# Find the solution of the Differentional equation by using Laplace Transformation2y'-3y=e^{2t}, y(0)=1y"+y=t, y(0)=0 and y'(0)=2

Find the solution of the Differentional equation by using Laplace Transformation
$2{y}^{\prime }-3y={e}^{2t},y\left(0\right)=1$

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Step 1
1) Given equation is
$2{y}^{\prime }-3y={e}^{2t},y\left(0\right)=1$
Taking Laplace transform , we get
$2L\left({y}^{\prime }\right)-3L\left(y\right)=L\left({e}^{2t}\right)$
$⇒sY\left(s\right)-y\left(0\right)-3Y\left(s\right)=1s-2$
$⇒\left(s+3\right)Y\left(s\right)-1=\frac{1}{\left(s-2\right)}\left[y\left(0\right)=1\right]$
$⇒Y\left(s\right)=\frac{1}{\left(s+3\right)\left(s-2\right)}+\frac{1}{\left(s+3\right)}$
Step 2
Using partial fraction decomposition,
$\frac{1}{\left(s+3\right)\left(s-2\right)}=\frac{1}{5\left(s-2\right)}-\frac{1}{5\left(s+3\right)}$
So,
$Y\left(s\right)=\frac{1}{5\left(s-2\right)}-\frac{1}{5\left(s+3\right)}+\frac{1}{\left(s+3\right)}$
$⇒Y\left(s\right)=\frac{1}{5\left(s-2\right)}+\frac{4}{5\left(s+3\right)}$
On taking inverse Laplace Transform, we get
$y\left(t\right)=\frac{1}{5}{e}^{2t}+\frac{4}{5}{e}^{-3t}\left[{L}^{-1}\left\{\frac{1}{\left(s-a\right)}\right\}={e}^{at}\right]$
Step 3
$y"+y=t,y\left(0\right)=0,{y}^{\prime }\left(0\right)=2$
Taking Laplace transform , we get $L\left(y"\right)+L\left(y\right)=L\left(t\right)$
$⇒{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+Y\left(s\right)=\frac{1}{\left({s}^{2}\right)}\left[L\left(tn\right)=\frac{\left(n!\right)}{{s}^{n+1}}\right]$
$⇒\left({s}^{2}+1\right)Y\left(s\right)-2=\frac{1}{\left({s}^{2}\right)}\left[y\left(0\right)=0,{y}^{\prime }\left(0\right)=2\right]$
$⇒Y\left(s\right)=\frac{1}{\left({s}^{2}+1\right){s}^{2}}+\frac{2}{\left({s}^{2}+1\right)}\left(1\right)$
Using partial fraction decomposition,
$\frac{1}{\left({s}^{2}+1\right){s}^{2}}=\frac{1}{{s}^{2}}-\frac{1}{\left({s}^{2}+1\right)}$
So, from (1), we get
$Y\left(s\right)=\frac{1}{{s}^{2}}-\frac{1}{\left({s}^{2}+1\right)}+\frac{2}{\left({s}^{2}+1\right)}$
$⇒Y\left(s\right)=\frac{1}{{s}^{2}}+\frac{1}{\left({s}^{2}+1\right)}$
On taking inverse Laplace Transform, we get
$y\left(t\right)=t+\mathrm{sin}\left(t\right)\left[{L}^{-1}\left\{\frac{1}{{s}^{n+1}}\right\}=\frac{{t}^{n}}{\left(n!\right)},{L}^{-1}\left\{\frac{a}{\left({s}^{2}+{a}^{2}\right)}\right\}=\mathrm{sin}\left(at\right)\right]$