1. The problem statement, all variables and given/known data 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 2. Relevant equations 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. fn(x)y(n)+fn-1(x)y(n-1)+...+f1(x)y'+f0(x)y=0 General solution: yc=c1y1(x)+c2y2(x)+...+cnyn(x) 3. 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: y=(1/4)-(1/4)cos(2x)+(1/8)x2-(1/2)x.