 # use the Laplace transform to solve the given initial-value problem.y"+y=f(t)y(0)=0 , y'(0)=1 wheredisplaystyle f{{left({t}right)}}={leftlbracebegin{matrix}{1}&{0}le{t}<frac{pi}{{2}}{0}&{t}gefrac{pi}{{2}}end{matrix}right.} bobbie71G 2021-03-04 Answered

use the Laplace transform to solve the given initial-value problem.
$y"+y=f\left(t\right)$
$y\left(0\right)=0,{y}^{\prime }\left(0\right)=1$ where
$f\left(t\right)=\left\{\begin{array}{cc}1& 0\le t<\frac{\pi }{2}\\ 0& t\ge \frac{\pi }{2}\end{array}$

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Step 1
Given, the initial value problem
$y"+y=f\left(t\right)$
$y\left(0\right)=0,{y}^{\prime }\left(0\right)=1$ where
$f\left(t\right)=\left\{\begin{array}{cc}1& 0\le t<\frac{\pi }{2}\\ 0& t\ge \frac{\pi }{2}\end{array}$
step 2
The Laplace transform of the two derivatives
$L\left\{{y}^{\prime }\right\}=sY\left(s\right)-y\left(0\right)$
$L\left\{y\right\}={s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left\{1\right\}=\frac{1}{s}$
$L\left\{{e}^{at}\right\}=\frac{1}{s-a}$
Where y(0) and y’(0) are initial conditions.
Step 3
Now,
when $0\le t\le \frac{\pi }{2}$
$y"+{y}^{\prime }=1,y\left(0\right)=0.{y}^{\prime }\left(0\right)=1$
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+sY\left(s\right)-y\left(0\right)=1$
$\left\{{s}^{2}+s\right\}Y\left(s\right)-s\cdot 0-1-0=1$
$\left\{{s}^{2}+s\right\}Y\left(s\right)=2$
$Y\left(s\right)=\frac{2}{{s}^{2}+s}=\frac{2}{s\left(s+1\right)}=2\left\{\frac{1}{s}-\frac{1}{s+1}\right\}$
$Y\left(s\right)=2\left\{1-{e}^{-t}\right\}$
when $t\ge \frac{\pi }{2}$
$y"+{y}^{\prime }=0,y\left(0\right)=0,{y}^{\prime }\left(0\right)=1$
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+sY\left(s\right)-y\left(0\right)=1$
$\left\{{s}^{2}+s\right\}Y\left(s\right)-s\cdot 0-1-0=0$
$\left\{{s}^{2}+s\right\}Y\left(s\right)=1$
$Y\left(s\right)=\frac{1}{{s}^{2}+s}=\frac{1}{s\left(s+1\right)}=\left\{\frac{1}{s}-\frac{1}{s+1}\right\}$
$Y\left(s\right)=\left\{1-{e}^{-t}\right\}$
Step 4
Hence the solution is