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TERM PAPER

Engineering Mathematics II

MTH 102

TOPIC: Write about the method of undetermined coefficients and methodof variation of parameters. Discuss and compare advantage and disadvantage

of these methods. Illustrate your findings with examples.

Submitted by-: Submitted To-:

SUHAIL ASIF Miss. SOFIA SINGLA

SECTION-B5901 Deptt of Maths

Roll no-: A 13

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CONTENTS

S.No Topic

1. Acknowledgement

2. Method of variation of parameter

3. Complementary Solution

4. Method of undetermined coefficients

5. References

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ACKNOWLEDGEMENT

First and foremost I thank my teacher Miss. SOFIA SINGLA who has given me this Term

Paper to bring out my creative capabilities. I am also thankful to him for their valuable

suggestions on my term paper.

I express my gratitude to my parents for being continuous source of encouragement and for

their entire financial ad given to me.

I would like to acknowledge the assistance provided to me by the library staff of L.P.U.

My heartfelt gratitude to my friends, for helping me morally to complete my work in time

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Method of variation of parameters:-

This method gives P.I.:-

Where w=w(y1,y2) is the Wronskian of y1 andy2.

Y1 Y2

W(y1, y2)(x) = This is called the Wronskian* of y1 and y2

Y11 Y2

1

This method is quite general and applies to equation of the form

Y11 + q(x)Y1+ r(x)Y=f(x)

Ln(y)= f(x)

Where P, Q and f are the function of the x or t etc.

Complementary solution:-

Y(x) = C1Y1(x) +C2Y2(x) Where C1 and C2 are constant

And y1 (x) , y2 (x)are the solution of auxiliary equation

Y11 + q(x) Y1+ r(x) Y=0 .(1)

Is a solution of (eq.1)for any constants and We now ``vary" and to functions of so

that

.(2)

is a solution of the equation

(3)

Proof: - Let u(x) and v(x)be continuously differentiable functions (to be determined) such

that

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yp=uy1+vy2 ..(4)

Is a particular solution of (3). Differentiation of (5) leads to

yp=uy1+vy1+uy1+vy2 (6)We choose and so that

uy1+vy2=0 (7)

Substituting (7) in (6), we have

..(8)

Since is a particular solution of (3), substitution of (5) and (8) in (3), we get

As and are solutions of the homogeneous equation (1), we obtain the condition

.(9)

We now determine and from (7) and (9). By using the Cramer's rule for a linear system of

Equations, we get

.(10)

(Note that and are linearly independent solutions of (1) and hence the Wronskian,

For any ). Integration of (10) gives us

(11)

From eq.(4)

Example1:- Find a general solution to the following differential equation.

http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.3http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.5http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.7http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.6http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.3http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.8http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.3http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.7http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.9http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.1http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.10http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.5http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.7http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.6http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.3http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.8http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.3http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.7http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.9http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.1http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.10http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html#eqn:9.3

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Solution:

First, since the formula for variation of parameters requires a coefficient of a one in front of th

second derivative lets take care of that before we forget. The differential equation that weactually be solving is

Well leave it to you to verify that the complimentary solution for this differential equation is

So, we have

The Wronskian of these two functions is

The particular solution is then,

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The general solution is ,

Method of undetermined coefficient:-

This method is using for finding Particular Integral(P.I.) for a

equation

An equation is

Y11 + PY1+ Q=f(x)

Ln(y)=f(x) ..(1

Where P , Q and f(x) are the function of the x

f(x) should be these types of function:-

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1. :-

2.:-

3.:-

So the solution is

Complete solution is C.S.= Complementary solution +Particular integral

C.S = C.F. + P.I

Complementary solution is the solution of auxiliary equation(A.E.)

A.E. :- Ln(y) = 0

Y11 + PY1+ Q=0

Finding P.I.:-

Case I.

:-

We first assume that is not a root of the characteristic equation, i.e., Note th

Therefore, let us assume that a particular solution is of the form

Where A an unknown, is an undetermined coefficient. Thus

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since

we can choose

to obtain

Thus, is a particularsolution of

Modification Rule: If is a root of the characteristic equation, i.e., with multiplicity

(i.e., and ) then we take, of the form

and obtain the value of by substituting in

Case 2.

.

We first assume that is not a root of the characteristic equation, i.e., Here, w

assume that is of the form

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and then comparing the coefficients of and (why!) in obtain th

values of and

Modification Rule: If is a root of the characteristic equation, i.e., wi

multiplicity then we assume a particular solution as

and then comparing the coefficients in obtain the values of and

Case 3.

:-

Suppose Then we assume that

and then compare the coefficient of in to obtain the values of

for

Modification Rule: If is a root of the characteristic equation, i.e., with

multiplicity then we assume a particular solution as

and then compare the coefficient of in to obtain the values of

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for

Ex:- Find a particular solution of

Solution: Here, and Thus which is not a root of the

characteristic equation Note that the roots of are

Thus, let us assume This gives us

Comparing the coefficients of and on both sides, we get an

On solving for and we get

So, a particular solution is

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References:-

1. http://www.sosmath.com/diffeq/second/variation/variation.html

2. http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.html

3. http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node78.html

4. http://en.wikipedia.org/wiki/Method_of_variation_of_parameters

5. http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients

http://www.sosmath.com/diffeq/second/variation/variation.htmlhttp://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.htmlhttp://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node78.htmlhttp://en.wikipedia.org/wiki/Method_of_variation_of_parametershttp://en.wikipedia.org/wiki/Method_of_undetermined_coefficientshttp://www.sosmath.com/diffeq/second/variation/variation.htmlhttp://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node76.htmlhttp://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node78.htmlhttp://en.wikipedia.org/wiki/Method_of_variation_of_parametershttp://en.wikipedia.org/wiki/Method_of_undetermined_coefficients