Question

Find the general solution of y"+9y=t^2*e^3t+6.

Answer 1

first off, solve for homogeneous part \(y_h\); then, solve for partial part. This partial part separate into 2 parts, \(y_{p1}= A\). solve for A, I get A = \(\frac{2}{3}\); the second part is
\[y_{p2}= (at^2+bt+c)e^{3t}\]solve for this \(y_{p2}\), At the end, add them together. I mean y = \(y_h +y_{p1} +y_{p2}\)

Answer 2

(I had this typed out a few hours ago but my connection broke down for long time)
\[y''+9y=t^2e^{3t}+6~~?\]
Homogeneous solution:
\[r^2+9=0~\Rightarrow~r=\pm3i\]
thus yielding
\[y_h=C_1\cos3t+C_2\sin3t\]
Nonhomogeneous solution: As a guess, I'd try \(y_p=e^{3t}(At^2+Bt+C)+D\)
Solving for the constants will take some work.