Q16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Solve the given integral equation or integro-differential equation for y(t)$\mathit{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{3}\underset{0}{\overset{t}{\int }}{e}^{t-v}y\left(v\right)dv\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{t}$

Step-by-Step Solution

Answer

The integral equation or integro-differential equation for y(t) is. $\cos t+\sin t-1$

Step 1: Given the integral equation is,

$\begin{array}{r}y\left(t\right)+{\int }_0^t{e}^{t-v}y\left(v\right)dv=\sin t\\ y\left(t\right)+\left[e^t\star y\left(t\right)\right]=\sin t\end{array}$                               $[ U\sin g$      $g\star h={\int }_0^{t}g(t-v)h\left(v\right)dv$

Step 2: Taking Laplace transform on both sides, we get

$\begin{array}{c}y\left(s\right)+\left[y\left(s\right)\cdot \frac{1}{s-1}\right]=\frac{1}{s^2+1}\\ [1+\frac{1}{s-1}]y\left(s\right)=\frac{1}{s^2+1}\\ y\left(s\right)=\frac{s-1}{s(s^2+1)}\dots \dots \left(1\right)\end{array}$

Step 3: Using the partial function, we get

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

Hence, the first equation becomes

$\begin{array}{c}y\left(s\right)=\frac{s+1}{s^2+1}-\frac{1}{s}\\ =\frac{s}{s^2+1}+\frac{1}{s^2+1}-\frac{1}{s}\end{array}$

Step 3: Taking inverse Laplace transform, we get

$y\left(t\right)=\cos t+\sin t-1$