FROBENIUS SERIES SOLUTION OF A SECOND ORDER HOMOGENEOUS LINEAR DIFFERENTIAL EQUATION C0NTINUED

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FROBENIUS SERIES SOLUTION OF A SECOND ORDER
HOMOGENEOUS LINEAR DIFFERENTIAL EQUATION
(C0NTINUED)
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Assume that x 0 is a regular singular point of
the second order homogeneous l.d.e.
and
Hence
are both analytic at x 0.
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Let us assume that the Power Series expansions of
p(x) and q(x) both converge at least for all x
x lt R. That is
( x lt R )
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Multiplying the given d.e. by x2, we get
We assume a solution to the above equation as
a0 ? 0, m a real number.
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i.e.
Hence
Note As m need not be zero, both
for y?, y? start from n 0.
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Substituting for y, y? , y? in the above d.e. we
get
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Dividing by xm throughout, we get
..()
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Note that
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And
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Hence equation () becomes
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Equating the coefficients of like powers of x on
both sides we get
(Constant term)
Since a0 ? 0, we get
Indicial equation
i.e.
(say)
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(Coefficient of x)
i.e.
(Continued on the next slide)
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(Coefficient of x2 )
i.e.
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(Coefficient of xn )
i.e.
n 1, 2, 3,
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Let us assume that the roots of the indicial
equation are real and are m1, m2 where m1 m2.
Thus corresponding to the bigger root (
exponent) m1, we will always get a Frobenius
Series solution
where all ans (n gt 0) can be expressed in terms
of a0.
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If m1 ? m2 and if m1 - m2 is not a positive
integer (i.e. the roots do not differ by an
integer), then corresponding to the smaller root
( exponent) m2 also, we will get a second L.I.
Frobenius Series solution
where all ans (n gt 0) can be expressed in terms
of a0.
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Problem 4(a) Page 191
For the d.e.
find the indicial equation and its roots
corresponding to the regular singular point x
0.
Here
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Now
and
are both analytic at x 0. Hence x 0 is a
regular singular point.
Also note that p(0)
q(0) 0.
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Hence the indicial equation is
Note that p0 p(0)
and q0 q(0) 0
Hence the indicial equation is
Roots m 0,
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Problem 3(a) Page 191
For the d.e.
find the indicial equation and its roots
corresponding to the regular singular point x
0.
Here
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Now
and
are both analytic at x 0. Hence x 0 is a
regular singular point.
q(0) 2.
Also note that p(0)
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Hence the indicial equation is
Note that p0 p(0)
and q0 q(0) 2
Hence the indicial equation is
Roots m
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Problem 3(a) Page 198
Solve the d.e.
Here
Hence the only singular point is x 0.
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Since
and
are both analytic at x 0, x 0 is a regular
singular point.
Note that p0 p(0) 2 and q0 q(0) 0
Hence the indicial equation is
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i.e.
Hence the exponents are m 0, -1 (Roots of the
indicial equation)
Let us find the solution corresponding to m
0.
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Assume a solution as
i.e.
as m 0.
Hence
Substituting for y, y?, y? in the given d.e. we
get
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i.e.
We replace n by n1 in the first two
And replace n by n-1 in the third
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We get
i.e.
Equating the coefficients of like powers of x, we
get
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or
n1 gives
n3 gives
.
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Also n2 gives
Hence
Thus the solution corresponding to m 0 is
Hence one solution of the given d.e. is
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A second LI solution is y2 v y1, where
Hence
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Since the given d.e. is homogeneous, y2 is a
solution implies y2 is also a solution.
Hence
and
are two LI solutions. Hence the general solution
is
c1, c2 arbitrary constants.
valid over any interval not containing x 0.
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Problem 1 Page 198
The equation
has only one Frobenius series solution. Find it.
Note that p0 p(0) -3 and q0 q(0) 4
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Hence the indicial equation is
i.e.
Roots are
2, 2 ( Repeated roots )
Hence the given equation has only one Frobenius
series solution.
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Let the solution be
i.e.
Substituting for y, y ?, y ? in the given d.e. we
get
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i.e.
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Dividing throughout by x2 we get
replacing n by n-1, we get
In the third
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Note the terms corresponding to n 0 become
zero. We thus get
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or
We thus have the recurrence relation
or
n 1 gives
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n 1 gives
n 2 gives
n 3 gives
And so on.
Hence the Frobenius Series solution is
a0 an arbitrary constant