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

Step 1
We know that$L[y^{″}]=s^{2}y(s)-sy(0)-y^{\prime }(0)$
$L[y^{\prime }]=sy(s)-y(0)$
Step 2
Given that
$y^{″}+5y^{\prime }+4y=0$
taking Laplace transform both sides,
$L[y^{\prime }{]}^{\prime }+5L[y^{\prime }]+4L[y]=0$
$(s^{2}y(s)-sy(0)-y^{\prime }(0))+5(sy(s)-y(0))+4y(s)=0$
Now use the given initial conditions y(0)=1 and y'(0)=0, so
$(s^{2}y(s)-s)+5(sy(s)-1)+4y(s)=0$
$s^{2}y(s)-s+5sy(s)-5+4y(s)=0$
$y(s)(s^2+5s+4)=s+5$
$y(s)=\frac{(s+5)}{(s^2+5s+4)}$
$y(s)=\frac{(s+5)}{(s^2+4s+s+4)}$
$y(s)=\frac{(s+5)}{s(s+4)+1(s+4)}$
$y(s)=\frac{(s+5)}{(s+1)(s+4)}$
Step 3
Now simplify $\frac{(s+5)}{(s+1)(s+4)}$ using partial fraction method,$\frac{(s+5)}{(s+1)(s+4)}=\frac{A}{(s+1)}+\frac{B}{(s+4)}$
$s+5=A(s+4)+B(s+1)$
$\text{Put }s=-1,4=3A\Rightarrow A=\frac{4}{3}$
$\text{Put }s=-4,1=-3B\Rightarrow B=-\frac{1}{3}$
So
$\frac{(s+5)}{(s+1)(s+4)}=\frac{(\frac{4}{3})}{(s+1)}-\frac{(\frac{1}{3})}{(s+4)}$
Step 4
So
$y(s)=\frac{(\frac{4}{3})}{(s+1)}-\frac{(\frac{1}{3})}{(s+4)}$
Now taking inverse Laplace transform,
$y=L^{-1}\left[\frac{\frac{4}{3}}{(s+1)}\right]-L^{-1}\left[\frac{\frac{1}{3}}{(s+4)}\right]$
$y=\frac{4}{3}{L}^{-1}\left[\frac{1}{(s+1)}\right]-\frac{1}{3}{L}^{-1}\left[\frac{1}{(s+4)}\right]$
$y=\frac{4}{3}{e}^{-t}-\frac{1}{3}{e}^{-4t}[L^{-1}\left(\frac{1}{(s-a)}\right)=e^{at}]$
Step 5
Hence using the Laplace transform solution of given initial value problem is
$y=\frac{4}{3}{e}^{-t}-\frac{1}{3}{e}^{-4t}$