Q25E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

solve the given initial value problem using the method of Laplace transforms.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{u}\mathbf{(}\mathit{t}\mathbf{-}\mathbf{2}\mathbf{\pi }\mathbf{)}\mathbf{-}\mathbf{-}\mathit{u}\mathbf{(}\mathit{t}\mathbf{-}\mathbf{4}\mathbf{\pi }\mathbf{)}\mathbf{;}\mathit{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

On solving the given initial value problem using the method of Laplace transforms,the solution is

$y\left(t\right)=e^{-t}\cos t+2e^{-t}\sin t+\frac{1}{2}\left[1-e^{-t+2\pi }[\cos t+\sin t]\right]u(t-2\pi )-\frac{1}{2}\left[1-e^{-t+4\pi }[\cos t+\sin t]\right]u(t-4\pi )$

Step 1: Definition

 The Laplace transform, is an integral transform that converts a function of a real variable usually t, the time domain to a function of a complex variable s.

  

Step 2: Laplace transform function

Given initial value problem,

$y\text{'}\text{'}+2y\text{'}+2y=u(t-2\pi )-u(t-4\pi )$

Where $y\left(0\right)=1$ and $y\text{'}\left(0\right)=1$

Taking Laplace transform for the initial value problem

$$\begin{gathered} \mathcal{L}y\text{'}\text{'}\left(s\right)+2\mathcal{L}y\text{'}\left(s\right)+2\mathcal{L}y\left(s\right)=\mathcal{L}\left[u\right(t-2\pi )-u(t-4\pi \left)\right] \\ {s}^2\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+2s\mathcal{L}y\left(s\right)-2y\left(0\right)+2\mathcal{L}y\left(s\right)=\frac{e^{-2\pi s}}{s}-\frac{e^{-4\pi s}}{s} \end{gathered}$$

$$\begin{gathered} s^2\mathcal{L}y\left(s\right)-s-1+2s\mathcal{L}y\left(s\right)-2+2\mathcal{L}y\left(s\right)=\frac{e^{-2\pi s}}{s}-\frac{e^{-4\pi s}}{s} \\ \left(s^2+2s+2\right)\mathcal{L}y\left(s\right)-(s+3)=\frac{e^{-2\pi s}}{s}-\frac{e^{-4\pi s}}{s} \end{gathered}$$

$$\begin{gathered} \mathcal{L}y\left(s\right)=\frac{s+3}{\left(s^2+2s+2\right)}+\frac{e^{-2\pi s}}{s\left(s^2+2s+2\right)}-\frac{e^{-4\pi s}}{s\left(s^2+2s+2\right)} \\ =\frac{s+1+2}{{(s+1)}^2+1}+\frac{e^{-2\pi s}}{s\left(s^2+2s+2\right)}-\frac{e^{-4\pi s}}{s\left(s^2+2s+2\right)} \end{gathered}$$

 

 

Step 3: By partial function method

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

$\mathcal{L}y\left(s\right)=\frac{s+1}{{(s+1)}^2+1}+\frac{2}{{(s+1)}^2+1}+\frac{\frac{e^{-2\pi s}}{2}}{s}-\frac{\frac{e^{-2\pi s}}{2}s+e^{-2\pi s}}{s^2+2s+2}-\frac{\frac{e^{-4\pi s}}{2}}{s}+\frac{\frac{e^{-4\pi s}}{2}s+e^{-4\pi s}}{s^2+2s+2}$

Step 4: Taking inverse Laplace transform

$$\begin{gathered} y\left(t\right)=e^{-t}\cos t+2e^{-t}\sin t+\frac{1}{2}u(t-2\pi )-\frac{1}{2}{e}^{-t+2\pi }\cos (t-2\pi )u(t-2\pi ) \\ +\frac{1}{2}{e}^{-t+2\pi }\sin (t-2\pi )u(t-2\pi )-e^{-t+2\pi }\sin (t-2\pi )u(t-2\pi ) \\ -\frac{1}{2}u(t-4\pi )+\frac{1}{2}{e}^{-t+4\pi }\cos (t-4\pi )u(t-4\pi ) \\ -\frac{1}{2}{e}^{-t+4\pi }\sin (t-4\pi )u(t-4\pi )+e^{-t+4\pi }\sin (t-4\pi )u(t-4\pi ) \end{gathered}$$

$$\begin{gathered} y\left(t\right)=e^{-t}\cos t+2e^{-t}\sin t+\frac{1}{2}u(t-2\pi )-\frac{1}{2}{e}^{-t+2\pi }\cos tu(t-2\pi )-\frac{1}{2}{e}^{-t+2\pi }\sin tu(t-2\pi ) \\ -\frac{1}{2}u(t-4\pi )+\frac{1}{2}{e}^{-t+4\pi }\cos tu(t-4\pi )+\frac{1}{2}{e}^{-t+4\pi }\sin tu(t-4\pi ) \\ =e^{-t}\cos t+2e^{-t}\sin t+\frac{1}{2}\left[1-e^{-t+2\pi }[\cos t+\sin t]\right]u(t-2\pi )-\frac{1}{2}\left[1-e^{-t+4\pi }[\cos t+\sin t]\right]u(t-4\pi ) \end{gathered}$$

hence

$y\left(t\right)=e^{-t}\cos t+2e^{-t}\sin t+\frac{1}{2}\left[1-e^{-t+2\pi }[\cos t+\sin t]\right]u(t-2\pi )-\frac{1}{2}\left[1-e^{-t+4\pi }[\cos t+\sin t]\right]u(t-4\pi )$