Module 2

Ordinary Differential Equations-2

2.1. Second order linear differential equations with variable coefficients

Linear differential equation is those in which the independent variable and its derivative occur only in the first degree and are not multiplied together.

Thus, the general linear differential equation of the n’th order is of the form

Where and X are function of x.

Linear differential equation with constant co-efficient are of the form-

Where are constants.

Rules to find the complementary function-

To solve the equation-

This can be written as in symbolic form-

Or-
 

It is called the auxiliary equation.

Let be the roots-

Case-1: If all the roots are real and distinct, then equation (2) becomes,

Now this equation will be satisfied by the solution of

This is a Leibnitz’s linear and I.F. =

Its solution is-

The complete solution will be-

Case-2: If two roots are equal

Then complete solution is given by-

Case-3: If one pair of roots be imaginary, i.e. then the complete solution is-

Where and

Case-4: If two points of imaginary roots be equal-

Then the complete solution is-

Example-Solve

Sol.

Its auxiliary equation is-

Where-

Therefore, the complete solution is-

Inverse operator-

is that function of x, not containing arbitrary constants which when operated upon gives X.

So that satisfies the equation f(D)y = X and is, therefore, its P.I.

f(D) and 1/f(D) are inverse operator.

Note-

1.

2.

Rules for finding the particular integral-

Let us consider the equation-

Or in symbolic form-

So that-

Now-

Case-1: When X =

In case f(a) = 0, then we see that the above rule will not work,

So that-

Example: Find the P.I. Of (D + 2)

Sol.

P.I. =

Now we will evaluate each term separately-

And

Therefore-

Example: Solve (D – D’ – 2) (D – D’ – 3) z =

Sol.

The C.F. Will be given by-

Particular integral-

Therefore, the complete solution is-

Case-2: when X = sin(ax + b) or cos (ax + b)

In case then the above rule fails.

Now-

And if

Similarly-

Example: Find the P.I. Of

Sol.

Example: Find the P.I. Of (D + 1) (D + D’ – 1) z = sin (x + 2y)

Sol.

Example: Find P.I. Of

Sol. P.I =

Replace D by D+1

Put

2.2. Method of variation of parameters

Method of variation of parameters-

Consider a second order LDE with constant co-efficient given by

Then let the complimentary function is given by

Then the particular integral is

Where u and v are unknown and to be calculated using the formula

u=

Example-1: Solve the following DE by using variation of parameters-

Sol. We can write the given equation in symbolic form as-

To find CF-

It’s A.E. Is

So that CF is-     

To find PI-

Here

Now

Thus PI = 

= 

= 

= 

= 

So that the complete solution is-

Example-2: Solve the following by using the method of variation of parameters.

Sol. This can be written as-

C.F.-

Auxiliary equation is-

So that the C.F. Will be-

P.I.-

Here

Now

Thus PI = 

= 

= 

So that the complete solution is-

Cauchy’s and Legendre’s DE

An equation of the form

Here X is the function of x, is called Cauchy’s homogeneous linear equation.

Example-1: Solve

Sol. As it is a Cauchy’s homogeneous linear equation.

Put

Then the equation becomes [D(D-1)-D+1] y = t   or

Auxiliary equation-

So that-

C.F.=

Hence the solution is- , we get-

Example-2: Solve

Sol. On putting so that,

and

The given equation becomes-

Or it can be written as-

So that the auxiliary equation is-

C.F. =

Particular integral-

Where

It’s a Leibnitz’s linear equation having I.F.=

Its solution will be-

P.I. =

=

So that the complete solution is-

An equation of the form-

Is called Legendre’s linear equation.

Example-3: Solve

Sol. As we see that this is a Legendre’s linear equation.

Now put

So that-

And

Then the equation becomes- D (D – 1) y+ Dy + y = 2 sin t

Its auxiliary equation is-

And particular integral-

P.I. =

Note -

Hence the solution is -

2.3. Power series solutions

We know that the solution of the differential equation-

Are

These are the power series solutions of the given differential equations.

Ordinary Point-

Let us consider the equation-

Here are polynomial in x.

X = a is an ordinary point of the above equation if does not vanish for x = a.

Note- If vanishes for x = a, then x= a is a singular point.

