Lennie Carroll

2021-09-06

Solve the initial value problem using Laplace transforms.
$y"+y=f\left(t\right),y\left(0\right)=0,{y}^{\prime }\left(0\right)=1$
Here
$f\left(t\right)=\left\{\begin{array}{ll}0& 0\le t<3\pi \\ 1& t\ge 3\pi \end{array}$

Arnold Odonnell

Given differential equation
$y"+y=f\left(t\right),y\left(0\right)=0,{y}^{\prime }\left(0\right)=1$
Here
$f\left(t\right)=\left\{\begin{array}{ll}0& 0\le t<3\pi \\ 1& t\ge 3\pi \end{array}$
To find: Solve the initial value problem using Laplace Transform
We begin with taking Laplace transformation on both sides,
$L\left(yy\right)=L\left(f\left(t\right)\right)$
$L\left(y+L\left(y\right)=L\left(f\left(t\right)\right)$
Note: $L\left(y={s}^{2}L\left(y\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)$
Substitute in the equation,
$\left({s}^{2}L\left(y\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right)+L\left(y\right)=L\left(f\left(t\right)\right)$
We begin with finding the Laplace for RHS,
$L\left(f\left(t\right)\right)={\int }_{0}^{3\pi }{e}^{-st}\left(0\right)dt+{\int }_{3\pi }^{\mathrm{\infty }}{e}^{-st}\left(1\right)dt$
$={\int }_{3\pi }^{\mathrm{\infty }}{e}^{-st}\left(1\right)dt$
$=\frac{{e}^{-st}}{-s}{\mid }_{3\pi }^{\mathrm{\infty }}$
$=\frac{{e}^{-3\pi s}}{s}$
Now, use the conditions given,
${s}^{2}L\left(y\right)-1+L\left(y\right)=L\left(f\left(t\right)\right)$
$\left({s}^{2}+1\right)L\left(y\right)-1=\frac{{e}^{-3\pi s}}{s}$
$\left({s}^{2}+1\right)L\left(y\right)=\frac{{e}^{-3\pi s}}{s}+1$
$L\left(y\right)=\frac{{e}^{-3\pi s}}{s\left({s}^{2}+1\right)}+\frac{1}{{s}^{2}+1}$
Now , find inverse Laplace transform
$y\left(t\right)={L}^{-1}\left(\frac{{e}^{-3\pi s}}{s\left({s}^{2}+1\right)}\right)+{L}^{-1}\left(\frac{1}{{s}^{2}+1}\right)$
Note:

Do you have a similar question?