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