Module 4

First Order Ordinary Differential Equations

Solve (

)

 Solution 

 We have and

  and

 Hence, the given equation is exact.

 Now,

 And 

 The Solution is 

2.     Solve

Solution:

We have and

and 

=

I.F.

Multiplying by we get, which is exact.

And

3.     Solve

Solution:

 We have and

 I.F

 Multiplying the given equation by, we get

 which is exact

 The solution is

4.     Solve

Solution:

 We have, 

 and

  And

 I.F

 Multiplying the given equation by I.F, we get

    which is exact

  The solution is

5.     Solve

 Solution:

 The equation can be written as

 The equation is of the above form and

 Dividing the equation by , we get

 Now, 

 And 

 The solution is

6.     Solve

Solution:

The equation is homogeneous and

Hence, is an integrating factor.

Dividing by  ,  we get

Now this is an exact differential equation.

(By partial fraction treating y constant. Let )

(Terms in N free from x)

The solution is

7.     Solve

Solution:

The equation is homogeneous and

Hence, is an integrating factor.

Dividing by we get

By partial fraction as in(A) of Ex.3 above)

(Terms in N free from x)

The solution is

8.     Solve

Solution:

Equation can be written as

  and 

Solution is

 To solve put

9.     Solve

Solution:

The equation can be written as , which is linear.

Now,

The solution is

The solution is 

10. Solve

Solution:

We have

This is linear of the form

I.F

The solution is

Put     

The solution is