In Problems $1-14$, solve the given initial value problem using the method of Laplace transforms

$\mathbf{}\mathbf{10}\mathbf{.}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{t}\mathbf{-}\mathbf{8}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}$

• The Laplace transform is a strong integral transform used in mathematics to convert a function from the time domain to the s-domain.
• In some circumstances, the Laplace transform can be utilized to solve linear differential equations with given beginning conditions.
• $\mathbf{F}\left(\mathbf{s}\right)\mathbf{=}{\int }_{\mathbf{0}}^{\infty }\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{t}\mathbf{\text{'}}$
  

Applying the Laplace transform and using its linearity as follows:

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}-4y\right\}=\mathcal{L}\left\{4t-8e^{-2t}\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}-4\mathcal{L}\left\{y\right\}=\frac{4}{s^2}-\frac{8}{s+2} \end{gathered}$$

Solve the Laplace transform as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-4Y\left(s\right)=\frac{-8s^2+4s+8}{s^2\left(s+2\right)} \\ {s}^{2}Y\left(s\right)-5-4Y\left(s\right)=\frac{-8s^2+4s+8}{s^2\left(s+2\right)} \\ {s}^{2}Y\left(s\right)-4Y\left(s\right)=\frac{-8s^2+4s+8}{s^2(s+2)}+5 \\ \left(s^2-4\right)Y\left(s\right)=\frac{5s^3+2s^2+4s+8}{s^2\left(s+2\right)} \end{gathered}$$

Solve further as:

$Y\left(s\right)=\frac{5s^3+2s^2+4s+8}{s^2{(s+2)}^2(s-2)}$

Using partial fractions solve as:

$\frac{5s^3+2s^2+4s+8}{s^2{(s+2)}^2(s-2)}=\frac{A}{s^2}+\frac{B}{s+2}+\frac{C}{{(s+2)}^2}+\frac{D}{s-2}$

Simplify the partial fractions as:

$$\begin{gathered} 5s^3+2s^2+4s+8=\left\{A{\left(s+2\right)}^2\left(s-2\right)+B{s}^2\left(s+2\right)\left(s-2\right) \\ +C{s}^2\left(s-2\right)+S-D{s}^2{\left(s+2\right)}^2\right\} \end{gathered}$$

Using s=0,2,1,-1 , respectively, gives

$$\begin{gathered} s=0:8=-8A\Rightarrow A=-1 \\ s=2:64=64D\Rightarrow D=1 \\ s=1:19=18-3B-C\Rightarrow 1=-3B-C \\ s=-1:1=4-3B-3C\Rightarrow 3=3B+3C \end{gathered}$$

Find B and C from the system.

$\left\{\begin{array}{l}\begin{array}{l}3B-C=1\\ 3B+3C=3\end{array}\\ \end{array}\right.\Rightarrow \left\{\begin{array}{l}\begin{array}{l}C=2\\ B=-1\end{array}\\ \end{array}\right.$

$$\begin{gathered} Therefore,Y\left(s\right)=-\frac{1}{s^2}-\frac{1}{s+2}+\frac{2}{{(s+2)}^2}+\frac{1}{s-2} \\ U\sin gtheinverseLaplacetransformweobtainthesolutionofgivendifferentialequation \\ y\left(t\right)={\mathcal{L}}^{-1}\left\{-\frac{1}{s^2}-\frac{1}{s+2}+\frac{2}{{(s+2)}^2}+\frac{1}{s-2}\right\}\left(t\right) \\ =-\mathcal{L}\left\{\frac{1}{s^2}\right\}-\mathcal{L}\left\{\frac{1}{s+2}\right\}+2\mathcal{L}\left\{\frac{1}{{(s+2)}^2}\right\}+{\mathcal{L}}^{-1}\left\{\frac{1}{s-2}\right\} \\ =-t-e^{-2t}+2t{e}^{-2t}+e^{2t} \\ Therefore,theinitialvaluefory\text{'}\text{'}-4y=4t-8e^{-2t}isy\left(t\right)=-t-e^{-2t}+2t{e}^{-2t}+e^{2t} \end{gathered}$$