# Solve the initial value problem below using the method of Laplace transforms. y"-8y'+41y=29e^{2t} ,y(0)=1 , y'(0)=7

Solve the initial value problem below using the method of Laplace transforms.
$y8{y}^{\prime }+41y=29{e}^{2t}$
$y\left(0\right)=1,{y}^{\prime }\left(0\right)=7$

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Neelam Wainwright

The given IVP is
$y8y+41y=29{e}^{2t}$
$y\left(0\right)=1,{y}^{\prime }\left(0\right)=7$
Taking Laplace in both side we get
$L\left[y\right]-8L\left[{y}^{\prime }\right]+41L\left[y\right]=29L\left[{e}^{2t}\right]$
$⇒{s}^{2}y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)-8\left(sy\left(s\right)-y\left(0\right)\right)+41y\left(s\right)=\frac{29}{s-2}$
$⇒\left({s}^{2}-8s+41\right)y\left(s\right)-s\left(1\right)-7-8\left(-1\right)=\frac{29}{\left(s-2\right)}$
$⇒\left({s}^{2}-8s+41\right)y\left(s\right)-s-7+8=\frac{29}{\left(s-2\right)}$
$⇒\left({s}^{2}-8s+41\right)y\left(s\right)-s+1=\frac{29}{\left(s-2\right)}$
$⇒\left({s}^{2}-8s+41\right)y\left(s\right)=\frac{29}{\left(s-2\right)}+s-1$
$⇒\left({s}^{2}-8s+41\right)y\left(s\right)=\frac{29+{s}^{2}-3s+2}{\left(s-2\right)}$
$⇒y\left(s\right)=\frac{{s}^{2}-3s+31}{\left(s-2\right)\left({s}^{2}-8s+41\right)}$
$⇒y\left(s\right)=\frac{{s}^{2}-3s-5s+5s+31+10-10}{\left(s-2\right)\left({s}^{2}-8s+41\right)}$
$⇒y\left(s\right)=\frac{{s}^{2}-8s+5s+41-10}{\left(s-2\right)\left({s}^{2}-8s+41\right)}$