**Overview**

An exact differential equation is formed by differentiating its solution directly without any other process,

Is called an exact differential equation if it satisfies the following condition-

Here

is the differential co-efficient of M with respect to y keeping x constant and

is the differential co-efficient of N with respect to x keeping y constant.

**Step by step method to solve an exact differential equation**

1. Integrate M w.r.t. x keeping y constant.

2. Integrate with respect to y, those terms of N which do not contain x.

3. Add the above two results as below-

Provided

**Example-3: Determine whether the differential function ydx –xdy = 0 is exact or not.**

Solution. Here the equation is the form of M(x , y)dx + N(x , y)dy = 0

But, we will check for exactness,

These are not equal results, so we can say that the given diff. eq. is not exact.

**Example: Solve-**

Sol.

Here,

So that-

Thus the equation is exact and its solution is-

Which means-

Or

**Equation reducible to the exact form**

**1. If M dx + N dy = 0 be an homogenous equation in x and y, then 1/ (Mx + Ny) is an integrating factor (M dx + N dy **

**Example: Solve-**

**Sol.**

We can write the given equation as-

Here,

Multiply equation (1) by 1/x^4 we get-

This is an exact differential equation-

**2. I.F. for an equation of the type **

**IF the equation Mdx + Ndy = 0 be this form, then 1/(Mx – Ny) is an integrating factor.**

** Example: Solve-**

**Sol.**

**Here we have-**

**Now divide by xy, we get-**

**Which is an exact differential equation-**

**3. In the equation M dx + N dy = 0,**

**(i) If ****be a function of x only = f(x), then**** is an integrating factor.**

**(ii) If be a function of y only = F(x), then is an integrating factor.**

**Example: Solve-**

**Sol.**

**Here given,**

**M = 2y and N = 2x log x – xy**

**Then-**

**Here**

**Then**

**Now multiplying equation (1) by 1/x, we get-**

**4. For the following type of equation-**

**An I.F. is **

**Where-**

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