Answer
See below
Work Step by Step
Given:
$y''-9y=13\sin 2t$
and $y(0)=3\\
y'(0)=1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]-9Y(s)=\frac{13\times 2}{s^2+4}=\frac{26}{s^2+4}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-9)-3s-1=\frac{26}{s^2+4}$
That is,
$Y(s)(s^2-9)=\frac{26}{s^2+4}+3s+1=\frac{3s^3+s^2+12s+39}{s^2+4}$
Thus, we have solved for the Laplace transform of y(t). To find y itself, we must take the inverse Laplace transform. We first decompose the right-hand side into partial fractions to obtain:
$Y(s)=\frac{3s^3+s^2+12s+30}{(s^2+4)(s-3)(s+3)}=\frac{2}{s-3}+\frac{1}{s+3}-\frac{2}{s^2+4}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=2e^{3t}+e^{-3t}-\sin 2t$
