Q29E - 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.

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{;}\mathbf{ }\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{ }\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,} \\ \mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{ }\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}sint, 0⩽t⩽2\pi ,\\ 0, 2\pi <t\end{array} \end{gathered}$$

Step-by-Step Solution

Answer

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

$y\left(t\right)=\cos 2t+\frac{1}{3}[1-u(t-2\pi \left)\right]\sin t+\frac{1}{6}[8+u(t-2\pi \left)\right]\sin 2t$

Step 1: Define Laplace Transform

The use of Laplace transformation is to convert differential equation into algebric equations. the formula for laplace transformation is    

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

Where, F(s)= laplace transformation

S= Complex number

t= real number>=0

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

 

Step 2: Apply Laplace transform

Given initial value problem

$y\text{'}\text{'}+4y=g\left(t\right)$

Where.$y\left(0\right)=1 and y\text{'}\left(0\right)=3$ Also

$$\begin{gathered} g\left(t\right)=\sin t,0⩽t⩽2\pi \\ =0,2\pi <t \end{gathered}$$

Using rectangular and unit function we can write

$g\left(t\right)=\sin t-\sin tu(t-2\pi )$

Taking Laplace transform of initial value problem is

$\mathcal{L}y\text{'}\text{'}\left(s\right)+4\mathcal{L}y\left(s\right)=\mathcal{L}\left[g\right(l\left)\right]$

 $$\begin{gathered} s^2\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+4y\left(s\right)=\mathcal{L}[\sin t-\sin tu(t-2\pi \left)\right] \\ {s}^2\mathcal{L}y\left(s\right)-s-3+4\mathcal{L}y\left(s\right)=\frac{1}{s^2+1}-\frac{e^{-2\pi s}}{s^2+1} \\ \left(s^2+4\right)\mathcal{L}y\left(s\right)-(s+3)=\frac{1}{s^2+1}-\frac{c^{2\pi s}}{s^2+1} \end{gathered}$$

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

Using partial fraction

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

Equation first becomes as

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

 

Step 3: Take inverse Laplace transform we get

$$\begin{gathered} y\left(t\right)=\cos 2t+\frac{3}{2}\sin 2t+\frac{1}{3}\sin t-\frac{1}{6}\sin 2t-\frac{1}{3}\sin (t-2\pi )u(t-2\pi )+\frac{1}{6}\sin 2(t-2\pi )u(t-2\pi ) \\ =\cos 2t+\frac{3}{2}\sin 2t+\frac{1}{3}\sin t-\frac{1}{6}\sin 2t-\frac{1}{3}\sin tu(t-2\pi )+\frac{1}{6}\sin 2tu(t-2\pi ) \\ =\cos 2t+\frac{1}{3}\sin t+\frac{4}{3}\sin 2t-\frac{1}{3}\sin tu(t-2\pi )+\frac{1}{6}\sin 2tu(t-2\pi ) \\ =\cos 2t+\frac{1}{3}[1-u(t-2\pi \left)\right]\sin t+\frac{1}{6}[8+u(t-2\pi \left)\right]\sin 2t \end{gathered}$$

Hence,

$y\left(t\right)=\cos 2t+\frac{1}{3}[1-u(t-2\pi \left)\right]\sin t+\frac{1}{6}[8+u(t-2\pi \left)\right]\sin 2t$