Use Laplace transform to solve the following initial value problem y" + 4y = 4, y(0) = -5, y'(0) = –4 Express your answer as the Laplace transform of y and use partial fraction decomposition where possible.

Given differential equation is $y^{\text{'}\text{'}}+4y=4$ and $y\left(0\right)=-5,y^{\text{'}}\left(0\right)=-4$

We have to solve the given initial value problem by using Laplace transformation.

Use $L\left\{y^{\text{'}\text{'}}\right\}=s^{2}Y\left(s\right)-sy\left(0\right)-y^{\text{'}}\left(0\right)$ and $L\left\{y\right\}=Y\left(s\right)$

Now, use Laplace transform of both side in $y^{\text{'}\text{'}}+4y=4$

$\begin{array}{c}L\left\{y^{\text{'}\text{'}}+4y\right\}=L\left\{4\right\}\\ L\left\{y^{\text{'}\text{'}}\right\}+4L\left\{y\right\}=\frac{4}{s}\left(\text{Since, }L\left\{4\right\}=\frac{4}{s}\right)\\ s^{2}Y\left(s\right)-sy\left(0\right)-y^{\text{'}}\left(0\right)+4Y\left(s\right)=\frac{4}{s}\\ s^{2}Y\left(s\right)-s\left(-5\right)-\left(-4\right)+4Y\left(s\right)=\frac{4}{s}\left(\text{Use }y\left(0\right)=-5,y^{\text{'}}\left(0\right)=-4\right)\\ s^{2}Y\left(s\right)+5s+4+4Y\left(s\right)=\frac{4}{s}\\ \left(s^2+4\right)Y\left(s\right)+5s+4=\frac{4}{s}\\ \left(s^2+4\right)Y\left(s\right)=\frac{4}{s}-\left(5s+4\right)\\ Y\left(s\right)=\frac{4}{s\left(s^2+4\right)}-\frac{\left(5s+4\right)}{\left(s^2+4\right)}\end{array}$

Hence, $Y\left(s\right)=\frac{4}{s\left(s^2+4\right)}-\frac{\left(5s+4\right)}{\left(s^2+4\right)}$

Now, taking a pretrial fraction of $\frac{4}{s\left(s^2+4\right)}$, which is given below

$\frac{4}{s\left(s^2+4\right)}=\frac{1}{s}-\frac{s}{s^2+4}$

And taking a pretrial fraction of $\frac{\left(5s+4\right)}{\left(s^2+4\right)}$, which is given below

$\frac{\left(5s+4\right)}{\left(s^2+4\right)}=\frac{5s}{s^2+4}+\frac{4}{s^2+4}$

Therefore, 

$\begin{array}{c}Y\left(s\right)=\frac{1}{s}-\frac{s}{s^2+4}-\left(\frac{5s}{s^2+4}+\frac{4}{s^2+4}\right)\\ Y\left(s\right)=\frac{1}{s}-\frac{s}{s^2+4}-\frac{5s}{s^2+4}-\frac{4}{s^2+4}\end{array}$

Now, taking inverse Laplace transform, which is given below

$\begin{array}{c}{L}^{-1}\left\{Y\left(s\right)\right\}=L^{-1}\left\{\frac{1}{s}-\frac{s}{s^2+4}-\frac{5s}{s^2+4}-\frac{4}{s^2+4}\right\}\\ y\left(t\right)=L^{-1}\left\{\frac{1}{s}\right\}-L^{-1}\left\{\frac{s}{s^2+4}\right\}-5L^{-1}\left\{\frac{s}{s^2+4}\right\}-L^{-1}\left\{\frac{4}{s^2+4}\right\}\left(\text{Since, }{L}^{-1}\left\{Y\left(s\right)\right\}=y\left(t\right)\right)\\ y\left(t\right)=\text{H}\left(t\right)-\cos \left(2t\right)-5\cos \left(2t\right)-2\sin \left(2t\right)\end{array}$

Hence, $y\left(t\right)=\text{H}\left(t\right)-\cos \left(2t\right)-5\cos \left(2t\right)-2\sin \left(2t\right)$