91Ó°ÊÓ

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

Since the initial conditions are given at $t=\mathrm{Ï€}$we need to shift them to $t=0$so let

$z\left(t\right)=y(t+\mathrm{Ï€})$

Solve for the initial value problem.

$z\text{'}\text{'}\left(t\right)+z\left(t\right)=t+\mathrm{Ï€},z\left(0\right)=0,z\text{'}\left(0\right)=0$

Applying the Laplace transform and using its linearity as:

$$\begin{gathered} \mathcal{L}\left\{z\text{'}\text{'}+z\right\}=\mathcal{L}\left\{t+\mathrm{Ï€}\right\} \\ \mathcal{L}\left\{z\text{'}\text{'}\right\}+\mathcal{L}\left\{z\right\}=\frac{1}{s^2}+\frac{\mathrm{Ï€}}{s} \end{gathered}$$

Solve for the transfer function as:

$$\begin{gathered} \left[s^{2}Z\left(s\right)-sz\left(0\right)-z\text{'}\left(0\right)\right]+Z\left(s\right)=\frac{1+s\mathrm{Ï€}}{s^2} \\ {s}^{2}Z\left(s\right)+Z\left(s\right)=\frac{1+s\mathrm{Ï€}}{s^2} \\ \left(s^2+1\right)Z\left(s\right)=\frac{1+s\mathrm{Ï€}}{s^2} \\ Z\left(s\right)=\frac{1+s\mathrm{Ï€}}{s^2\left(s^2+1\right)} \end{gathered}$$

Using partial fractions as follows:

$\frac{1+s\mathrm{Ï€}}{s^2\left(s^2+1\right)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+1}$

Solve the partial fraction as:

$$\begin{gathered} 1+s\mathrm{Ï€}=As\left(s^2+1\right)+B\left(s^2+1\right)+\left(Cs+D\right)s^2 \\ =\left(A+C\right)s^3+\left(B+D\right)s^2+As+B \end{gathered}$$

This equation give us systems as:

$\left\{\begin{array}{l}\begin{array}{l}A+C=0\\ B+B=0\end{array}\\ A=\mathrm{π}\\ B=1\end{array}\right.⇒\left\{\begin{array}{ll}\begin{array}{ll}A=\mathrm{π}& \\ B=1& \end{array}& \\ C=-\mathrm{π}& \\ D=-1& \end{array}\right.$

Therefore, solve for the transfer function as:$Z\left(s\right)=\frac{\mathrm{Ï€}}{s}+\frac{1}{s^2}-\frac{\mathrm{Ï€}s+1}{s^2+1}$

Using the inverse Laplace transform solve the initial problem as:

$$\begin{gathered} z\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{\mathrm{Ï€}}{s}+\frac{1}{s^2}-\frac{\mathrm{Ï€}s+1}{s^2+1}\right\}\left(t\right) \\ =\mathrm{Ï€}\mathcal{L}\left\{\frac{1}{s}\right\}+\mathcal{L}\left\{\frac{1}{s^2}\right\}-\mathrm{Ï€}\mathcal{L}\left\{\frac{s}{s^2+1}\right\}-\mathcal{L}\left\{\frac{1}{s^2+1}\right\} \\ =\mathrm{Ï€}+t-\mathrm{Ï€}\cos t-\sin t \end{gathered}$$

Since, $y\left(t\right)=z\left(t-\mathrm{Ï€}\right)$obtain the solution of given IVP

$y\left(t\right)=t+\mathrm{Ï€}\cos t+\sin t$

Consider the below formulas used:

$$\begin{gathered} \cos (t-\mathrm{Ï€})=\cos (\mathrm{Ï€}-t)=-\cos t \\ \sin (t-\mathrm{Ï€})=-\sin (\mathrm{Ï€}-t)=-\sin t \end{gathered}$$

Therefore, the initial value for$y\text{'}\text{'}+y=t$is$y\left(t\right)=t+\mathrm{Ï€}\cos t+\sin t$