials. Examples: 1) 2) 3).

d2 y dy + 5 +6 y=0 2 dx dx dy=( 2 x + y ) dx ∂z ∂z +2 x =5 ∂x ∂y

Types of differential equations: 1. Ordinary – one that contains only one independent variable. (Examples 1 & 2) 2. Partial – one that contains two or more independent variables. (Example 3) The order of a differential equation is the order of the highest-ordered derivative appearing in the equation. In the above examples, the first is of order two and the second is of order one. The degree of a differential equation is the power of the highest-ordered derivative occurring in the equation. In the above examples, all the three equations are of degree one. The differential equation

y right )} ^ {{2} / {3}} =2+3y' ¿ ¿

can be rationalized by cubing both sides to obtain

y right )} ^ {2} = {left (2+3y' right )} ^ {3} ¿ ¿

First Order Differential Equations: I. Variable Separable A first order equation is variable separable if it can be transformed into the form A(x)dx + B(y)dy = 0 The solution is obtained by integrating each term of the above equation. II. Homogeneous A function f(x, y) is homogeneous of degree k if there is a number λ such that f(λx, λy) = λkf(x, y) Note that all monomials are homogeneous while a polynomial is homogeneous only if all of its terms are of the same degree.The equation M(x, y)dx + N(x, y)dy = 0 belongs to this type if M(x, y) and N(x, y) are both homogeneous of the same degree. The solution can be obtained by replacing x by vy or y by vx which will consequently transform the equation into one that is variable separable. III. Exact The equation M(x,y)dx + N(x,y)dy = 0 is exact if

∂M ∂N = ∂y ∂x

. Thus, it is of degree two.

The equation is considered linear if it is of the first degree in the dependent variable and its derivatives.

The solution is obtained by using the equation

∫ DI +∫ NDI =c

where A solution to a differential equation is an equation, free of derivatives, that satisfies the given equation. If the solution contains at least one arbitrary constant, it is classified as general solution; if the solution doesn’t

∫ DI

is the sum of directly integrable integrals such as

∫ g ( y ) dy

∫ NDI

∫ f ( x ) dx

or

of the equation and

is the sum of non-directly

the

integrable integrals i.e integrals of the form ∫ f ( x , y ) dx ∨∫ g ( x , y ) dy of equation taken from either the first or the second term of the given equation.

IV. Linear Equation of Order One A first order equation belongs to this type if it can be transformed into the form

dy + P ( x ) y=Q ( x) dx

The solution is obtained by first solving for the integrating factor (IF) using IF = e∫ Pdx and substituting into the equation

y ( IF ) =∫ Q ( IF ) dx V. Bernoulli’s Equation A first order equation belongs to this type if it can be transformed into the form

dy + P ( x ) y=Q (x) y n dx

The solution can be obtained by dividing both sides of the equation by yn thus getting

y−n

dy + P ( x ) y 1−n=Q(x) dx

Using the substitution z = y1 – n the equation is transformed into a linear equation of order one. Example: Solve the equation y’ = y – xy3e-2x Solution: Transforming the equation into the standard form, we get

dy 3 −2 x − y=−x y e dx

Dividing the equation by y3, we get

dy y − y −2=−x e−2 x dx If we let z=− y−2 , then dz=2 y−3 dy .

dz +2 z=−2 x e−2 x dx

in z where P(x) = 2 and the integrating factor is IF = e∫ Pdx =e∫ 2 dx =e 2 x Thus, the resulting equation becomes 2x −2 x 2 x z e =−∫ x e e dx

−1 2 c x− 2 2 −1 Replacing z by , 2 y −1 2 x −1 2 c e = x− 2 2 y2 ze 2 x =

or

2 e2 x = y 2 ( x2 + c)

dy 2y −2 y −2 =−2 x e−2 x dx −3

→ Answer

VI. Equations where the integrating factor can be determined Consider the equation M(x, y)dx + N(x, y)dy = 0. Case I. For the case where

