1. The problem statement, all variables and given/known data y''+16y=9xe^(4x) y(0)=0 y'(0)=0 find the solution, y(x) to the differential equation 3. The attempt at a solution I found the particular solution to the right side of the equation, which is correct, yp= .28125x-.0703125e^(4x) For the left hand side of the equation I ended up with +- 4i, so using 4 as the beta value plugged it into, y=Acos(4x)+Bsin(4x) But plugging back into y''+16y I found it was the complementary equation... but does it even matter because there are no sines or cosines on the right hand side of the equation? I tried writing out the solution as y = Acos(4x)+Bsin(4x) + .28125x-.0703125e^(4x) Then solving it like an initial value problem but the computer won't take my answer. Any help would be greatly appreciated.