Module 3

First order ordinary Differential Equation

3.1 Exact Differential Equation:-

An exact differential equation is formed by differentiating its primitive solution without any other process mdx + N dy =0 is said to be an exact differential equation if it is satisfies the following conditions

Method for solving exact differential equation

Step I :- Integrate m with respect to x keeping by constant

Step II :- Integrate with respect to y only those terms of N which do not contain x.

Step III :- Result of step I + Result of step II= constant

Solve.

Ans. Here m=

N=

Since,     the given equation is exact.

Now, (Term of N is not containing x) d y=c

3.2 Linear differential equation

A differential equation of the form is called linear differential equation where p and q are function of x(but not of y) or constant

Note:-   is called integrating factor

Working rule

Step 1:- Convert the given equation to the standard form of linear differential equation i.e.

Step 2:- Find the integrating factor

Step 3:- then the solution is

Example:- Solve (x+1)

Ans.

IF=

Then solution is

3.3 Bernoulli equation

The equation of the form where p and q are constant or function of x can be reduced to the linear form on dividing by and substituting

Example. Solve

Ans. Given

Dividing by xy we get

Equation (1) becomes

Where,

Its solution is

3.4 Euler's equation

Differential equation of the form. around is called Eulor’s equation.

3.5 Equation not of first degree

The differential equation will involve in higher degree and will be denoted by P. The differential equation will be of the form f(x,y,P)=0

Equation solvable for P

Solve:

Ans.

On integrating we get

Equation solvable for y

Working rule:-

(i)                 Differentiating the given equation with respect to x

(ii)               Eliminate P from the given equation and equation obtained as above

(iii)            The eliminant is the required solution

Solve.

Ans. Given,

Differentiating 1 with respect to x we get

On integrating we get P=c

Putting the value of P in equation 1 we get

Equation solvable for x:-

Working rule

Step 1 Differentiate the given equation with respect to y

Step 2 solve the equation obtained as in (1) for P

Step 3 Eliminate P by putting the value of P in the given equation.

Step 4 the eliminant is the required solution

Solve.

Ans.

Differentiating equation (2) with respect to y we get

On integrating we get

Putting the value of P in equation 1 we get

3.6 Clairaut's equation

The equation y=Px+f(P) is known as clairaut’s equation.

Put P=a (constant)

Then y=ax+f(a) is the required solution

Working rule

In the clairaut’s equation on replacing P by a constant we get the solution of the equation.

Example. Solve

Ans. Given,

Which is clairaut’s equation

Hence the solution is

TEXTBOOKS/REFERENCES:

1. ERWIN KREYSZIG, ADVANCED ENGINEERING MATHEMATICS, 9TH EDITION, JOHN WILEY & SONS, 2006.
2. W. E. BOYCE AND R. C. DIPRIMA, ELEMENTARY DIFFERENTIAL EQUATIONS AND BOUNDARY VALUE PROBLEMS, 9TH EDITION, WILEY INDIA, 2009.
3. S. L. ROSS, DIFFERENTIAL EQUATIONS, 3RD ED., WILEY INDIA, 1984.
4. E. A. CODDINGTON, AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS, PRENTICE HALL INDIA, 1995.
5. E. L. INCE, ORDINARY DIFFERENTIAL EQUATIONS, DOVER PUBLICATIONS, 1958.
6. G.F. SIMMONS AND S.G. KRANTZ, DIFFERENTIAL EQUATIONS, TATA MCGRAW HILL, 2007.