Answer
See below
Work Step by Step
Given:
$y''+9y=7\sin 4t+14\cos 4t$
and $y(0)=1\\
y'(0)=2$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]+9Y(s)=\frac{28+14s}{s^2+16}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2+9)-s-2=\frac{28+14s}{s^2+16}$
That is,
$Y(s)(s^2+9)=\frac{28+14s}{s^2+16}+s+2=\frac{s^3+2s^2+30s+60}{s^2+16}$
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:
$Y(s)=\frac{s^3+2s^2+30s+60}{(s^2+16)(s^2+9)}=\frac{3s+6}{s^2+9}-\frac{2s+4}{s^2+16}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=3\cos 3t+2\sin 3t-2\cos 4t-\sin 4t$
