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
Just for clarification purposes, my professor uses yc for the general homogeneous solution; I've seen yh used and didn't want to cause confusion.
General solution: yc=c1y1(x)+c2y2(x)+...+cnyn(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 yc, I set the left side of the equation equal to zero and found the characteristic equation m3+4m=0, from which I found roots 0, 2i, -2i. I converted them into real solutions and ended up with yc=c1+c2cos(2x)+c3sin(2x).
To find c1 and c2, I took the first and second derivatives of yc and plugged in the initial values as given. I ended up with yc=(1/4)-(1/4)cos(2x).
Next, to find yp, I guessed yp=Ax3+Bx2+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 yp=(1/8)x2-(1/2)x.
I got the general solution to the heterogeneous differential equation y=yc+yp: