Answer
$y(t)=2e^{3t}$
Work Step by Step
Given:
$y'+y=8e^{3t}$
and $y(0)=2$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[sY(x)-y(0)]+Y(s)=\frac{8}{s-3}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s+1)-2=\frac{8}{s-3}$
That is,
$Y(s)(s+1)=2+\frac{8}{s-3}$
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:
$Y(s)=\frac{2s+2}{(s+1)(s-3)}=\frac{2}{s-3}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=2e^{3t}$
