Q17E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In problem 15-22, solve the given integral equation or integro differential equation for$\mathit{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

$\mathit{L}\underset{0}{\overset{t}{\int }}(t-v)\mathit{y}\mathbf{\left(}\mathbf{v}\mathbf{\right)}\mathit{d}\mathit{v}\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

The integral equation or integro-differential equation for y(t) is.$y\left(t\right)=\cos t$

Step 1: Definition of Laplace transform

A transformation of a function f(x) into the function is particularly useful for reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

Applying the Laplace transform we get

$\begin{array}{c}L\{y\left(t\right)+\underset{0}{\overset{t}{\int }}(t-v)y\left(v\right)dv\}\left(s\right)=\left\{1\right\}\\ L\left\{y\left(t\right)\right\}\left(s\right)+L\left\{\underset{0}{\overset{t}{\int }}(t-v)y\left(v\right)dv\right\}\left(s\right)=\frac{1}{s}\end{array}$

$\begin{array}{c}Y\left(s\right)=L\{t*y\left(t\right)\}\left(s\right)\\ =\frac{1}{s}\end{array}$

$\begin{array}{c}Y\left(s\right)=L\left\{t\right\}\left(s\right).L\left\{y\left(t\right)\right\}\\ =\frac{1}{s}\end{array}$

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

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

Step 2: Definition of inverse

Now, 

using inverse Laplace we get

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

Hence,

$y\left(t\right)=\cos t$