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