How to change Differential Equations into Integral Equations?

This post explains the basic method of converting an integral equation into a corresponding differential equation.

A differential equation can be easily converted into an integral equation just by integrating it once or twice or as many times, if needed. Let’s start with an example. Let $$\frac{dy}{dx} + 5y+1=0 \ldots (1)$$ be a simple first order differential equation. We can integrate it one time with respect to $x$ , to obtain $$\int \frac{dy}{dx} dx + 5 \int y dx +\int 1 \cdot dx=c$$

Or, $$y + 5 \int y dx +x =c \ldots (2)$$

If we arrange equation (2) in standard integral equation forms, as studied in very first part of this series, we get

$$y=(c-x)- 5 \int y dx$$
or, $$y(x)=(c-x)-5\int y(t) dt \ldots (3)$$

We can remove the arbitrary constant $c$ from the above integral equation by applying a boundary condition. For example, if we have $$y(0)=1$$ , then it can be easily seen that

$$y(0)=(c-0)-5\int y(0) dt$$
or, $$c=y(0)+5\int y(0) dt$$
$$ \Rightarrow c=1+5 \int 1 \cdot dt$$

$$\Rightarrow c=1+5 \int dt \ldots (4)$$
At this instance, we see that if the limits of the integration could have known, the value of $c$ should have been easier to interpret. Still we can convert the given differential equation into integral equation by substituting the value of $c$ in equation (3) above:

$$y(x)=(1-x+5 \int dt)-5\int y(t) dt $$

$$y(x)=(1-x)+5 \int (1-y(t)) dt \ldots (5)$$

Equation(5) is the resulting integral equation converted from equation (1). $\Box$

We see that there is one boundary condition required to obtain the single constant $c$ in First Order differential equation. In the same way, there are two boundary conditions needed in a second order differential equation.

Problems in second order differential equation with boundary conditions, are of two types.

Initial value problem

For some finite value of variable $x$, the value of function $y$ and its derivative $dy/dx$ is given in an initial value differential equation problem.
For example $$\frac{d^2y}{dx^2} +ky=tx $$ with $$y(0)=2$$ and $$y'(0)=5$$ is an initial value problem. Just try to see how, point $x=0$ is used for both $y$ and $y’$, which is called the initial value of the differential equation. This initial value changes into the lower limit when we try to derive the integral equation. And, also, the integral equation derived from an initial value problem is of Volterra type, i.e., having upper limit as variable $x$.

Boundary value problem

For different values of variable $x$, the value of function given in a boundary value condition.
For example $$ \frac{d^2y}{dx^2}+ly=mx$$ with $$y(a)=A$$ and $$y(b)=B$$ is a boundary value problem. Generally, we chose the lower limit of the integration as zero and integrate the differential equation within limit $(0,x)$. After the boundary values are substituted, we obtain a Fredholm integral equation, i.e., having upper limit as a constant $b$ (say).

All doubts may cleared by working out the following two examples

Converting initial value problem into a Volterra integral equation

Example 1 :  Convert the following differential equation into integral equation:
$$y”+y=0$$ when $$y(0)=y'(0)=0$$

Solution: Given
$$y”(x)+y(x)=0 \ldots (6)$$ with $$y(0)=0 \ldots (7)$$ and $$y'(0)=0 \ldots(8)$$
From (1), $$y”(x)=-y(x) \ldots (9)$$
Integrating (9) with respect to $x$ from $0$ to $x$.
$$\int_{0}^{x} y”(x) dx =-\int_{0}^{x} y(x) dx$$

$(y'(x))_0^x= -\int{0}^{x} y(x) dx$

$\Rightarrow y'(x)-y'(0) = -\int_{0}^{x} y(x) dx$

Since, $$y'(x)=0$$ ,

$$\Rightarrow y'(x)-0 =-\int_{0}^{x} y(x) dx$$

$$\Rightarrow y'(x) =-\int_{0}^{x} y(x) dx \ldots (10)$$
Integrating both sides of (10) with respect to $x$ from $0$ to $x$ –

$$\int_{0}^{x} y'(x) dx =-\int_{0}^{x} \left(\int_{0}^{x} y(x) dx\right) dx $$

$$\int_{0}^{x} y'(x) dx =-\int_{0}^{x} y(x) dx^2 $$

$$\Rightarrow (y(x))0^x=-\int{0}^{x} y(t) dt^2 $$

$$y(x)-y(0)=-\int_{0}^{x} (x-t) y(t) dt$$

$$\Rightarrow y(x)-0 =-\int_{0}^{x} (x-t) y(t) dt$$

$$\Rightarrow y(x)=-\int_{0}^{x} (x-t) y(t) dt \ldots (11)$$

This equation (11) is the resulting integral equation derived from the given second order differential equation. $\Box$



