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## Homework Statement

I managed to work this problem all the way through, but I am in no way certain of my answer. I'd greatly appreciate any insight!

Find the solution of the initial value problem.

y'''+4y'=x, y(0)=y'(0)=0, y''(0)=1

## Homework Equations

Just for clarification purposes, my professor uses y

_{c}for the general homogeneous solution; I've seen y

_{h}used and didn't want to cause confusion.

f

_{n}(x)y

^{(n)}+f

_{n-1}(x)y

^{(n-1)}+...+f

_{1}(x)y'+f

_{0}(x)y=0

General solution: y

_{c}=c

_{1}y

_{1}(x)+c

_{2}y

_{2}(x)+...+c

_{n}y

_{n}(x)

## The Attempt at a Solution

Note: m is used as the variable for characteristic equations rather than r, which I've also seen.

First, to find y

_{c}, I set the left side of the equation equal to zero and found the characteristic equation m

^{3}+4m=0, from which I found roots 0, 2i, -2i. I converted them into real solutions and ended up with y

_{c}=c

_{1}+c

_{2}cos(2x)+c

_{3}sin(2x).

To find c

_{1}and c

_{2}, I took the first and second derivatives of y

_{c}and plugged in the initial values as given. I ended up with y

_{c}=(1/4)-(1/4)cos(2x).

Next, to find y

_{p}, I guessed y

_{p}=Ax

^{3}+Bx

^{2}+Cx+D and plugged its first and third derivatives into the original differential equation. I ended up finding A=0, B=1/8, C=-1/2. That gave me y

_{p}=(1/8)x

^{2}-(1/2)x.

I got the general solution to the heterogeneous differential equation y=y

_{c}+y

_{p}:

y=(1/4)-(1/4)cos(2x)+(1/8)x

^{2}-(1/2)x.