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