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