In Maths, an **integrating factor** is a function used to solve differential equations. It is a function in which an ordinary differential equation can be multiplied to make the function integrable. It is usually applied to solve ordinary differential equations. Also, we can use this factor within multivariable calculus. When multiplied by an integrating factor, an inaccurate differential is made into an accurate differential (which can be later integrated to give a scalar field). It has a major application in thermodynamics where the temperature becomes the integrating factor that makes entropy an exact differential.

## What are Differential Equations?

Differential equations play a vital role in Mathematics. These are the equations that necessarily involve derivatives. There are various types of differential equations; such as – **homogeneous and non-homogeneous, linear and nonlinear, ordinary and partial**. The differential equation may be of the first order, second order and ever more than that. The n^{th} order differential equation is an equation involving nth derivative. The most common differential equations that we often come across are first-order linear differential equations.

The ordinary linear differential equations are represented in the following general form:

**y’ + P y=Q**

**or**

**dy/dx + P(x) y = Q(x)**

Where y’ or dy/dx is the first derivative. Also, the functions P and Q are the functions of x only.

There are mainly two methods which are utilized in order to solve the linear first-order differential equations:

- Separable Method
- Integrating Factor Method

In this article, we are going to discuss what is integrating factor method, and how the integrating factors are used to solve the first and second-order differential equations.

## Integrating Factor Method

Integrating factor is defined as the function which is selected in order to solve the given differential equation. It is most commonly used in ordinary linear differential equations of the first order.

When the given differential equation is of the form;

dy/dx + P(x) y = Q(x)

then the integrating factor is defined as;

Where P(x) (the function of x) is a multiple of y and Î¼ denotes integrating factor.

## Solving First-Order Differential Equation Using Integrating Factor

Below are the steps to solve the first-order differential equation using the integrating factor.

- Compare the given equation with differential equation form and find the value of P(x).
- Calculate the integrating factor Î¼.
- Multiply the differential equation with integrating factor on both sides in such a way;Â Î¼ dy/dx + Î¼P(x)y = Î¼Q(x)
- In this way, on the left-hand side, we obtain a particular differential form. I.eÂ d/dx(Î¼ y) = Î¼Q(x)
- In the end, we shall integrate this expression and get the required solution to the given equation: Î¼ y = âˆ«Î¼Q(x)dx+C

## Solving Second Order Differential Equation Using Integrating Factor

The second-order differential equation can be solved using the integrating factor method.

Let the given differential equation be,

y” + P(x) y’ = Q(x)

The second-order equation of the above form can only be solved by using the integrating factor.

- Substitute y’ = u; so that the equation becomes similar to the first-order equation as shown: u’ + P(x) u = Q(x)
- Now, this equation can be solved by integrating factor technique as described in the section above for first-order equations and we reach the equation:Â Î¼ u=âˆ«Î¼Q(x)dx+C
- Find the value of u from this equation.

Since u = y’, hence to find the value of y, integrate the equation. In this way, we get the required solution.

### Integrating Factor Example

**Example: **

Solve the differential equation using the integrating factor: (dy/dx) – (3y/x+1) = (x+1)^{4}

**Solution:**

Given:Â (dy/dx) – (3y/x+1) = (x+1)^{4}

First, find the integrating factor:

Î¼ = e^{ âˆ« p(x) dx}

Î¼ = e^{ âˆ«(-3/x+1) dx}

âˆ«(-3/ x+1)dx = -3 ln (x+1) = ln (x+1)^{-3}

Hence, we get

Î¼ =eÂ ^{ln (x+1)-3}

Î¼ = 1/ (x+1)^{3}

Now, multiply the integrating factor on both the sides of the given differential eqaution:

[1/ (x+1)^{3}] [dy/dx] – [3y/( (x+1)

^{4})] = (x+1)

Integrate both the sides, we get:

[y/(x+1)^{3}] = [(1/2)x

^{2}+x+c]

Here, c is a constant

Therefore, the general solution of the given differential eqaution is

y = [(x+1)^{3}] [(1/2)x^{2}+x+c].

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