An equation which involves unknown functions and their derivatives with respect to one or more independent variables is called a differential equation. A solution of a differential equation is a function (explicit or implicit) by means of which and the derivatives obtained therefrom, the equation is satisfied.

Differential equations in which the variables can be separated are called variable separable.

Such differential equations can be solved easily by first reducing them to variable separable form and then integrating them. However for certain forms of differential equations which are not variable separable we have to use integrating factors. In this article we will be discussing the Integrating Factor and its use for the solution of first and second degree differential equations.

Integrating Factor

There are many types of differential equations. Among them, one is Linear Differential Equation.

A differential equation of the form \( \frac{dy}{dx}+Py=Q \), where P and Q are constants or functions of x only, is called a linear differential equation of the first order in y. We solve this type of equation through the Integrating factor(also called I.F.).

In other words, when we are given a non exact differential equation, we make it exact by multiplying the given differential equation a function of x or y or both, such a function is called an Integrating factor to solving linear differential equation. If \( \frac{dy}{dx}+Py=Q \), where P and Q are functions of x only, then it has \( e^{\int_{ }^{Pdx}} \) as an integrating factor and its solution is given by \( y\left(I.F.\right)=\int_{ }^{ }Q\left(I.F.\right)dx+C \).

For example, if we have \( x\frac{dy}{dy}+y=3x^2-2 \), then after dividing the whole equation by x, we can write this equation as, \( \frac{dy}{dy}+\frac{y}{x}=3x-\frac{2}{x} \)

Comparing it with the standard form \( \frac{dy}{dy}+Py=Q \), we get

\( P=\frac{1}{x},\ Q=3x-\frac{2}{x} \),

Now we find the Integrating factor, which is

\( I.F.=e^{\int_{ }^{Pdx}}=e^{\int_{ }^{\frac{1}{x}dx}}=e^{\log x}=x \)

Therefore according to the solution formula, we can write,

\( y\ .\ x=\int_{ }^{ }\left(3x-\frac{2}{x}\right)\ .\ x\ dx=\int_{ }^{ }\left(3x^2-2x\right)dx=3\ .\ \frac{x^2}{3}-2x+C=x^3-2x+C \)

Thus the required differential equation is \( y=x^2-2+\frac{C}{x},\ \) where C is an arbitrary constant.

Integrating Factor Formula

The formula can be written for two conditions as mentioned below.

• If \( \frac{dy}{dx}+Py=Q \), where P and Q are functions of x only, then it has \( e^{\int_{ }^{Pdx}} \) as an integrating factor and its solution is given by \( y\left(I.F.\right)=\int_{ }^{ }Q\left(I.F.\right)dx+C \).
• Similarly, if \( \frac{dx}{dy}+Px=Q \), where P and Q are functions of y only, then it has \( e^{\int_{ }^{Pdy}} \) as an I.F., and the solution is given by \( x\left(I.F.\right)=\int_{ }^{ }Q\left(I.F.\right)dy+C \).

Integration Factor Method (Cover Steps)

The procedure to solve the linear differential equation is given below.

Case 1:

• Write the given differential equation in the form \( \frac{dy}{dx}+Py=Q \), where P and Q are constants or functions of x only.
• Then find the Integrating factor(I.F.), \( e^{\int_{ }^{Pdx}} \).
• Write the solution of differential equation as,

\( y\left(I.F.\right)=\int_{ }^{ }Q\left(I.F.\right)dx+C \)

Case 2:

• Write the given differential equation in the form \( \frac{dx}{dy}+Px=Q \), where P and Q are functions of y only.
• Then find the Integrating factor(I.F.), \( e^{\int_{ }^{Pdy}} \).
• Write the solution of differential equation as,

\( x\left(I.F.\right)=\int_{ }^{ }Q\left(I.F.\right)dy+C \).

Differential Equation using Integrating Factor (Cover What are Differential equation)

In engineering, physics, chemistry and sometimes in subjects like economics, biology etc., It becomes necessary to build a mathematical model to represent certain problems. It is often the case that these mathematical models involve the search of the unknown functions that satisfy the equation which contains the derivatives of unknown functions which follow derivative rule. Such functions are called differential equations.

The order of the differential gives us different methods of finding its solution using Integrating factor.

Solving First order differential equation

We use the above given formula to find the solution for the first order differential equation.

For example, \( 2x\frac{dy}{dx}+y=6x^3 \)

\( \Rightarrow\frac{dy}{dx}+\frac{y}{2x}=3x^2 \), which is a linear differential equation of first order in y.

Comparing it with the standard form \( \frac{dy}{dy}+Py=Q \), we get

\( P=\frac{1}{2x},\ Q=3x^2 \),

First we find the I.F., \( e^{\int_{Pdx}^{ }}=e^{\int_{\frac{1}{2x}dx}^{ }}=e^{\frac{1}{2}\log x}=e^{\log\sqrt{x}=\sqrt{x}} \)

Now we put these in the final equation for solution,

Therefore, \( y\left(I.F.\right)=\int_{ }^{ }Q\left(I.F.\right)dx+C=3x^2\ .\ \sqrt{x}dx+C=3\int_{ }^{ }x^{\frac{5}{2}}dx+C=3\ .\ \frac{x^{\frac{7}{2}}}{\frac{7}{2}}+C=\frac{6}{7}x^{\frac{7}{2}}+C \)

Solving second order differential equation

Let us consider that we are given, \( \frac{d^2y}{dx^2}+P\frac{dy}{dx}=Q \).

