Originally Posted by

**Jones** I have another question.

$\displaystyle y'' + 4y=2cos x + 3xsin x$

$\displaystyle \begin{aligned}y(0) = 1\\

y'(0) = 2\end{aligned}$

Show that $\displaystyle x~sin x $ is a solution to this equation, and then solve the initial value problem.

hmm, how do you go about this one?

I thougt the solution would be on the form $\displaystyle Ce^{-4t} + B +x~sin x$

since the solution to the homogenous equation $\displaystyle r^2+4r=0$

$\displaystyle \begin{aligned}

r_1 = -4 \\

r_2 = 0

\end{aligned}$