Solution of the differential equation when x = 0 is an ordinary point, which means does not vanish for x = 0.

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

2. Find

3.

4. Substitute the expressions of y, etc. in the given differential equations.

5. Calculate Coefficients of various powers of x by equating the coefficient to zero.

6. Put the values of In the differential equation to get the required series solution.

Example- Solve

Sol.

Here we have-

Let the solution of the given differential equation be-

Since x = 0 is the ordinary point of the given equation-

Put these values in the given differential equation-

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

Therefore, the solution is-

Example: Solve in series the equation-

Sol.

Here we have-

Let us suppose-

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

Then-

And

Put these values in equation (1)-

We get-

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

And so on….

In general, we can write-

Now putting n = 5,

Put n = 6-

Put n = 7,

Put n = 8,

Put n = 9,

Put n = 10,

Put the above values in equation (1), we get-

2.4. Legendre polynomials

The Legendre’s equations are-

Now the solution of the given equation is the series of descending powers of x is-

Here is an arbitrary constant.

If n is a positive integer and

The above solution is

So that-

Here is called the Legendre’s function of first kind.

Note- Legendre’s equations of second kind is and can be defined as-

The general solution of Legendre’s equation is-

Here A and B are arbitrary constants.

Rodrigue’s formula-

Rodrigue’s formula can be defined as-

Legendre Polynomials-

We know that by Rodrigue formula-

If n = 0, then it becomes-

If n = 1,

If n = 2,

Now putting n =3, 4, 5……..n we get-

…………………………………..

Where N = n/2 if n is even and N = 1/2 (n-1) if n is odd.

Example: Express in terms of Legendre polynomials.

Sol.

By equating the coefficients of like powers of x, we get-

Put these values in equation (1), we get-

Example: Let be the Legendre’s polynomial of degree n, then show that for every function f(x) for which the n’th derivative is continuous-

Sol.

We know that-

On integrating by parts, we get-

Now integrate (n – 2) times by parts, we get-

Recurrence formulae for -

Formula-1:

Fromula-2:

Formula-3:

Formula-4:

Formula-5:

Formula-6:

Generating function for

Prove that is the coefficient of in the expansion of in ascending powers of z.

Proof:

Now coefficient of in

Coefficient of in

Coefficient of in

And so on.

Coefficient of in the expansion of equation (1)-

The coefficients of etc. in (1) are

Therefore-

Example: Show that-

Sol.

We know that

Equating the coefficients of both sides, we have-

Orthogonality of Legendre polynomials

Proof: is a solution of

  …………………. (1)

And

is a solution of-

……………. (2)

Now multiply (1) by z and (2) by y and subtracting, we have-

Now integrate from -1 to +1, we get-

Example: Prove that-

By using Rodrigue formula for Legendre function.

On integrating by parts, we get-

Now integrating m – 2 times, we get-

2.5. Bessel functions for the first kind and their properties

The Bessel equation is-

The solution of this equations will be-

The Bessel function is denoted by and defined as-

If we put n = 0 then Bessel function becomes-

Now if n = 1, then-

The graph of these two equations will be-

General solution of Bessel equation-

Example: Prove that-

Sol.

As we know that-

Now put n = 1/2 in equation (1), then we get-

Hence proved.

Example: Prove that-

Sol.

Put n = -1/2 in equation (1) of the above question, we get-

Recurrence formulae –

Formula-1:

Proof:

As we know that-

On differentiating with respect to x, we obtain-

Putting r – 1 = s

Formula-2:

Proof:

We have-

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

Formula-3:

Proof:  We know that from formula first and second-

Now adding these two, we get-

Or

Formula-4:

Proof:

We know that-

On subtracting, we get-

Formula-5:

Proof:

We know that-

Multiply this by we get-

I.e.

Or

Formula-6:

Proof:

We know that-

Multiply by we get-

Or

Example: Show that-

By using recurrence relation.

Sol.

We know that-

The recurrence formula-

On differentiating, we get-

Now replace n by n -1 and n by n+1 in (1), we have-

Put the values of and from the above equations in (2), we get-

Example: Prove that-

Sol.

We know that- from recurrence formula

On integrating we get-

On taking n = 2 in (1), we get-

Again-

Put the value of from equation (2) and (3), we get-

By equation (1), when n = 1