Q26E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms. $$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathit{u}\mathbf{(}\mathit{t}\mathbf{-}\mathit{\pi }\mathbf{)}\mathbf{-}\mathit{u}\mathbf{(}\mathit{t}\mathbf{-}\mathbf{2}\mathit{\pi }\mathbf{)} \\ \mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

Step-by-Step Solution

Answer

The solution of the given initial value problem using the method of Laplace transforms is.

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

 

Step 1: Define Laplace Transform

The use of Laplace transformation is to convert differential equations into algebraic equations. The formula for Laplace transform is   

$\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{+}\mathbf{\infty }}\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{·}{\mathit{e}}^{\mathbf{-}\mathbf{s}\mathbf{·}\mathbf{t}}\mathbf{·}\mathit{d}\mathit{t}$

Where, F(s) = Laplace Transform

S  is complex number

t = real number >=0 

t’ = first derivative of the function f(t)

 

Step 2: Apply Laplace transform

Given initial value problem

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

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

Taking Laplace transform of initial value problem is

$\mathcal{L}y\text{'}\text{'}\left(s\right)+4\mathcal{L}y\text{'}\left(s\right)+4\mathcal{L}y\left(s\right)=\mathcal{L}\left[u\right(t-\pi )-u(t-2\pi \left)\right]$

$$\begin{gathered} s^2\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+2s\mathcal{L}y\left(s\right)-2y\left(0\right)+2\mathcal{L}y\left(s\right)=\frac{e^{-\pi s}}{s}-\frac{e^{-2\pi s}}{s} \\ {s}^2\mathcal{L}y\left(s\right)-0-0+4s\mathcal{L}y\left(s\right)-0+4\mathcal{L}y\left(s\right)=\frac{e^{-\pi s}}{s}-\frac{e^{-2\pi s}}{s} \end{gathered}$$

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

Using partial fraction

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

Equation first becomes as,

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

 

Taking inverse Laplace transform we get

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

Hence 

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