Q24E - 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.$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{3}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{t}\mathbf{-}\mathbf{-}\mathbf{3}\left(sin2t\right)\mathit{u}\mathbf{(}\mathit{t}\mathbf{-}\mathbf{2}\mathit{\pi }\mathbf{)}\mathbf{;}\mathit{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{2}$

Step-by-Step Solution

Answer

On solving the given initial value problem using the method of Laplace transforms, the solution is $y\left(t\right)=\cos t-\sin 2t-[2\sin t-\sin 2t]u(t-2\pi )$ 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 domainto a function of a complex variables.

 

 

Step 2: Taking Laplace Transform of initial value Problem

Given initial value problem

$y\text{'}\text{'}+y=3\sin 2t-3\left(\sin 2t\right)u(t-2\pi )$

Where $y\left(0\right)=1$ and $y\text{'}\left(0\right)=-2$

Laplace Transform for the initial value problem 

$$\begin{gathered} \mathcal{L}y\text{'}\text{'}\left(s\right)+\mathcal{L}y\left(s\right)=\mathcal{L}[3\sin 2t-3\sin 2tu(t-2\pi \left)\right] \\ {s}^2\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+\mathcal{L}y\left(s\right)=\frac{3·2}{s^2+4}-\frac{3·2e^{-2\pi s}}{s^2+4} \end{gathered}$$

$$\begin{gathered} s^2\mathcal{L}y\left(s\right)-s+2+\mathcal{L}y\left(s\right)=\frac{6}{s^2+4}-\frac{6e^{-2\pi s}}{s^2+4} \\ \left(s^2+1\right)\mathcal{L}y\left(s\right)-(s-2)=\frac{6}{s^2+4}-\frac{6e^{-2\pi s}}{s^2+4} \end{gathered}$$

$\mathcal{L}y\left(s\right)=\frac{s-2}{s^2+1}+\frac{6}{\left(s^2+4\right)\left(s^2+1\right)}-\frac{6e^{-2\pi s}}{\left(s^2+4\right)\left(s^2+1\right)}$

 

Step 3: By Partial function method

$\frac{1}{\left(s^2+4\right)\left(s^2+1\right)}=\frac{\frac{1}{3}}{s^2+1}-\frac{\frac{1}{3}}{s^2+4}$

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

Step 4: Laplace transform function

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

Hence,$y\left(t\right)=\cos t-\sin 2t-[2\sin t-\sin 2t]u(t-2\pi )$

and the graph is