Solve the initial value problem using Laplace transforms. y"+y=f(t) , y(0)=0 , y'(0)=1

Given differential equation
$y"+y=f(t),y(0)=0,y^{\prime }(0)=1$
Here
$f(t)=\{\begin{array}{ll}0& 0\le t<3\pi \\ 1& t\ge 3\pi \end{array}$
To find: Solve the initial value problem using Laplace Transform
We begin with taking Laplace transformation on both sides,
${\displaystyle L\left(yy\right)=L\left(f\left(t\right)\right)}$
${\displaystyle L(y+L\left(y\right)=L\left(f\left(t\right)\right)}$
Note: ${\displaystyle L(y=s^{2}L\left(y\right)-sy\left(0\right)-y^{\prime }\left(0\right)}$
Substitute in the equation,
${\displaystyle (s^{2}L\left(y\right)-sy\left(0\right)-y^{\prime }\left(0\right))+L\left(y\right)=L\left(f\left(t\right)\right)}$
We begin with finding the Laplace for RHS,
${\displaystyle L\left(f\left(t\right)\right)={\int }_0^{3\pi }{e}^{-st}\left(0\right)dt+{\int }_{3\pi }^{\mathrm{\infty }}{e}^{-st}\left(1\right)dt}$
${\displaystyle ={\int }_{3\pi }^{\mathrm{\infty }}{e}^{-st}\left(1\right)dt}$
${\displaystyle =\frac{e^{-st}}{-s}{\mid }_{3\pi }^{\mathrm{\infty }}}$
${\displaystyle =\frac{e^{-3\pi s}}{s}}$
Now, use the conditions given,
${\displaystyle s^{2}L\left(y\right)-1+L\left(y\right)=L\left(f\left(t\right)\right)}$
${\displaystyle (s^2+1)L\left(y\right)-1=\frac{e^{-3\pi s}}{s}}$
${\displaystyle (s^2+1)L\left(y\right)=\frac{e^{-3\pi s}}{s}+1}$
${\displaystyle L\left(y\right)=\frac{e^{-3\pi s}}{s(s^2+1)}+\frac{1}{s^2+1}}$
Now , find inverse Laplace transform
${\displaystyle y\left(t\right)=L^{-1}\left(\frac{e^{-3\pi s}}{s(s^2+1)}\right)+L^{-1}\left(\frac{1}{s^2+1}\right)}$
Note:

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