Answer
See below
Work Step by Step
Given:
$y''-3y'+2y=4$
and $y(0)=0\\
y'(0)=1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]-3[sY(x)-y(0)]+2Y(s)=\frac{4}{s}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-3s+2)-0-1+0=\frac{4}{s}$
That is,
$Y(s)(s^2-3s+2)=\frac{4}{s}+1=\frac{4+s}{s}$
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:
$Y(s)=\frac{4+s}{s^2-3s+2}=\frac{s+4}{s^2-3s+2}=\frac{s+4}{s(s-2)(s-1)}=\frac{2}{s}-\frac{5}{s-1}+\frac{3}{s-2}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=2-5e^t+3e^{2t}$
