# Use Laplace transforms to solve the initial value problems. y"+y'=\delta(t)-\delta(t-3), y(0)=1 , y'(0)=0

Use Laplace transforms to solve the initial value problems.
$y{y}^{\prime }=\delta \left(t\right)-\delta \left(t-3\right),y\left(0\right)=1,{y}^{\prime }\left(0\right)=0$
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$L\left\{\delta \left(t-a\right)\right\}={e}^{-as}$
${L}^{-1}\left\{\frac{1}{s-a}\right\}={e}^{at}$
${L}^{-1}\left\{\frac{{e}^{-as}}{s}\right\}=u\left(t-a\right)$
Given
$y{y}^{\prime }=\delta \left(t\right)-\delta \left(t-3\right),y\left(0\right)=1,{y}^{\prime }\left(0\right)=0$
Taking laplace transformation both sides:
$L\left\{y"\right\}+L\left\{{y}^{\prime }\right\}=L\left\{\delta \left(t\right)\right\}-L\left\{\delta \left(t-3\right)\right\}$
$\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right]+\left[sy\left(s\right)-y\left(0\right)\right]=1-{e}^{-3s}$
$\left({s}^{2}y\left(s\right)-s\right)+\left(sy\left(s\right)-1\right)=1-{e}^{-3s}$
$y\left(s\right)\left[{s}^{2}+s\right]-s-1=1-{e}^{-3s}$
$y\left(s\right)\left[{s}^{2}+s\right]=2-{e}^{-3s}+s$
$y\left(s\right)=\frac{2+s-{e}^{-3s}}{s\left(s+1\right)}$
$y\left(s\right)=\frac{2+s}{s\left(s+1\right)}-\frac{{e}^{-3s}}{s\left(s+1\right)}$
$y\left(s\right)=\frac{2}{s}-\frac{1}{s+1}-\frac{{e}^{-3s}}{s}+\frac{{e}^{-3s}}{s+1}$
Now taking inverse laplace transformation both sides:
${L}^{-1}\left\{y\left(s\right)\right\}=2{L}^{-1}\left\{\frac{1}{s}\right\}-{L}^{-1}\left\{\frac{1}{s+1}\right\}-{L}^{-1}\left\{\frac{{e}^{-3s}}{s}\right\}+{L}^{-1}\left\{\frac{{e}^{-3s}}{s+1}\right\}$
$y\left(t\right)=2×1-{e}^{-t}-u\left(t-3\right)+{e}^{3-t}u\left(t-3\right)$