TERM PAPER
MTH-102
TOPICMETHOD OF VARIATION OF UNDETERMINED
COFFICIENT AND METHOD OF VARIATION OF
PARAMETER & IT’S ADVANTAGES DISADVANTANTAGES
SUBMITTED TO,
SUBMTTED BY,
MR. GURPREET SINGH BHATIA
ROHIT KUMAR
DEPTT. OF MATHEMATICS
ACKNOWLEDGEMENT
I WOULD LIKE TO THANKS ALL THOSE WHO HAVE
ENCOURAGED ME TO SUBMIT THIS PROJECT RELATED TO
“METHOD OF VARIATION OF UNDETERMINED COEFFICIENTS”
AS A TERM PAPER IT IS AN EXTREMELY ORDOUS JOB TO
ACKNOWLEDGEMENT THE INVALUABLE HELP RENDERED
WHO DIRECTLY AND INDIRECT CONTRIBUTION TO MAKE
FEAT AS POSSIBLE FOR ME.
I CERTAINLY FEEL ELATED TO EXPRESS MY DEEP SENSE TO
GRATITUDE TO ALL OF THEM.
I THANK MY MATHS TEACHER TO EXTEND HIS HANDS FOR
HELP, CO-OPERATION AND KINDNESS
CONTENT:1.DEFINITION OF UNDETERMINED COFFICIENT
2.CONDITION
3.REMARK
4.SUMMARY
5.EXAMPLE
6.DEFINATION OF VARIATION OF PARAMETER
7.SUMMARY
8.EXAMPLE
9.REMARK
10.REFERENCES
1. DEFINATION OF METHOD OF VARIATION OF UNDETERMINED
COFFICIENT:This method is based on a guessing technique. That is, we will guess the form of and then
plug it in the equation to find it. However, it works only under the following two conditions:
2.CONDITION:Condition 1: the associated homogeneous equations has constant coefficients;
Condition 2: the nonhomogeneous term g(x) is a special form
where P(x) and L(x) are polynomial functions.
Note that we may assume that g(x) is a sum of such functions (see the remark below
for more on this).
Assume that the two conditions are satisfied. Consider the equation
where a, b and c are constants and
where
is a polynomial function with degree n. Then a particular solution
is given by
where
,
where the constants
and
have to be determined. The power s is equal to 0 if
not a root of the characteristic equation. If
double root.
is
is a simple root, then s=1 and s=2 if it is a
3.Remark: If the nonhomogeneous term g(x) satisfies the following
where
are of the forms cited above, then we split the original equation into N equations
then find a particular solution
. A particular solution to the original equation is given by
4.Summary:Let us summarize the steps to follow in applying this method:
(1)
First, check that the two conditions are satisfied;
(2)
If the equation is given as
,
where
or
, where
is a polynomial function with degree n, then split this equation into N equations
;
(3)
Write down the characteristic equation
(4)
, and find its roots;
Write down the number
. Compare this number to the roots of the
characteristic equation found in previous step.
(4.1)
If
(4.2)
is not one of the roots, then set s = 0;
If
(4.3)
is one of the two distinct roots, set s = 1;
If
is equal to both root (which means that the characteristic equation has
a double root), set s=2.
In other words, s measures how many times
equation;
(5)
Write down the form of the particular solution
is a root of the characteristic
where
(6)
Find the constants
and
by plugging
(7)
Once all the particular solutions
original equation is
5.EXAMPL:Find a particular solution to the equation
Solution: Let us follow these steps:
into the equation
are found, then the particular solution of the
(1)
First, we notice that the conditions are satisfied to invoke the method of undetermined
coefficients.
(2)
We split the equation into the following three equations:
(3)
The root of the characteristic equation
(4.1)
Particular solution to Equation (1):
Since
, and
, then
The particular solution is given as
are r=-1 and r=4.
, which is not one of the roots. Then s=0.
If we plug it into the equation (1), we get
,
which implies A = -1/2, that is,
(4.2)
Particular solution to Equation (2):
Since
, and
, then
The particular solution is given as
If we plug it into the equation (2), we get
,
which implies
, which is not one of the roots. Then s=0.
Easy calculations give
, and
, that is
(4.3)
Particular solution to Equation (3):
Since
, and
, then
The particular solution is given as
which is one of the roots. Then s=1.
If we plug it into the equation (3), we get
,
which implies
, that is
(5)
A particular solution to the original equation is
6.Method of Variation of Parameters
This method has no prior conditions to be satisfied. Therefore, it may sound more general
than the previous method. We will see that this method depends on integration while the
previous one is purely algebraic which, for some at least, is an advantage.
Consider the equation
In order to use the method of variation of parameters we need to know that
is a set
of fundamental solutions of the associated homogeneous equation y'' + p(x)y' + q(x)y = 0. We
know that, in this case, the general solution of the associated homogeneous equation is
. The idea behind the method of variation of parameters is to look for a
particular solution such as
where and
The functions
are functions. From this, the method got its name.
and are solutions to the system
,
which implies
,
where
is the wronskian of
and
. Therefore, we have
7.Summary:Let us summarize the steps to follow in applying this method:
(1)
Find
a set of fundamental solutions of the associated homogeneous equation
y'' + p(x)y' + q(x)y = 0.
;
(2)
Write down the form of the particular solution
;
(3)
Write down the system
;
(4)
Solve it. That is, find
and
;
(5)
Plug
and
into the equation giving the particular solution.
8.Example: Find the particular solution to
Solution: Let us follow the steps:
(1)
A set of fundamental solutions of the equation y'' + y = 0 is
;
(2)
The particular solution is given as
(3)
We have the system
;
(4)
We solve for
and
, and get
Using techniques of integration, we get
;
(5)
The particular solution is:
,
or
9.Remark: Note that since the equation is linear, we may still split if necessary. For example,
we may split the equation
,
into the two equations
then, find the particular solutions
the original equation by
for (1) and
for (2), to generate a particular solution for
There are no restrictions on the method to be used to find or . For example, we can use
the method of undetermined coefficients to find , while for , we are only left with the
variation of parameters.
ADVANTAGESBasically both method is used to find the particular integral of
differential equation. The main advantage of these method is that
without we finding the particular integral we can’t find complete
solution of differential equation.The finding the particular integral of
both method is different but aim is same.
DISADVANTAGE-
HoweveR when X=tanu or secu this method fails. Since the number of terms obtained by
differentiating X=tanu or secu is infinite, this method is only used to find the particular
integral.
10.REFERENCESS:1.B.S GREWALL
2.WWW.GOOGLE.COM
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