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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{8}\mathbf{cost}\mathbf{-}\mathbf{2}\mathbf{sint}\mathbf{;}\mathbf{y}\left(\frac{\mathbf{\pi }}{\mathbf{2}}\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\frac{\mathbf{\pi }}{\mathbf{2}}\right)\mathbf{=}\mathbf{0}$

• 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.
• $\mathbf{F}\left(\mathbf{s}\right)\mathbf{=}{\int }_{\mathbf{0}}^{\infty }\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{t}\mathbf{\text{'}}$
  

Since the initial conditions are given at $t=\frac{\pi }{2}$we need to shift them to $t=0$so let. $z\left(t\right)=y\left(t+\frac{\pi }{2}\right)$

Solve for the initial value problem as:

$$\begin{gathered} z\text{'}\text{'}\left(t\right)-z\text{'}\left(t\right)-2z\left(t\right)=-8\cos (t+\pi /2)-2\sin (t+\pi /2) \\ =8\sin t-2\cos t,z\left(0\right)=1,z\text{'}\left(0\right)=0 \end{gathered}$$

Applying the Laplace transform and using its linearity as follows:

$$\begin{gathered} \mathcal{L}\left\{z\text{'}\text{'}-z\text{'}-2z\right\}=\mathcal{L}\left\{8\sin t-2\cos t\right\} \\ \mathcal{L}\left\{z\text{'}\text{'}\right\}-\mathcal{L}\left\{z\text{'}\right\}-2\mathcal{L}\left\{z\right\}=\frac{8}{s^2+1}-\frac{2s}{s^2+1} \end{gathered}$$

Solve for the Laplace as:

$$\begin{gathered} \left[s^{2}Z-sz\left(0\right)-z\text{'}\left(0\right)\right]-\left[sZ\left(s\right)-z\left(0\right)\right]-2Z\left(s\right)=\frac{-2s+8}{s^2+1} \\ {s}^{2}Z\left(s\right)-s-sZ\left(s\right)+1-2Z\left(s\right)=\frac{-2s+8}{s^2+1} \\ {s}^{2}Z\left(s\right)-sZ\left(s\right)-2Z\left(s\right)=\frac{-2s+8}{s^2+1}+s-1 \\ \left(s^2-s-2\right)Z\left(s\right)=\frac{s^3-s^2-s+7}{s^2+1} \end{gathered}$$

Write the respective transfer function as:

$Z\left(s\right)=\frac{s^3-s^2-s+7}{\left(s^2+1\right)\left(s^2-s-2\right)}$

Using partial fractions solve as:

$$\begin{gathered} \frac{s^3-s^2-s+7}{\left(s^2+1\right)\left(s^2-s-2\right)}=\frac{s^3-s^2-s+7}{\left(s^2+1\right)(s-2)(s+1)}=\frac{As+B}{s^2+1}+\frac{C}{s-2}+\frac{D}{s+1} \\ {s}^3-s^2-s+7=\left(\begin{array}{c}\left(As+B\right)\left(s-2\right)\left(s+1\right)+C\left(s^2+1\right)\left(s+1\right) \\ +D\left(s^2+1\right)\left(s-2\right)\\ \end{array}\right) \end{gathered}$$

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

$$\begin{gathered} s=-1:6=-6D\Rightarrow D=-1 \\ s=2:9=15C\Rightarrow C=\frac{3}{5} \\ s=0:7=-2B+\frac{3}{5}+2\Rightarrow B=-\frac{11}{5} \\ s=1:6=-2A+2\frac{11}{5}+4\frac{3}{5}+2\Rightarrow A=\frac{7}{5} \end{gathered}$$

$$\begin{gathered} Therefore,thetransferfunctionis:Z\left(s\right)=\frac{\frac{7}{5}s-\frac{11}{5}}{s^2+1}+\frac{\frac{3}{5}}{s-2}-\frac{1}{s+1} \\ U\sin gtheinverseLaplacetransformobtainthesolutionofIVP1 \\ z\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{\frac{7}{5}s-\frac{11}{5}}{s^2+1}+\frac{\frac{3}{5}}{s-2}-\frac{1}{s+1}\right\}\left(t\right) \\ =\frac{7}{5}\mathcal{L}\left\{\frac{s}{s^2+1}\right\}-\frac{11}{5}\mathcal{L}\left\{\frac{1}{s^2+11}\right\}+\frac{3}{5}\mathcal{L}\left\{\frac{1}{s-2}\right\}-{\mathcal{L}}^{-1}\left\{\frac{1}{s+1}\right\} \\ =\frac{7}{5}\cos t-\frac{11}{5}\sin t+\frac{3}{5}{e}^{2t}-e^{-t} \end{gathered}$$

Since $y\left(t\right)=z\left(t-\frac{\pi }{2}\right)$obtain the solution of given IVP

$y\left(t\right)=\frac{7}{5}\sin t+\frac{11}{5}\cos t+\frac{3}{5}{e}^{2t-\pi }-e^{\frac{\pi }{2}-t}$

Consider the below trigonometric formulas as:

$$\begin{gathered} \cos \left(t-\frac{\pi }{2}\right)=\sin t \\ -\sin \left(t-\frac{\pi }{2}\right)=\cos t \end{gathered}$$

Therefore, the initial value for$y\text{'}\text{'}-y\text{'}-2y=-8\cos t-2\sin t$is$y\left(t\right)=\frac{7}{5}\sin t+\frac{11}{5}\cos t+\frac{3}{5}{e}^{2t-\pi }-e^{\frac{\pi }{2}-t}$