Unit – 2

Linear differential equations

2.1 Linear differential equations of second and higher order

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

2.2 Homogeneous equations with constant coefficients

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.

Solve:  Solve (4D² +4D -3)y = 

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  .

2.3 Euler Cauchy equations

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

2.4 Solution by undetermined coefficients

Look at the following case ,

, here p and q are coefficients.

We can find the solution of these type of equation by two types of solution-

(1) general solution of the homo. Equations:

(2)  particular solutions of the non- homo. Equations:

Solve: solve – y = 2x² - x – 3

Solution.: first we find general solution:

The characteristic function is:   r² - 1 = 0

( r-1)(r+1) = 0

R = 1, -1

General solution is -  A + B,

Now , let        

y = ax² + bx +c

  = 2ax + b

2a

Put these value in  – y = 2x² - x – 3,

2a – (ax² + bx +c) = 2x² - x – 3

2a – ax² - bx - c = 2x² - x – 3

Now compare coeff.

Coeff. Of x² ,    a = -2

Coeff. Of x ,      b = 1

Constant coeff.

2a – c = -3,       c =-1

So the particular solution will be,

y =  -2x² + x – 1

Complete solution is,

y = A + B- 2x² + x – 1

Solve : Solve

Here cf is  r² -6r +9 = 0

(r – 3)² = 0

r = 3

So the general solution is - A + B

Now we will find particular solution,

Lets, y =

=

Substitute these values,

+ 9 =

=

C = 1/5

The particular solution is ,

y = 1/5

Complete solution,

y = - A + B 1/5

2.5 Solution by 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

2.6 Modelling of electric circuits

Let us consider a resister R, an inductor L and a capacitor C which are connected in a series ,

Let ϵ(t) = ϵ’cosωt is an electromotive force to the circuit,

The equations for the voltage drops across a capacitor , a register and an inductor are

Where C is the capacitance, R is the resistance and L is the inductance , q is the charge and i is the current

We know that from Kirchhof’s law,

This is the second order non- homogeneous differential equation with constant coefficients.

Diagram for RLC circuit

Text Book:

1) Calculus: Gorakh Prasad

2) Advance Engineering Mathematics – E. Kreyszig, John Wiley & Sons Inc.