# Solving Integral Equations (4) : Integral Equations into Differential Equations

## Introduction

In earlier parts we discussed about the basics of integral equations and how they can be derived from ordinary differential equations. In second part, we also solved a linear integral equation using *trial method*. Now we are in a situation from where main job of **solving Integral Equations** can be started. But before we go ahead to that mission, it will be better to learn how can integral equations be converted into differential equations.

## Integral Equation ⇔ Differential Equation

The method of converting an integral equation into a differential equation is exactly opposite to what we did in last part where we converted **boundary value differential equations into respective integral equations**. In last workout, initial value problems$ always ended up as **Volterra Integrals$ ** and boundary value problems$ resulted as **Fredholm Integrals. $ **In converse process we will get **initial value problems$ from Volterra Integrals** and **boundary value problems$ from Fredholm Integral Equations**. Also, as in earlier conversion *we continuously integrated the differentials* within given boundary values, we will *continuously differentiate* provided integral equations and refine the results by putting all constant integration limits.

The above instructions can be practically understood by following two examples. First problem involves the conversion of *Volterra Integral Equation into differential equation *and the second problem displays the conversion of *Fredholm Integral Equation into differential equation*.

### Problem 1: Converting Volterra Integral Equation into Ordinary Differential Equation with initial values

*Convert *$$y(x) = – \int_{0}^x (x-t) y(t) dt$$ *into initial value problem.*

*( See Problem 1 of Part 3 )*

#### Solution:

We have, $$y(x) = – \int_{0}^x (x-t) y(t) dt \ldots (1)$$

**Differentiating (1) **with respect to $x$ will give

$$y'(x) = -\frac{d}{dx} \int_{0}^x (x-t) y(t) dt$$

$$ \Rightarrow y'(x)=-\int_{0}^x y(t) dt \ldots (2)$$

Again **differentiating (2)** w.r.t. $x$ will give

$$ y”(x)=-\frac{d}{dx}\int_{0}^x y(t) dt$$

$$ \Rightarrow y”(x)=-y(x) \ldots (3′)$$

$$ \iff y”(x)+y(x)=0 \ldots (3) $$

Putting *the lower limit* $x=0$ (i.e., the initial value) in equation (1) and (2) will give, respectively the following:

$$y(0) = – \int_{0}^0 (0-t) y(t) dt$$

$$y(0)=0 \ldots (4)$$

And, $$ y'(0)=-\int_{0}^0 y(t) dt$$

$$y'(0)=0 \ldots (5)$$

These equations (3), (4) and (5) form the ordinary differential form of given integral equation. $\Box$

### Problem 2: Converting Fredholm Integral Equation into Ordinary Differential Equation with boundary values

*Convert *$$ y(x) =\lambda \int_{0}^{l} K(x,t) y(t) dt$$ *into boundary value problem where* $$ K(x,t)=\frac{t(l-x)}{l} \qquad \mathbf{0<t<x} $$ *and* $$ K(x,t)=\frac{x(l-t)}{l} \qquad \mathbf{x<t<l}$$

#### Solution:

The given integral equation is $$ y(x) =\lambda \int_{0}^{l} K(x,t) y(t) dt \ldots (1)$$ or $$y(x) =\lambda (\int_{0}^{x} \frac{(l-x)t}{l} y(t) dt + \int_{x}^{l} \frac{x(l-t)}{l} y(t) dt) \ldots (2)$$

**Differentiating (2) **with respect to $x$ will give $$ y'(x) = -\frac{\lambda}{l} \int_{0}^x t y(t) dt + \frac{\lambda}{l} \int_{x}^l (l-t) y(t) dt \ldots (3)$$

Continued differentiation of (3) will give $$ y”(x) = -\lambda y(x)$$ That’s $$ y”(x) +\lambda y(x) =0 \ldots (4)$$

To get the boundary values, we place $x$ equal to both integration limits in (1) or (2).

$x =0 \Rightarrow$ $$y(0)=0 \ldots (5)$$

$x=l \Rightarrow$ $$y(l)=0 \ldots (6)$$

The ODE (4) with boundary values (5) & (6) is the exact conversion of given integral equation. $\Box$

If y =a x^2+(1-a)x then find I =∫(πy^2) dx

integral from 0 to 1

find dI/dx

If y =a x^2+(1-a)x then find I =∫(πy^2) dx

integral from 0 to 1