In Problems $\mathbf{1}\mathbf{-}\mathbf{14}$, solve the given initial value problem using the method of Laplace transforms.

$\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

• 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 we get

$\mathcal{L}\left\{y\text{'}\text{'}-y\text{'}-2y\right\}=0$

$\mathcal{L}\left\{y\text{'}\text{'}\right\}-\mathcal{L}\left\{y\text{'}\right\}-2\mathcal{L}\left\{y\right\}=0$

Solve for the Laplace transform as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-\left[sY\right(s)-y(0\left)\right]-2Y\left(s\right)=0 \\ {s}^{2}Y\left(s\right)+2s-5-sY\left(s\right)-2-2Y\left(s\right)=0 \\ {s}^{2}Y\left(s\right)-sY\left(s\right)-2Y\left(s\right)=7-2s \\ \left(s^2-s-2\right)Y\left(s\right)=7-2s \\ Y\left(s\right)=\frac{7-2s}{s^2-s-2} \end{gathered}$$

Using partial fractions solve as:

$$\begin{gathered} \frac{7-2s}{s^2-s-2}=\frac{7-2s}{(s-2)(s+1)}=\frac{A}{s-2}+\frac{B}{s+1} \\ 7-2s=A(s+1)+B(s-2) \end{gathered}$$

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

$$\begin{gathered} s=-1:7+2=-3B\Rightarrow B=-3 \\ s=2:7-4=3A\Rightarrow A=1 \end{gathered}$$

Therefore

$Y\left(s\right)=\frac{1}{s-2}-\frac{3}{s+1}$

Using the inverse Laplace transform we obtain the solution of given differential equation

$$\begin{gathered} y\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{1}{s-2}-\frac{3}{s+1}\right\}\left(t\right) \\ =\mathcal{L}\left\{\frac{1}{s-2}\right\}-3\mathcal{L}\left\{\frac{1}{s+1}\right\} \\ =e^{2t}-3e^{-t} \end{gathered}$$

Therefore

$y\left(t\right)=e^{2t}-3e^{-t}$