## 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 workoutinitial 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.

Please note that this was the same integral equation we obtained after converting initial value problem: $$y”+y=0$$ when $$y(0)=y'(0)=0$$ ( 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:

Please see Example 2 of Part 3.

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. $\Box$

Feel free to ask questions, send feedback and even point out mistakes. Great conversations start with just a single word. How to write better comments?
1. jot says:

If y =a x^2+(1-a)x then find I =∫(πy^2) dx
integral from 0 to 1

1. jot says:

find dI/dx
If y =a x^2+(1-a)x then find I =∫(πy^2) dx
integral from 0 to 1

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