Answer
See below
Work Step by Step
Given:
$y''+4y=0$
and $y(0)=5\\
y'(0)=1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]+4[sY(x)-y(0)]=0$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2+4)-5s-1=0$
That is,
$Y(s)(s^2+4)=5s+1$
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:
$Y(s)=\frac{5s+1}{s^2+4}=\frac{5s+1}{s^2+4}+\frac{1}{s^2+4}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=5\cos 2t+\frac{1}{2}\sin 2t$
