Answer
See below
Work Step by Step
We are given that $y''+y=t-u_1(t)(t-1) $
and $y(0)=2\\
y'(0)=1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(s)-sY(0)-Y'(0)]+Y(s)=\frac{1}{s^2}-\frac{e^{-s}}{s^2}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2+1)-2s-1=\frac{1}{s^2}-\frac{e^{-s}}{s^2}\\
Y(s)(s^2+1)=\frac{1}{s^2}-\frac{e^{-s}}{s^2}+2s+1$
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:
$Y(s)=\frac{1}{s^2}-\frac{e^{-s}}{s^2}+\frac{2s}{s^2+1}+\frac{1}{s^2+1}\\
=(\frac{1}{s^2}-\frac{1}{s^2+1})-e^{-s}(\frac{1}{s^2}-\frac{1}{s^2+1})+\frac{2s}{s^2+1}+\frac{1}{s^2+1}$
Now,$f(t)= L^{-1}[(\frac{1}{s^2}-\frac{1}{s^2+1})-e^{-s}(\frac{1}{s^2}-\frac{1}{s^2+1})+\frac{2s}{s^2+1}+\frac{1}{s^2+1}]\\
=t+2\cos t-[(t-1)-\sin (t-1)]u_1(t)$
