suhail rb5901 a13
TRANSCRIPT
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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