Q8E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathit{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathit{L}\underset{0}{\overset{t}{\int }}(t-v)\mathit{y}\mathbf{\left(}\mathbf{v}\mathbf{\right)}\mathit{d}\mathit{v}\mathbf{=}{\mathit{t}}^{\mathbf{2}}$

Step-by-Step Solution

Answer

The solved equation is,

$\begin{array}{c}L\underset{0}{\overset{t}{\int }}(t-v)y\left(v\right)dv=t^2\\ Y\left(t\right)+L[t*y\left(t\right)]=t^2\end{array}$

$\begin{array}{c}y\left(s\right)=\frac{2s^2}{s^3(s^2+1)}\\ =\frac{2}{s(s^2+1)}\end{array}$

$y\left(t\right)=2-2\cos t$

Step 1: Definition of integral equation

In mathematics, an integral equation is an equation in which the unknown function to be found is contained within an integral sign.

Given,

Integral equation

$\begin{array}{c}L\underset{0}{\overset{t}{\int }}(t-v)y\left(v\right)dv=t^2\\ Y\left(t\right)+L[t*y\left(t\right)]=t^2\end{array}$

Taking Laplace transform on both sides

We get,

$\begin{array}{c}Y\left(s\right)+[y\left(s\right).\frac{1}{s^2}]=\frac{2}{s^3}\\ [1+\frac{1}{s^2}]y\left(s\right)=\frac{2}{s^3}\end{array}$

$\begin{array}{c}y\left(s\right)=\frac{2s^2}{s^3(s^2+1)}\\ =\frac{2}{s(s^2+1)}\end{array}$

Step 2: Definition of inverse

The inverse is a opposite in order, nature, effect an inverse relationship then a mathematical operation that is opposite in effect to another operation Multiplication is the inverse operation of division.

Use the partial fraction 

We get,

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

Equation first becomes

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

Applying inverse Laplace transform, we get,

$y\left(t\right)=2-2\cos t$

Hence,

$y\left(t\right)=2-2\cos t$