# Solve the Laplace transforms dot x-2ddot x+x=e^t t text{ given } t=0 text{ and } x=0 text{ and } x=1

Solve the Laplace transforms
$\stackrel{˙}{x}-2\stackrel{¨}{x}+x={e}^{t}t$
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Given :
The given Laplace Transform
$-2\stackrel{¨}{x}+\stackrel{˙}{x}+x={e}^{t}t$
with the given conditions
$x\left(0\right)=1,\stackrel{˙}{x}\left(0\right)=0$
Step 2
Solution :
The given Laplace Transform
$-2\stackrel{¨}{x}+\stackrel{˙}{x}+x={e}^{t}t\dots \left(1\right)$
with the given conditions
$x\left(0\right)=1,\stackrel{˙}{x}\left(0\right)=0$
Taking Laplace transform of both sides in equation (1) , we have
$-2\left[{s}^{2}\overline{x}-sx\left(0\right)-{x}^{\prime }\left(0\right)\right]+\left[s\overline{x}-x\left(0\right)\right]+\overline{x}=\frac{1}{\left(s-1{\right)}^{2}}$
$-2\left[{s}^{2}\overline{x}-s\right]+\left[s\overline{x}-1\right]+\overline{x}=\frac{1}{\left(s-1{\right)}^{2}}$
$\left[-2{s}^{2}+s+1\right]\overline{x}+\left[2s-1\right]=\frac{1}{\left(s-1{\right)}^{2}}$
$\left[-2{s}^{2}+2s-s+1\right]\overline{x}=\frac{1}{\left(s-1{\right)}^{2}}-\left[2s-1\right]$
$\left[-2s\left(s-1\right)-\left(s-1\right)\right]\overline{x}=\frac{1}{\left(s-1{\right)}^{2}}-\left[2s-1\right]$
$\left[\left(-2s-1\right)\left(s-1\right)\right]\overline{x}=\frac{1}{\left(s-1{\right)}^{2}}-\left[2s-1\right]$
$\overline{x}=\frac{1}{\left(s-1{\right)}^{2}\left[\left(-2s-1\right)\left(s-1\right)\right]}+\frac{\left[2s-1\right]}{\left[\left(2s+1\right)\left(s-1\right)\right]}$
$\overline{x}=\frac{1}{\left(s-1{\right)}^{3}\left(-2s-1\right)}+\frac{\left[2s-1\right]}{\left[\left(2s+1\right)\left(s-1\right)\right]}$
Solving By Partial fractions ,
$\frac{\left[2s-1\right]}{\left[\left(2s+1\right)\left(s-1\right)\right]}=\frac{A}{\left(s-1\right)}+\frac{B}{\left(2s+1\right)}$
$\frac{\left[2s-1\right]}{\left[\left(2s+1\right)\left(s-1\right)\right]}=\frac{A\left(2s+1\right)+B\left(s-1\right)}{\left(s-1\right)\left(2s+1\right)}$
$2A+B=2,A-B=-1$
$3A=1⇒A=\frac{1}{3}$
$\frac{1}{3}-B=-1⇒B=\frac{4}{3}$
$\frac{\left[2s-1\right]}{\left[\left(2s+1\right)\left(s-1\right)\right]}=\frac{1}{3\left(s-1\right)}+\frac{4}{3\left(2s+1\right)}$
Now , another term ,
$-\frac{1}{\left(s-1{\right)}^{3}\left(2s+1\right)}=\frac{A}{\left(2s+1\right)}+\frac{B}{\left(s-1\right)}+\frac{C}{\left(s-1{\right)}^{2}}+\frac{D}{\left(s-1{\right)}^{3}}$