In this problem you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+4y=12te−2t−(8t+12) with initial values y(0)=−2andy′(0)=1.

Question

In this problem, you will use undetermined coefficients to solve the nonhomogeneous equation y"+4y'+4y=12te^(-2t)-(8t+12) 

with initial values y(0)=-2 and y'(0)=1. 

 Write the form of the particular solution and its derivatives. (Use A, B, C, etc. for undetermined coefficients.

Y =

Y' =

Y" =

D. Write the general solution. (Use c1 and c2).

Y =

E. Plug in the initial values and solve for c1 and c2 to find the solution to the initial value problem.

Y =

Answer 1

Below is the solution to the problem asked. If you need to find the A,B,C coefficients, simply plug the particular solutions into the governing equation and solve.

y'' + 4y' + 4y = 12te^(-2t) - 8t - 12

y(t) = y_p + y_h

Starting with the particular solution: y_p

G(t) = 12te^(-2t) - 8t - 12

Let: y_p(t) = A t^3 e^(-2t) + Bt + C

First derivative using chain rule:

y'_p(t) = [3A t^2 e^(-2t) - 2A t^3 e^(-2t)] + B

Second derivative using chain rule:

y''_p(t) = [6A t e^(-2t) - 6A t^2 e^(-2t)] + [-6A t^2 e^(-2t) + 4A t^3 e^(-2t)]

Solution to first part:

y = A t^3 e^(-2t) + Bt + C

y' = -2A t^3 e^(-2t) + 3A t^2 e^(-2t) + B

y'' = 4A t^3 e^(-2t) - 12A t^2 e^(-2t) + 6A t e^(-2t)

General Solution:

y'' + 4y' + 4y = 0

r^2 + 4r + 4 = 0

(r + 2)^2 = 0

r = -2,-2

y_h = C_1 e^(-2t) + C_2 t e^(-2t)

Solving for coefficients:

y_h = C_1 e^(-2t) + C_2 t e^(-2t)

y'_h = -2 C_1 e^(-2t) + C_2 e^(-2t) - 2 C_2 t e^(-2t)

-2 = C_1 e^(0) + 0

C_1 = -2

1 = -2(-2) e^(0) + C_2 e^(0) + 0

C_2 = -3

General Solution:

y_h = -2 e^(-2t) - 3 t e^(-2t)