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Solution of Assignment # 3
MTH401 (Spring 2012)
Total marks: 30 Lecture # 28-
Due date: 04-07-
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Question#1 Marks 15
Solve the following initial value problem using a power series representation of the
solution around x 0. Find the recurrence relation and the first five nonzero terms of the
series solution.
'
2
2
d y dy x y y y dx dx
Solution:
The equation is already in the standard form with p x 2 x and q x 4. These
functions are polynomials and hence analytic at x 0. Further x 0 is an ordinary
point, so we can look for a power series solution of the above initial value problem
having the form
0
n n n
y x C x
Further notice that the initial conditions will imply that
C 0 (^) 1, C 1 0
Also
1
0
2
0
2
2
n n n n n n dy n C x dx
d y n n C x dx
Substituting the above relations into the given differential equation, it becomes
2 1
0 0 0
2 1 1
0 0 0 0
2 1
0 0 0 0
n n n n n n n n n
n n n n n n n n n n n n
n n n n n n n n n n n n
n n C x x n C x C x
n n C x n C x n C x x C x
n n C x n C x n C x C x
Now we need to make each of the series in 1 of the form
n n n
C x by an appropriate
shift of indices. The index in first term can start at n 2 , since the terms for n 0 and
n 1 are both zero. Then letting k n 2 we see that k can start at zero and the series
can be written as
2 2
0 2
2 0
n n n n n n
k k k
n n C x n n C x
k k C x
0 1 0 2
C C
C
Similarly for n 1 , we can compute C 3 as
1 1 1 3
2
C C
C
C
Similarly for n 2 , we can compute C 4 as
2 1 2 4
3 2
C C
C
C C
^
Similarly for n 3 , we can compute C 5 as
3 1 3 5
4 3
C C
C
C C
Therefore, the first five nonzero terms of the series solution around x = 0 are
2 3 4 5 0 2 3 4 5
2 3 4 5
2 3 4 5
y x C C x C x C x C x
x x x x
x x x x
Question#2 Marks 15
Suppose the differential equation.
2
2
2
d y dy x x x x y dx dx
- Find the ordinary points and the singular points of the above differential equation.
- Classify each singular point you find in 1. as a regular singular point or an
irregular singular point.
- It is desired to solve this equation using a power series centered at 7. Find the
radius of convergence of such a power series. (It is not required to find this power
series solution!)
Solution:
Writing the given differential equation in standard form as
(^)
2
2
2
d y dy y dx x^ x^ dx x
Where
p x x x
and (^)
2
q x x
Direct inspection of the coefficient functions p (^) x (^) and q x (^) shows that they are
discontinuous at x 1 (^) 1 and x (^) 2 4. Therefore the singular points of (^) 1 are
x 1 (^) 1, x 2 4.
The set of all ordinary points of (^) 1 is represented by the set
