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

P.I. = f(x)

 General solution = C.F. +P.I.

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  .

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

Let be the CF of

Putting the value of in equation (1) we get

If are the roots the CF is

If are the roots then

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

Case 1:

If,

If,

Example:Solve

Ans. Given,

Auxiliary equation is

Case2:

Expand by the

Case 3:

Or,

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 solutio0n 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

1. The second order linear differential equation with constant coefficients is,

Where a,b,c are the constants.

2.     General solution = C.F. +P.I.

3.    

4.    

5.     Roots Real and Equal-

6.     Roots Real and Different-

7.     If roots are in complex form i.e.

Or

Or

The solution will be-

Example:

Solve

Auxiliary equation  

Complementary function

Complete Solution is  

Example:

Solve

Solution:

Auxiliary equation

C.F is

    []

   The Complete Solution is  

Example

Solve 

Solution:

The Auxiliary equation is

  The C.F is  

P.I

   The Complete Solution is 

Example

Solve

Solution:

The auxiliary equation is

The C.F is

   [Put ]

   The Complete Solution is

Example

Solve

Solution:

The auxiliary equation is     

The C.F is

But

   The Complete Solution is

Example

Solve

Solution:

The auxiliary equation is   

The C.F is

 [By parts]

   The Complete Solution is

Example

Solve

Solution:

The auxiliary equation is  

The C.F is

Now, 

And 

   The Complete Solution

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=

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-

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-

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 -

Cauchy’s homogeneous linear equation-

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-

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-

We know that-

On integrating by parts, we get-

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

Formula-1:

Fromula-2:

Formula-3:

Formula-4:

Formula-5:

Formula-6:

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-

Sol.

We know that

Equating the coefficients of both sides, we have-

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-

1. The Legendre’s equations is-
    
2. Rodrigue’s formula-

3.     Orthogonality of Legendre polynomials-