Converting boundary value problem into a Fredholm integral equation

Example 2: Reduce the following boundary value problem into an integral equation $$\frac{d^2y}{dx^2} +\lambda y =0$$ with $$y(0)=0$$ and $$y(l)=0$$
Solution: Given differential equation is $$y”(x)+\lambda y(x)=0 \ldots (12)$$ with $$y(0)=0 \ldots (13)$$ and $$y(l)=0 \ldots (14)$$
Since, $$(12) \Rightarrow y”(x) = -\lambda y(x) \ldots (15)$$
Integrating both sides of (15) w.r.t. $x$ from $0$ to $x$
$$\int_{0}^{x}y”(x) dx = -\lambda \int_{0}^{x}y(x) dx \ldots (16)$$

$$ {(y'(x))}0^x=-\lambda \int{0}^{x}y(x) dx$$

$$ y'(x) -y'(0)=-\lambda \int_{0}^{x}y(x) dx \ldots (17)$$

Let $y'(0)=\mathbf{ constant}=c$, then

$$ y'(x) -c=-\lambda \int_{0}^{x}y(x) dx $$

$$ y'(x)=c-\lambda \int_{0}^{x}y(x) dx \ldots (18)$$

Integrating (18) again with respect to $x$ from 0 to $x$

$$\int_{0}^{x} y'(x)dx=c\int_{0}^{x} dx-\lambda \int_{0}^{x} \left({ \int_{0}^{x}y(x) dx}\right) dx$$

or, $$ (y(x))0^x=cx-\lambda \int{0}^{x} y(x) dx^2$$

$$y(x)-y(0)=cx-\lambda \int_{0}^{x}y(t) dt^2$$

Putting $y(0)=0$

$$y(x)=cx-\lambda \int_{0}^{x} (x-t) y(t) dt \ldots (19)$$

Now, putting $x=l$ in (19):

$$y(l)=cl-\lambda \int_{0}^{l} (l-t) y(t) dt$$

$$0=cl-\lambda \int_{0}^{l} (l-t) y(t) dt$$

$$c=\frac{\lambda}{l} \int_{0}^{l} (l-t) y(t) dt \ldots (20)$$

Putting this value of $c$ in (19), (19) reduces to:

$$y(x) =\frac{\lambda x}{l} \int_{0}^{l} (l-t) y(t) dt-\lambda \int_0^x (x-t) y(t) dt \ldots (21) $$
On simplifying (21) we get

$$y(x) =\lambda (\int_{0}^{x} \frac{(l-x)t}{l} y(t) dt + \int_{x}^{l} \frac{x(l-t)}{l} y(t) dt) \ldots (22)$$

Which is the required integral equation derived from the given differential equation.

The solution can also be written as $$y(x) =\lambda \int_{0}^{l} K(x,t) y(t) dt$$

where $$K(x,t)=\frac{t(l-x)}{l} \qquad \mathbf{0<t<x}$$ and $$K(x,t)=\frac{x(l-t)}{l} \qquad \mathbf{x<t<l}$$


We can now define a strategy for changing the ordinary differential equations of second order into an integral equation.

Step 1: Write the differential equation and its boundary conditions.

Step 2: Now re-write the differential equation in its normal form, i.e., highest derivatives being on one side and other, all values on the other side. For example, $ y”=-\frac{\alpha}{2} xy’ +ny$ is the normal form of $ 2y”+\alpha xy’ -2ny=0$ .

Step 3: Integrate the normal form of the differential equation, from 0 to $x$. Use applicable rules and formulas to simplify it.

Step 4: If substitutable, substitute the values of the boundary conditions. In boundary value problems, take $y'(0)=c$ a constant.

Step 5: Again integrate, the, so obtained differential-integral equation, within the limits $(0,x)$ with respect to $x$.

Step 6: Substitute the values of given boundary conditions.

Step 7: Simplify using essential integration rules, change the variable inside the integration sign to $t$ . Use the  ‘multiple integral‘ rules to change multiple integral into linear integral, as we discussed in Part(1).

Why should I convert a differential equation into an integral equation?

DoTheOrdersStilStand asks:

Hi Gaurav,

This is all good, but it would help if you added some context on why you’d want to convert differential equations into integral equations.

Finding analytical or numerical solutions in the former case is often easier, also qualitative analysis of the asymptotic and singularity behavior in the phase space.

Possible Answer:

Returning to basics of differential equation, we know that the values of $y(x)$ which satisfy above differential equations are their solutions. Performing a conversion from differential equation in $y$ to integral equation in $y$ is nothing but solving the differential equation for $y$. After converting an initial value or boundary value problem into an integral equation, we can solve them by  shorter methods of integration. This conversion may also be treated as another representation formula for the solution of an ordinary differential equation.

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4 comments add your comment

  1. With both of this system students are able to do massive and complicated calculations
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  2. With both of this system students are able to do massive and complicated calculations
    of their head with out want of any modern day electronic devices .
    In case you are supposed to train your youngsters, you want to know
    what you are endeavor, means too.

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