# Solve y"+4y=f(t) text{ subject to } y(0)=0 , y'(0)=2 text{ and } f(t)=begin{cases}0 & t< pi1 & tgeqpiend{cases}

Solve
$f\left(t\right)=\left\{\begin{array}{ll}0& t<\pi \\ 1& t\ge \pi \end{array}$
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Step 1 Given differential equation $y"+4y=f\left(t\right)$ subject to $y\left(0\right)=0,{y}^{\prime }\left(0\right)=2$ and $f\left(t\right)=\left\{\begin{array}{ll}0& t<\pi \\ 1& t\ge \pi \end{array}$
$Lf\left(t\right)={\int }_{0}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt$
$={\int }_{0}^{\pi }\left(0\right){e}^{-st}dt+{\int }_{\pi }^{\mathrm{\infty }}\left(1\right){e}^{-st}dt$
$=0+{\int }_{\pi }^{\mathrm{\infty }}\left(1\right){e}^{-st}dt$
$=|\frac{{e}^{-st}}{-s}{|}_{\pi }^{\mathrm{\infty }}$
$=|\frac{{e}^{-s\left(\mathrm{\infty }\right)}}{-s}-\frac{{e}^{-s\left(\pi \right)}}{-s}|$
$=\frac{{e}^{-\pi t}}{s}$
Step 2
Now taking Laplace transform on both sides of the given differential equation
$L\left\{{y}^{″}\right\}+4L\left\{y\right\}=L\left\{f\left(t\right)\right\}$
${s}^{2}Y\left(s\right)-s.y\left(0\right)-{y}^{\prime }\left(0\right)+4Y\left(s\right)=\frac{{e}^{-\pi s}}{s}$
Applying the conditions
${s}^{2}Y\left(s\right)-s.0-2+4Y\left(s\right)=\frac{{e}^{-\pi s}}{s}$
$Y\left(s\right)\left({s}^{2}+4\right)=\frac{{e}^{-\pi s}}{s}+2$
$Y\left(s\right)=\frac{{e}^{-\pi s}}{s\left({s}^{2}+4\right)}+\frac{2}{{s}^{2}+4}$
Now taking Inverse Laplace transform Step 3
${L}^{-1}\left\{Y\left(s\right)\right\}={L}^{-1}\frac{{e}^{-\pi s}}{s\left({s}^{2}+4\right)}+\frac{2}{{s}^{2}+4}$
$={L}^{-1}\left\{{e}^{-\pi s}\frac{1}{s\left({s}^{2}+4\right)}\right\}+2{L}^{-1}\left\{\frac{1}{{s}^{2}+4}\right\}$
By partial fraction
$\frac{1}{s\left({s}^{2}+4\right)}=\frac{1}{4s}-\frac{s}{4\left({s}^{2}+4\right)}$
${L}^{-1}\left\{Y\left(s\right)\right\}={L}^{-1}\left\{{e}^{-\pi s}\left(\frac{1}{4s}-\frac{s}{4\left({s}^{2}+4\right)}\right)\right\}+2{L}^{-1}\left\{\frac{1}{{s}^{2}+{2}^{2}}\right\}$
Step 4
Apply the rule then ${L}^{-1}\left\{{e}^{-as}F\left(s\right)\right\}=u\left(t-a\right)f\left(t-a\right)$ and from laplace transform table
${L}^{-1}\left\{\frac{1}{4s}-\frac{s}{4\left({s}^{2}+4\right)}\right\}=\frac{1}{4}-\frac{1}{4}\mathrm{cos}\left(2t\right)$
$y\left(t\right)=u\left(1-\pi \right)\left(\frac{1}{4}-\frac{1}{4}\mathrm{cos}\left(2\left(t-\pi \right)\right)\right)+2\left(\frac{1}{2}\right)\mathrm{sin}\left(2t\right)$