So first we reduce the second order differential equation to the first order differential equation.

Let us consider \( \frac{dy}{dx}=w\Rightarrow\frac{d^2y}{dx^2}=\frac{dw}{dx} \).

Now the reduced equation becomes, \( \frac{dw}{dx}+Pw=Q \).

We can solve this by using the method of Integrating factor. We get \( I.F.=e^{\int_{ }^{Pdx}} \). We will hence get, \( w\left(I.F.\right)=\int_{ }^{ }Q\left(I.F.\right)dx+C \).

Thus we will form this equation and then equate it with \( \frac{dy}{dx} \) to calculate the value of y and integrate again to get the final result.

Properties of Integrating Factor (I.F.)

• Helps Make the Equation Solvable:
  • The main purpose of an integrating factor is to convert a non-exact differential equation into an exact one or make it easier to solve.
• ​Used in Linear Differential Equations:
  • Integrating factors are most commonly used in solving first-order linear differential equations of the form:

\(dy/dx + P(x)·y = Q(x)\)

The Integrating Factor (I.F.) is:
I.F. = e^(∫P(x) dx)

• Depends on the Form of the Equation:
  • The integrating factor changes depending on whether the equation is in terms of x or y. For example:
  • If the equation is linear in y, the I.F. is usually a function of x.
  • If the equation is linear in x, the I.F. is usually a function of y.
• Makes the Left-Hand Side a Product Derivative:
  • After multiplying the differential equation by the integrating factor, the left-hand side becomes the derivative of a product like:
  • d/dx [y · I.F.]
  • This helps simplify the integration process.
• Uniquely Defined Up to a Constant Multiple:
  • The integrating factor is not unique; multiplying it by a non-zero constant still gives a valid solution.

Important Points About Integrating Factor

1. What is an Integrating Factor (I.F.):
An integrating factor is a special expression (usually a function of x or y) that we multiply with both sides of a differential equation to make the left-hand side look like the derivative of a product of two functions (often involving y and x). This helps us to solve the equation easily.

2. Changing Variables to Make the Equation Linear:
In some cases, a differential equation is not linear in the given form. But if we switch the roles of x and y (i.e., take y as the independent variable and x as the dependent one), the equation may turn into a linear form.
For example, an equation like:

y dx – (x + 2y²) dy = 0

can be rearranged as:

dx/dy – (1/y) x = 2y

Integrating Factor Examples

Problem 1:
Solve the following differential equation.
\( ydx-\left(x+2y^2\right)dy=0 \)

Solution:

We can write, \( ydx-\left(x+2y^2\right)dy=0\Rightarrow y\frac{dy}{dx}=x+2y^2\Rightarrow\ \frac{dx}{dy}-\frac{1}{y}\ .\ x=2y \).

It is of the form \( \frac{dx}{dy}+Px=Q \), where \( P=-\frac{1}{y}\ and\ Q=2y \) which are functions of y only, so it is a linear differential equation in x.

Therefore, \( I.F.=e^{\int_{ }^{Pdy}}=e^{\int_{ }^{-\frac{1}{y}dy}}=e^{-\log y}=e^{\log y^{-1}}=y^{-1}=\frac{1}{y} \).

Now according to the formula of case 2,

\( x\left(I.F.\right)=\int_{ }^{ }Q\left(I.F.\right)dy+C\Rightarrow\frac{x}{y}=\int_{ }^{ }\frac{2y}{y}dy+C=\int_{ }^{ }2dy+C=2y+C \)

Thus the required solution is, \( x=2y^2+Cy \). where C is an arbitrary constant.

Problem 2:Solve the following differential equation.\( \cos x\frac{dy}{dx}+y=\sin x \)

Solution:

Given, \( \cos x\frac{dy}{dx}+y=\sin x [\latex] \(\) \Rightarrow\frac{\cos x}{\cos x}\frac{dy}{dx}+\ \frac{y}{\cos x}=\frac{\sin x}{\cos x}\Rightarrow\sec x\frac{dy}{dx}+\ \sec x\ .\ y=\tan x \)

It is a linear differential equation in y.

\( I.F.=e^{\int_{ }^{Pdx}}=e^{\int_{ }^{\sec x\ dx}}=e^{\log\left(\sec x+\tan x\right)}=\sec x+\tan x \).

Therefore according to the formula of case 1,

\( y\left(I.F.\right)=\int_{ }^{ }Q\left(I.F.\right)dx+C\Rightarrow y\ .\ \left(\sec x+\tan x\right)=\int_{ }^{ }\tan x\left(\sec x+\tan x\right)dx+C \)

\( =\int_{ }^{ }\left(\sec x\tan x+\tan^2x\right)dx+C \)

\( =\int_{ }^{ }\left(\sec x\tan x+\sec^2x-1\right)dx+C \)

\( =\sec x+\tan x-x+C \)

Thus the required equation is, \( \left(y-1\right)\left(\sec x+\tan x\right)+x=C \). Where C is an arbitrary constant.