Question

use the laplace transform to solve the given initial - value problem. use the table of laplace transforms in appendix c as needed.
$y + y = f(t)$, $y(0)=1$, $y(0)=0$, where
$f(t)=\

$$\begin{cases}1, & 0\\leq t<\\frac{\\pi}{2}\\\\ \\sin(t), & t\\geq\\frac{\\pi}{2}\\end{cases}$$

$
$y(t)=(\quad)+(\quad)+(\quad)$

Explanation:

Step1: Find Laplace transform of \(f(t)\)

We know that \(f(t)=1 - u(t-\frac{\pi}{2})+ \sin(t)u(t-\frac{\pi}{2})\).
Using the formula \(L\{u(t - a)g(t)\}=e^{-as}L\{g(t + a)\}\), \(L\{1\}=\frac{1}{s}\), \(L\{\sin(t)\}=\frac{1}{s^{2}+1}\).
\(L\{f(t)\}=\frac{1}{s}-e^{-\frac{\pi s}{2}}\frac{1}{s}+e^{-\frac{\pi s}{2}}\frac{\cos(\frac{\pi}{2})+s\sin(\frac{\pi}{2})}{s^{2}+1}=\frac{1}{s}-e^{-\frac{\pi s}{2}}\frac{1}{s}+e^{-\frac{\pi s}{2}}\frac{s}{s^{2}+1}\)

Step2: Take Laplace transform of the differential equation

Given \(y''+y = f(t)\), \(y(0)=1\), \(y'(0)=0\).
Using \(L\{y''\}=s^{2}Y(s)-sy(0)-y'(0)\) and \(L\{y\}=Y(s)\)
\(s^{2}Y(s)-s\times1 - 0+Y(s)=\frac{1}{s}-e^{-\frac{\pi s}{2}}\frac{1}{s}+e^{-\frac{\pi s}{2}}\frac{s}{s^{2}+1}\)
\(Y(s)(s^{2}+1)=s+\frac{1}{s}-e^{-\frac{\pi s}{2}}\frac{1}{s}+e^{-\frac{\pi s}{2}}\frac{s}{s^{2}+1}\)
\(Y(s)=\frac{s}{s^{2}+1}+\frac{1}{s(s^{2}+1)}-e^{-\frac{\pi s}{2}}\frac{1}{s(s^{2}+1)}+e^{-\frac{\pi s}{2}}\frac{s}{(s^{2}+1)^{2}}\)

Step3: Use partial - fraction decomposition

\(\frac{1}{s(s^{2}+1)}=\frac{1}{s}-\frac{s}{s^{2}+1}\)
\(Y(s)=\frac{s}{s^{2}+1}+\frac{1}{s}-\frac{s}{s^{2}+1}-e^{-\frac{\pi s}{2}}(\frac{1}{s}-\frac{s}{s^{2}+1})+e^{-\frac{\pi s}{2}}\frac{s}{(s^{2}+1)^{2}}\)
\(Y(s)=\frac{1}{s}-e^{-\frac{\pi s}{2}}\frac{1}{s}+e^{-\frac{\pi s}{2}}\frac{s}{s^{2}+1}+e^{-\frac{\pi s}{2}}\frac{s}{(s^{2}+1)^{2}}\)

Step4: Take inverse Laplace transform

Using \(L^{-1}\{\frac{1}{s}\}=1\), \(L^{-1}\{e^{-as}F(s)\}=u(t - a)f(t - a)\), \(L^{-1}\{\frac{s}{s^{2}+1}\}=\cos(t)\), \(L^{-1}\{\frac{s}{(s^{2}+1)^{2}}\}=\frac{1}{2}\sin(t)\)
\(y(t)=1 - u(t-\frac{\pi}{2})+u(t-\frac{\pi}{2})\cos(t-\frac{\pi}{2})+\frac{1}{2}u(t-\frac{\pi}{2})\sin(t-\frac{\pi}{2})\)
Since \(\cos(t-\frac{\pi}{2})=\sin(t)\) and \(\sin(t-\frac{\pi}{2})=-\cos(t)\)
\(y(t)=1 - u(t-\frac{\pi}{2})+u(t-\frac{\pi}{2})\sin(t)-\frac{1}{2}u(t-\frac{\pi}{2})\cos(t)\)

Answer:

\(y(t)=1+( \sin(t)-1-\frac{1}{2}\cos(t))u(t - \frac{\pi}{2})\)