Q21E - 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.

 $\mathbf{ }\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathit{u}\left(t--3\right)\mathbf{;}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}$

 

Step-by-Step Solution

Answer

On solving the given initial value problem using the method of Laplace transforms, the solution is $y\left(t\right)=\sin t+[1-\cos (t-3\left)\right]u(t-3)$ 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

Given initial value problem

$y\text{'}\text{'}+y=u(t-3)$ 

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

Laplace Transform for the initial value problem 

$$\begin{gathered} \mathcal{L}y\text{'}\text{'}\left(s\right)+\mathcal{L}y\left(s\right)=\mathcal{L}\left[u\left(t-3\right)\right] \\ {s}^2\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+\mathcal{L}y\left(s\right)=\frac{e^{-3s}}{s} \\ {s}^2\mathcal{L}y\left(s\right)-0-1+\mathcal{L}y\left(s\right)=\frac{e^{-3s}}{s} \end{gathered}$$

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

 

 

Step 3: By Partial fraction method

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

Equation first becomes,

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

 

Step 4: Taking inverse Laplace transform

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

Hence,

 $y\left(t\right)=\sin t+[1-\cos (t-3\left)\right]u(t-3)$

The graph is given below