Harlen Pritchard

2020-11-10

Solve by Laplace transforms
$y"-3{y}^{\prime }+2y=2\delta \left(t-1\right)$
$y\left(0\right)=1$
${y}^{\prime }\left(0\right)=0$

### Answer & Explanation

Bertha Stark

$\text{Step 1}$

$y"-3{y}^{\prime }+2y=2\delta \left(t-1\right)$

${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)-3sY\left(s\right)-y\left(0\right)+2Y\left(s\right)=2{e}^{-s}$
$\text{Step 2}$

${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)-3sY\left(s\right)+y\left(0\right)+2Y\left(s\right)=2{e}^{-s}$

$\text{Step 3}$

$\frac{2{e}^{-s}+s-1}{\left(s-1\right)\left(s-2\right)}=\frac{A}{s-1}+\frac{B}{s-2}$

$\text{Comparing the coefficients:}$

$\text{Solving the above two equations for the value of A and B, we get}$

$\text{Step 4}$

$\frac{2{e}^{-s}+s-1}{\left(s-1\right)\left(s-2\right)}=\frac{-1}{s-2}$
$\text{taking inverse Laplace transform both sides, we get}$
${L}^{-1}\left[\frac{2{e}^{-s}+s-1}{\left(s-1\right)\left(s-2\right)}\right]={L}^{-1}\left[\frac{-1}{s-2}\right]$

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