Q22E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

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

Step-by-Step Solution

Answer

On solving the given initial value problem using the method of Laplace transforms,the solution is$w\left(t\right)=\cos t+[1-\cos (t-2\left)\right]u(t-2)-[1-\cos (t-4\left)\right]u(t-4)$  and the corresponding graph is

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: Taking Laplace Transform of initial value Problem

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

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

 $$\begin{gathered} \mathcal{L}w\text{'}\text{'}\left(s\right)+\mathcal{L}w\left(s\right)=\mathcal{L}\left[u\right(t-2)-u(t-4\left)\right] \\ {s}^2\mathcal{L}w\left(s\right)-s\mathcal{L}w\left(0\right)-w\text{'}\left(0\right)+\mathcal{L}w\left(s\right)=\frac{e^{-2s}}{s}-\frac{e^{-48}}{s} \\ {s}^2\mathcal{L}w\left(s\right)-s-0+\mathcal{L}w\left(s\right)=\frac{e^{-2s}}{s}-\frac{e^{-48}}{s} \end{gathered}$$

$$\begin{gathered} \left(s^2+1\right)\mathcal{L}y\left(s\right)-s=\frac{e^{-28}}{s}-\frac{e^{-18}}{s} \\ \mathcal{L}w\left(s\right)=\frac{s}{s^2+1}+\frac{e^{-2s}}{s\left(s^2+1\right)}-\frac{e^{4s}}{s\left(s^2+1\right)} \end{gathered}$$

 

 

Step 3: By Partial fraction

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

Equation 1 becomes,

 

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

Step 4: Taking inverse Laplace transform

$$\begin{gathered} w\left(t\right)={\mathcal{L}}^{-1}\frac{s}{s^2+1}+{\mathcal{L}}^{-1}\frac{e^{-2s}}{s}-{\mathcal{L}}^{-1}\frac{s{e}^{-2s}}{s^2+1}-{\mathcal{L}}^{-1}\frac{e^{-4s}}{s}+{\mathcal{L}}^{-1}\frac{s{e}^{-4s}}{s^2+1} \\ =\cos t+u(t-2)-\cos (t-2)u(t-2)-u(t-4)+\cos (t-4)u(t-4) \\ =\cos t+[1-\cos (t-2\left)\right]u(t-2)-[1-\cos (t-4\left)\right]u(t-4) \end{gathered}$$

Hence, $w\left(t\right)=\cos t+[1-\cos (t-2\left)\right]u(t-2)-[1-\cos (t-4\left)\right]u(t-4)$

 , the graph is given below