# use the Laplace transform to solve the given initial-value problem y''+5y'+4y=0 y(0)=1 y'(0)=0

use the Laplace transform to solve the given initial-value problem
${y}^{″}+5{y}^{\prime }+4y=0$
$y\left(0\right)=1$
${y}^{\prime }\left(0\right)=0$
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Step 1
We know that$L\left[{y}^{″}\right]={s}^{2}y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left[{y}^{\prime }\right]=sy\left(s\right)-y\left(0\right)$
Step 2
Given that
${y}^{″}+5{y}^{\prime }+4y=0$
taking Laplace transform both sides,
$L\left[{y}^{\prime }{\right]}^{\prime }+5L\left[{y}^{\prime }\right]+4L\left[y\right]=0$
$\left({s}^{2}y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right)+5\left(sy\left(s\right)-y\left(0\right)\right)+4y\left(s\right)=0$
Now use the given initial conditions y(0)=1 and y'(0)=0, so
$\left({s}^{2}y\left(s\right)-s\right)+5\left(sy\left(s\right)-1\right)+4y\left(s\right)=0$
${s}^{2}y\left(s\right)-s+5sy\left(s\right)-5+4y\left(s\right)=0$
$y\left(s\right)\left({s}^{2}+5s+4\right)=s+5$
$y\left(s\right)=\frac{\left(s+5\right)}{\left({s}^{2}+5s+4\right)}$
$y\left(s\right)=\frac{\left(s+5\right)}{\left({s}^{2}+4s+s+4\right)}$
$y\left(s\right)=\frac{\left(s+5\right)}{s\left(s+4\right)+1\left(s+4\right)}$
$y\left(s\right)=\frac{\left(s+5\right)}{\left(s+1\right)\left(s+4\right)}$
Step 3
Now simplify $\frac{\left(s+5\right)}{\left(s+1\right)\left(s+4\right)}$ using partial fraction method,$\frac{\left(s+5\right)}{\left(s+1\right)\left(s+4\right)}=\frac{A}{\left(s+1\right)}+\frac{B}{\left(s+4\right)}$
$s+5=A\left(s+4\right)+B\left(s+1\right)$

So
$\frac{\left(s+5\right)}{\left(s+1\right)\left(s+4\right)}=\frac{\left(\frac{4}{3}\right)}{\left(s+1\right)}-\frac{\left(\frac{1}{3}\right)}{\left(s+4\right)}$
Step 4
So
$y\left(s\right)=\frac{\left(\frac{4}{3}\right)}{\left(s+1\right)}-\frac{\left(\frac{1}{3}\right)}{\left(s+4\right)}$
Now taking inverse Laplace transform,
$y={L}^{-1}\left[\frac{\frac{4}{3}}{\left(s+1\right)}\right]-{L}^{-1}\left[\frac{\frac{1}{3}}{\left(s+4\right)}\right]$
$y=\frac{4}{3}{L}^{-1}\left[\frac{1}{\left(s+1\right)}\right]-\frac{1}{3}{L}^{-1}\left[\frac{1}{\left(s+4\right)}\right]$
$y=\frac{4}{3}{e}^{-t}-\frac{1}{3}{e}^{-4t}\left[{L}^{-1}\left(\frac{1}{\left(s-a\right)}\right)={e}^{at}\right]$
Step 5
Hence using the Laplace transform solution of given initial value problem is
$y=\frac{4}{3}{e}^{-t}-\frac{1}{3}{e}^{-4t}$