Unit - 1

Higher order linear differential equations and applications

1.1 Basic definition, differential operator, complimentary functions, particular integral, Shortcut methods for standard functions like , sin(ax+ b), cos(ax+ b),,, V and xV

Linear differential equations 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

Polynomials in x,

Let us consider the equation-

Or in symbolic form-

So that-

Now-

Case-1: When , V is the function of x.

Case-2: When X is any other function of x.

Example: Find P.I. Of

Sol.

Put
 

Working method to find the complete solution of an equation-

Example: Solve

Sol.

Here first we will find the C.F.-

Its auxiliary equation will be-

Here we get-

Now we will find P.I.-

Now the complete solution is-

Example: Solve-

Sol.

The given equation can be written as-

Its auxiliary equation is-

We get-

So that the C.F. Will be-

Now we will find P.I.-

Therefore the complete solution is-

Method of variation of parameters-

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-

Key takeaways-

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

Where are constants.

2.    

3.    

4.    

1.2 Particular integral by general method (without method of variation of parameters) for other functions

A general method of finding the PI of any function

Or

Or

It is the LDE.

The solution will be-

Example:

Solve

Solution:

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

P.I

Now,  

   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

1.3 Homogeneous Linear Differential equations

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

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 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

Key takeaways-

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.

1.4 Legendre’s Linear equation

Where, are constant is called Cauchy’s homogenous linear 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

Legendre’s differential equation-

A differential equation of the form-

Where k is the constants and X is a function of x, is known as Legendre’s linear differential equation.

We can reduce such type of equations to linear equations with constant coefficient by the substitution as-

i.e.

Example: Solve

Sol. As we see that this is 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 -

Key takeaways-

1. Cauchy’s linear equation-

2.     Legendre’s linear differential equation-

1.5 Civil Engineering Applications

Introduction

In “real-world,” there are many physical quantities that can be represented by functions. Involving only one of the four variables e.g., (x, y, z, t). Equations involving highest order derivatives of order one = 1st order differential equations. Examples Function σ(x) = the stress in a uni – axial stretched tapered metal rod.

Function v(x) =the velocity of fluid flowing a straight channel with varying cross-section.

Solution Method of First Order ODEs Solution of Linear (Homogeneous Equation)

Typical form of the equation

The solution u(x) in eq(1) is given by,

U(x) =

Where k= constant to be determined by given condition and the function F(x) has the form:

F(x)=...(3)

In which p(x) is given in differential equation.

Type 2

Solution Method of First Order ODEs Solution of Linear (Non-Homogeneous Equation)

Typical form of the equation

The appearance of g(x) in eq(4) tends to non-homogenous

The solution u(x) in eq(4) is given by,

U(x) =

Where k= constant to be determined by given condition and the function F(x) has the form:

F(x)=

Some important terms to be considered to solve the electrical circuit problems:

*If a current i(t) is flowing through a resistor R ohms, then the voltage   across

*the resistor is given by

*for an inductor of L Hentry, the voltage-current relationship is given by

*For a capacity of c faraday, the voltage current relationship is given by

Example-1:

Finding the optimal current of an electrical circuit(RL circuits) in which the initial

Condition is i=0 at t=0

Solution:

By Kirchhoff voltage law(KVL) method, we get

The differential equation for the RL circuit will be

In which initial conditions are i=0 at t=0

The standard form of the equation is,

Dividing the differential equation by L to obtain

The integrating factor is

Multiplying the above equation with standard form gives rise to

By applying integration on both sides we get

Now applying i=0 at t-0 gives us

0=

C=-

NOW

i=

  = t

Therefore by finding current of the RL circuits is   i=

Hence we complete the solution by first order differential equation of first order and even several types of networking circuits and fluid mechanics uses this method.

Example-2:

Resistors of R1= 10Ω, R2 = 4Ω and R3 = 8Ω are connected up to two batteries (of negligible resistance) as shown. Find the current through each resistor.

Solution:

Assume currents to flow in directions indicated by arrows

Apply KCL on junctions C and A

Therefore, current in mesh ABC = i1

Current in Mesh CA = i2

Then current in mesh is = i1-  i2

Now, apply KVL on mesh ABC,20V are acting in clockwise direction, equating the sum of IR products, we get

10i1 +4i2 = 20.....(1)

In mesh ACD 12 volts are acting in clock-wise direction, then

8( i1 -  i2)-4i2 = 12

8 i1 - 8 i2-4i2 = 12

8 i1 - 12 i2 = 12...(2)

Multiplying eq(1) by 3 we get

30 i1 + 12 i2 = 60

By solving equation 1 and 2 we get,

38i1 = 72

The above equation can be also simplified by elimination or cramer’sruke

  I1 = 72/38 = 1.895 Amperes = current in 10 ohms resistor

Substituting this value in (1) we get,

10(1.895)+4i2 = 20

4i2 = 20-18.95

I2 = 0.263 Amperes=current in 4 ohms resistors

Now,

i1-  i2 = 1.895-0.263=1.632 Amperes

Note-

L-C circuit-

Suppose we have an electric circuit which has an inductance L and capacitance C.

Let i is the current and q is the charge in the condenser plate at time t, so that the voltage drop across-

And the voltage drop across-

By Kirchhoff’s first law-

  ………….. (1)

Divide by L and replacing 1/LC =, equation (1) becomes-

It represents the free electric oscillations of the current having period .

L-C-R circuit-

Let us consider the discharge of a condenser C through an inductance L and the resistance R.

Since the voltage drop across L,C and R are respectively.

So that by Kirchhoff’s law- we have

On replacing R/L  by 2λ and 1/LC by μ², we get-

Key takeaways-

1. The voltage current relationship is given by

2.     Kirchhoff’s first law-

3.     Free electric oscillations of the current having period .

4.     For an inductor of L Hentry, the voltage-current relationship is given by

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
7. T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010