Question

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

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

Short 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-by-step solution

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,

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

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

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,

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