Module 3

Ordinary Differential Equation of higher orders

3.1 Second order Linear Differential equation with variable coefficient

The general form of linear differential equation of second order is

Where p and q are constants and R is a function of x or constant.

Differential Operator

D stands for operation of differential i.e.

stands for the operator of integration.

stands for operation of integration twice.

Thus,

Note:- Complete solution = complementary function + Particular integral

i.e. y=CF + PI

Method for finding the CF

Step1:- In finding the CF right hand side of the given equation is replaced by zero.

Step 2:- Let be the CF of

Putting the value of in equation (1) we get

It is called auxiliary equation.

Step 3:- Roots Real and Different

If are the roots the CF is

If are the roots then

Step 4- Roots Real and Equal

If both the roots are then CF is

If roots are

Example: Solve

Ans. Given,

Here Auxiliary equation is

Solve:

Or,

Ans. Auxiliary equation are

Note: If roots are in complex form i.e.

Solve:

Ans. Auxiliary equation are

Solve.

Ans. Its auxiliary equation is

Solution is

Solve.

Ans. The auxiliary equation is

Hence the solution is

Rules to find Particular Integral

Case 1:

If,

If,

Solve:

Ans. Given,

Auxiliary equation is

Case2:

Expand by the binomial theorem in ascending powers of D as far as the result of operation on is zero.

Solve.

Given,

For CF,

Auxiliary equation are

For PI

Case 3:

Or,

Solve:

Ans. Auxiliary equation are

Case 4:

Solve.

Ans. AE=

Complete solution is

Solve

Ans. The AE is

Complete solution y= CF + PI

Solve.

Ans. The AE is

Complete solution = CF + PI

Solve.

Ans. The AE is

Complete solutio0n is y= CF + PI

Find the PI of

Ans.

Solve

Ans. Given equation in symbolic form is

Its Auxiliary equation is

Complete solution is y= CF + PI

Solve.

Ans. The AE is

We know,

Complete solution is y= CF + PI

Solve. Find the PI  of  (D2-4D+3)y=ex cos2x

Ans.

Solve. (D3-7D-6) y=e2x (1+x)

Ans. The auxiliary equation i9s

Hence complete solution is y= CF + PI

3.2 Method of variation of Parameters

Working Rule

Step1: Find out the CF i.e.

Step 2: Particular integral

Step 3: Find u and v by formula

Solve.

Ans. AE is

Where,

General solution = CF + PI

Solve. (d2 y)/dx2+x2 y=sec nx

Ans. The AE is

The Wronskian of is

Solve.

Ans. Given,

AE is

CF is

Complete solution is

3.3 Cauchy Euler Equation

Where, are constant is called homogenous equation.

Put,

Solve.

Ans. Put,

AE is

Solve.  

Ans. Putting,

AE is

CS = CF + PI

Solve.

Ans. Let,

AE is

y= CF + PI

Solve.

Ans. Let, so that z = log x

AE is

3.4 Power Series solution

Working Rule

Step 1 Let be the solution of the given differential equation.

Step 2: Find etc.

Step 3: Substitute the expression of in the given differential equation.

Step 4: Calculate coefficient of various powers of x by equating coefficients to zero.

Step 5: Substitute the values of in the differential equation to get the required series solution.

Example. Solve in series the equation

Ans. 

Since x=0 is the ordinary point of the equation (1)

Then

Substituting in (1) we get

Equating to zero the coefficient of the various powers of x we obtain

Substituting these values in (2) we get

Solve.

Ans. Let,

Substituting the value of   in the given equation we get

Where the first summation extends over all values of K from 2 to

And the second from K =

Now equating the coefficient of equal to zero we have

For K =4

Solve.

Ans. Let

Substituting for in the given differential equation

Equating the coefficients of various powers of x to zero we get

3.5 Legendre’s Equation

Legendre’s equation is

And

Prove that

Ans. We know that

Put n=2

Prove that .

Ans. We know

+

Put x = 1 both sides we get

Equating the coefficient of on both sides we get

Prove that

Ans. We know

Differentiating with respect to z we get

Multiplying both sides by we get

Equating the coefficient of from both sides we get

Solve. Statement

Proof. Let is a solution of

is the solution of

Multiplying (1) by z and (2) by y and subtracting we get

Now integrative -1 to 1 we get

Now we have to prove that

We know that,

Squaring both sides we get

Integrating both sides between -1 to +1 we get

on both sides we get

here n = m

Prove that

Ans. The Recurrence formula is

Pn+1+nPn-1

Replacing n by (n+1) and (n-1) we have

Multiplying (1) and (2) and integrating in the limits -1 to 1 we get

(By orthogonality property)

3.6 Bessel function

The Bessel equation is

Bessel function of first kind

Bessel function of second kind

Recurrence Formula

1)     xJn'=nJn-xJn+1

2)     

3)     

4)     

5)     

6)     

Prove that (1)

Ans. We know

(b) Prove that

Ans. We know that

(3) Prove that

Ans. We know that

Jn(x)=

If n = 0

If n = 1

Note General solution of Bessel Equation

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.