Answer
See below
Work Step by Step
Given:
$y''-y'-12y=36$
and $y(0)=0\\
y'(0)=12$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]-[sY(x)-y(0)]-12Y(s)=\frac{36}{s}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-s-12)-0-12+0=\frac{36}{s}$
That is,
$Y(s)(s^2-s-12)=\frac{36}{s}+12=\frac{12s+36}{s}$
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:
$Y(s)=\frac{12s+36}{s(s^2-s-12)}=\frac{12s+36}{s(s+3)(s-4)}=\frac{3}{s-4}-\frac{3}{s}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=3e^{4t}-3$
