Answer
See below
Work Step by Step
Given:
$y'+y=5e^t \sin t$
and $y(0)=1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[sY(x)-y(0)]+Y(s)=\frac{5}{(s-1)^2+1}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s+1)-1=\frac{5}{(s-1)^2+1}$
That is,
$Y(s)(s+1)=\frac{5}{(s-1)^2+1}+1=\frac{s^2-2s+7}{s^2-2s+2}$
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:
$Y(s)=\frac{s^2-2s+7}{(s+1)(s^2-2s+2)}=\frac{5}{3(s-1)^2+1}+\frac{5}{6(s+1)}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=\frac{5}{3}e^t\sin t+\frac{5}{6}e^{-t}$