1 ∂M ∂N − =f ( x) N ∂ y ∂x

(

)

the desired integrating factor is IF = e∫ f (x )dx Case II. For the case where

1 ∂M ∂ N − =g ( y) M ∂ y ∂x

(

)

the desired integrating factor is IF = e−∫ g ( y ) dy Example: Solve the equation y(8x – 9y) dx + 2x(x – 3y) dy = 0. Solution:

∂M =8 x−18 y ∂y

∂N =4 x−6 y ∂x Taking the difference, we get

∂M ∂N − =4 x−12 y=4 (x−3 y) ∂y ∂x

−3

Hence, the above equation becomes

→ linear

which is divisible by N. Hence

4 ( x−3 y ) 2 1 ∂M ∂N − = = N ∂ y ∂x 2 x ( x−3 y ) x

(

)

The integrating factor is then equal to

IF =

e

∫ 2x dx

=e 2 ln x =x 2

Multiplying the given equation by x2, we get x2y(8x – 9y) dx + 2x3(x – 3y) dy = 0 which is now an exact equation. Note that no term is directly integrable. Taking the non-directly integrable integrals from the first term, we get

8 x3 y (¿−9 x2 y 2)dx=c ∫¿ 2 x 4 y −3 x 3 y 2=c

A. B. C. D.

2ydx – xdy = 0 2xdx – ydy = 0 ydx - 2xdy = 0 ydx + 2xdy = 0

6. Which of the following is a solution of the differential equation 4y” + 9y = 0? A. y = c1ex + c2e–x B. y = c1e(2/3)x + c2ex C. y = c1 cos x + c2 sin x D. y = c1 cos (3x/2) + c2 sin (3x/2) 7. Solve the differential equation dy=x √ x 2 +9 dx , if the curve passes through (–4, 0).

3/ 2 1 2 y= ( x +9 ) +13 3 3 → Answer x y ( 2 x−3 y ) =c 3/ 2 1 y= ( x 2 +9 ) −13 B. 3 Review Exercises: ( x 2 +9 ) [¿ ¿3 /2+ 125] 1. The equation C. 2 y = {left [1+ {left (y' right )} ^ {2} right ]} ^ {{3} / {2}} 1 y= ¿ is of 3 A. second order, first degree 2 ( x + 9) B. second order, second degree [¿ ¿3 / 2−125] C. second order, third degree D. 1 D. first order, third degree y= ¿ 2 3 2. The equation y = cx is the general

which simplifies into

solution of A. y’ = 2y/x B. y’ = 2x/y

C. y’ = y/2x D. y’ = x/2y

3. Find the differential equation whose x general solution is y = c 1 x +c 2 e . A. (x – 1)y” – xy’ + y = 0 B. (x + 1)y” – xy’ + y = 0 C. (x – 1)y” + xy’ + y = 0 D. (x + 1)y” + xy’ + y = 0 4. Determine the differential equation of lines passing through (h, k). A. (y – k)dx – (x – h)dy = 0 B. (x – h) + (y – k) =dy/dx C. (x – h)dx – (y – k)dy = 0 D. (x + h)dx – (y – k)dy = 0 5. What is the differential equation of the family of parabolas having their vertices at the origin and their foci on the y-axis?

A.

8. Solve the differential equation: y’ = 1 + x + y + xy A. y = x + (1/2)x2 + xy + (1/2)x2y + c B. y = y ln |1+ x| + c C. y = x + (1/2)x2 + c D. ln |1+ y| = x + (1/2)x2 + c 9. Solve the differential equation 3xy’ + y = 12x A. y = 3x + cx– 1/3 B. y = 3x + c C. y = 4x + cx1/3 D. y = 4x + cx– 1/3 10. A certain radioactive substance has a half-life of 38 hr. Find how long it takes for 90% of the radioactivity to be dissipated. A. 162 hr. C. 145 hr. B. 126 hr. D. 154 hr.

11. A bacterial population is known to have a rate of growth proportional to the population itself. If between noon and 2 P.M. the population triples, at what time, no controls being exerted, should the population become 100 times what it was at noon. A. 8:05 P.M. C. 8:23 P.M. B. 8:16 P.M. D. 8:45 P.M.

Answer Key

1. B 2. C 3. A 4. A 5. A 6. D 7. D 8. D 9. A 10. B 11. C