UNIT-3

Ordinary differential equations higher order

A homogeneous linear ordinary differential equation with constant coefficients is an ordinary differential equation in which coefficients are constants, all terms are linear, and the entire differential equation is equal to zero,

The form of second order linear differential equation with constant coefficients is,

Where a,b,c are the constants.

Let, aD²y+bDy+cy = f(x), where d² = , D =

∅(D)y = f(x) , where ∅(D)y = aD²y+bDy+cy

Here first we solve, ∅(D)y = 0, which is called complementary function(C.F)

Then we find particular integral (P.I)

P.I. = f(x)

General solution = C.F. +P.I.

Let’s do an examples to understand the concept,

Example1: Solve (4D² +4D -3)y =

Solution: Auxiliary equation is 4m² +4m – 3 = 0

We get, (2m+3)(2m – 1) = 0

m = ,

Complementary function: CF is A+ B

Now we will find particular integral,

P.I. = f(x)

= .

= .

= .

= . = .

General solution is y = CF + PI

= A+ B .

Differential operators

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

Example: Solve

Or,

Ans. Auxiliary equation are

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

Example : Solve

Ans. Auxiliary equation are

Rules to find Particular Integral

Case 1:

If,

If,

Example: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.

Example .

Given,

For CF,

Auxiliary equation are

For PI

Case 3:

Or,

Example :

Ans. Auxiliary equation are

Case 4:

Example: Solve

Ans. AE=

Complete solution is

Example: Solve

Ans. The AE is

Complete solution y= CF + PI

Example: Solve

Ans. The AE is

Complete solution = CF + PI

Example: Solve

Ans. The AE is

Complete solution is y= CF + PI

Example: Find the PI of

Ans.

Example: solve

Ans. Given equation in symbolic form is

Its Auxiliary equation is

Complete solution is y= CF + PI

Example: Solve

Ans. The AE is

We know,

Complete solution is y= CF + PI

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

Ans.

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

Ans. The auxiliary equation i9s

Hence complete solution is y= CF + PI

The equation involving the derivatives of two or more dependent variables, which are the functions of a single independent variable, is called simultaneous equation.

Such as-

Example 1: Solve the following simultaneous differential equations-

....(2)

Solution:

Consider the given equations

....(2)

Consider eq(1),(2)

Dx+2y = et....(1)

Dx +2x =e-t....(2)

Eliminating ‘x’ from both the equations we get,

12 2Dx + 2y = 2et

D 2Dx +D2y = e-t

y = Ae2t + Be-2t +

Example 2: Solve the following simultaneous differential equations-

Given that x(0)=1 and y(0)= 0

Solution:

Consider the given equations,

Dy +2x = sin2t

Dx -2y = cos2t

By solving the above equations we get,

(D2 +4)Y =0

X(0) = 1, y(0) = 0

A =0, B=-1

Example-3: Solve the following simultaneous differential equations-

It is given that x = 0 and y = 1 when t = 0.

Sol. Given equations can be written as-

Dx + 2y = - sin t ………. (1) and -2x + Dy = cos t ……… (2)

Eliminate x by multiplying (1) by 2 and (2) by D then add-

Here A.E =

So that C.F. =

And P.I. =

So that- …………. (3)

And ………….. (4)

Substitute (3) in (2), we get-

2x = Dy – cos t =

………… (5)

When t = 0, x = 0, y = 1. (3) and (5) gives-

Hence

Linear differential equation are 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

Let be the differential equation where P,Q,R are the functions of x.

If y = u(x) is the solution we know, then put y = uv in it.

It reduces the diff. Equation to one of first order in dv/dx which can be solved completely.

Note-

1 if 1+P+Q = 0, Then is a solution.

2. If 1-P+Q = 0, Then is a solution.

3. If P+Qx = 0, Then is a solution.

Example: Solve

It is given that y = x is a solution.

Sol.

Suppose y = xv so that-

And

Now put these values in the given equation, we get-

Or

Or

Where p = dv/dx

On integrating, we get-

Or

Again integrating,

Or

Hence the required solution-

Equation which can be solved by changing the independent variable-

Consider the equation-

Lets change the independent variable x to z and z = f(x)

Now put these values in the equation, we get-

........ (1)

Here

,

Equation-1 can be solved by taking

Example: Solve-

Sol.

Here P = cot x and Q =

Choosing z so that

Changing the independent variable x to z, we get-

......(1)

Where-

Equation(1) becomes-

Its sol. Is-

i.e.

Which is the required solution

Consider a second order LDE with constant co-efficients 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-

Where, are constant is called homogenous equation.

Put,

Example.

Ans. Putting,

AE is

CS = CF + PI

Example: Solve

Ans. Let,

AE is

y= CF + PI

Example: Solve

Ans. Let, so that z = log x

AE is

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-

LEGENDRE POLYNOMIALS

The Legendre’s equations is-

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-

BESSEL DIFFERENTIAL EQUATION

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