Answer
$2u_{\frac{\pi}{4}}(t)e^{t-\frac{\pi}{4}}+e^t-2u_{\frac{\pi}{4}}(t)\cos (t-\frac{\pi}{4})+2u_{\frac{\pi}{4}}(t)\sin (t-\frac{\pi}{4})$
Work Step by Step
We are given that $y'-y=4u_{\frac{\pi}{4}}(t)\cos (t-\frac{\pi}{4})$
and $y(0)=1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[sY(s)-y(0)]-Y(s)=\frac{4e^{-\frac{\pi}{4}s}}{s^2+1}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-1)-1=\frac{4e^{-\frac{\pi}{4}s}}{s^2+1}\\
Y(s)(s^2-1)=\frac{4e^{-\frac{\pi}{4}s}}{s^2+1}+1$
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:
$Y(s)=\frac{4e^{-\frac{\pi}{4}s}}{(s^2+1)(s^2-1)}+\frac{1}{s^2-1}\\
=\frac{2e^{-\frac{\pi}{4}s}}{s-1}+\frac{1}{s-1}-2\frac{e^{-\frac{\pi}{4}s}}{s^2+1}+\frac{2e^{-\frac{\pi}{4}s}}{s^2+1}$
Now,$f(t)= L^{-1}[\frac{2e^{-\frac{\pi}{4}s}}{s-1}+\frac{1}{s-1}-2\frac{e^{-\frac{\pi}{4}s}}{s^2+1}+\frac{2e^{-\frac{\pi}{4}s}}{s^2+1}]\\
=2u_{\frac{\pi}{4}}(t)e^{t-\frac{\pi}{4}}+e^t-2u_{\frac{\pi}{4}}(t)\cos (t-\frac{\pi}{4})+2u_{\frac{\pi}{4}}(t)\sin (t-\frac{\pi}{4})$
