Amari Flowers

2021-02-06

Solve the following equation with Laplace Transform Method (Inverse Laplace the equation to find the solution)
$y"-3{y}^{\prime }-4y=3{e}^{2x}$
$y\left(0\right)=1,{y}^{\prime }\left(0\right)=0$

pattererX

$\text{Step 1}$
$\text{Given differential equation,}$
$y"-3{y}^{\prime }-4y=3{e}^{2x}$
$y\left(0\right)=1,{y}^{\prime }\left(0\right)=0$
$\text{Solve this differential equation by Laplace Transform method.}$
$\text{Step 2}$
$y"-3{y}^{\prime }-4y=3{e}^{2x}$
$\text{Take Laplace Transform of both sides,}$
$L\left[y"-3{y}^{\prime }-4y\right]=L\left[3{e}^{2x}\right]$
$L\left[y"\right]-3L\left[{y}^{\prime }\right]-4L\left[y\right]=3L\left[{e}^{2x}\right]$

$L\left[y"\right]=s2L\left[y\right]-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left[{y}^{\prime }\right]=sL\left[y\right]-y\left(0\right)$
$L\left[{e}^{ax}\right]=\frac{1}{s-a}$
$\text{Then from (1),}$
$\text{Step 3}$
${s}^{2}L\left[y\right]-sy\left(0\right)-{y}^{\prime }\left(0\right)-3\left[sL\left[y\right]-y\left(0\right)\right]-4L\left[y\right]=3×\frac{1}{s-2}$
${s}^{2}L\left[y\right]-s×1-0-3\left[sL\left[y\right]-1\right]-4L\left[y\right]=\frac{3}{s-2}$
$\left\{{s}^{2}-3s-4\right\}L\left[y\right]=\frac{3}{s-2}+s-3$
$L\left[y\right]=\frac{3}{\left(s-2\right)\left({s}^{2}-3s-4\right)}+\frac{s-3}{{s}^{2}-3s-4}$
$y={L}^{-1}\left[\frac{3}{\left(s-2\right)\left({s}^{2}-3s-4\right)}+\frac{s-3}{{s}^{2}-3s-4}\right]$
$y=3{L}^{-1}\left[\frac{1}{\left(s-2\right)\left({s}^{2}-3s-4\right)}\right]+{L}^{-1}\left[\frac{s-3}{{s}^{2}-3s-4}\right]...\left(2\right)$
$\text{Step 4}$
$\text{Now}$
$\frac{1}{\left(s-2\right)\left({s}^{2}-3s-4\right)}=-\frac{1}{6\left(s-2\right)}+\frac{1}{15\left(s+1\right)}+\frac{1}{10\left(s-4\right)}$
$⇒{L}^{-1}\left[\frac{1}{\left(s-2\right)\left({s}^{2}-3s-4\right)}\right]={L}^{-1}\left[-\frac{1}{6\left(s-2\right)}\right]+{L}^{-1}\left[\frac{1}{15\left(s+1\right)}\right]+{L}^{-1}\left[\frac{1}{10\left(s-4\right)}\right]$

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