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.

Example-1: Solve

Sol.

Here M = and N =

Then the equation is exact and its solution is-

Sol. We can write the equation as below-

Here M = and N =

So that-

The equation is exact and its solution will be-

Or

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

   But, we will check for exactness,

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

Example: Solve-

Sol.

We can write the given equation as-

Here,

M =

Multiply equation (1) by 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-

Multiply (1) by , 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-

Example: Solve-

Sol.

We can write the equation as below-

Now comparing with-

We get-

a = b = 1, m = n = 1, a’ = b’ = 2, m’ = 2, n’ = -1

I.F. =

Where-

On solving we get-

h = k = -3

Multiply the equation by , we get-

It is an exact equation.

So that the solution is-

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

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

3. 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.

(1)Convert the equation to the standard form  

(2) Find the integrating factor.

(3) Then the solution will be y (I.F) =

Sol. We can write the given equation as-

So that-

I.F. =

The solution of equation (1) will be-

Or

Or

Or

Example-2: Solve-

Sol.

We can write the equation as-

We see that it is a Leibnitz’s equation in x-

So that-

Therefore the solution of equation (1) will be-

Or

Example-3: Solve sin x )

Solution:  here we have,

sin x )

which is the linear form,

Now,

             Put tan   so that  sec² dx =  dt, we get

 Which is the required solution.

Divide both sides of the equation -
 

By, so that

Put so that

Then equation (1) becomes-

)

Example: Solve

Sol.

We can write the equation as-

On dividing by , we get-

Put so that

Equation (1) becomes,

Here,

Therefore the solution is-

Or

Now put

Integrate by parts-

Or
 

Example: Solve

Sol. here given,

Now let z = sec y, so that dz/dx = sec y tan y dy/dx

Then the equation becomes-

Here,

Then the solution will be-

Example: Solve-

Sol. here given-

We can re-write this as-

Which is a linear differential equation-

The solution will be-

Put

1. A differential equation of the form

Is called linear differential equation

2.     is called the integrating factor.

3.     y (I.F) =

4.     The equation  called Bernoulli’s equation.

Example- Solve-

Sol. Here we have-

or

On integrating, we get-

Example: Solve

Sol.

Here we have-

Now differentiate it with respect to x, we get-

Or

This is the Leibnitz’s linear equation in x and p, here

Then the solution of (2) is-

Or

Or

Put this value of x in (1), we get

Example: Solve-

Sol. Here we have-

On solving for x, it becomes-

Differentiating w.r.t. y, we get-

or

On solving it becomes

Which gives-

Or

On integrating

Thus eliminating from the given equation and (1), we get

Which is the required solution

y = px + f(p) ...... (2)

Differentiating (1) w.r.t. x, we get-

Put the value of p in (1) we get-

y = ax + f(a)

Which is the required solution.

Sol.

Put

So that-

Then the given equation becomes-

Or

Or

Which is the Clairaut’s form.

Its solution is-

i.e.

1. Clairaut’s equation- y = px + f(p)