Mathematical Methods for Physicists: A concise introduction - Site Map

ORDINARY DIFFERENTIAL EQUATIONS
Integrating, we have
Multiplying this expression by the series
u ˆ 1
px p ‡k p ln x ‡ k p‡1 x ‡:
y 1 …x† ˆx r 1
…a 0 ‡ a 1 x ‡ a 2 x 2 ‡†
and remembering that r 1 p ˆ r 2 we see that y 2 ˆ uy 1 is of the form
y 2 …x† ˆk p y 1 …x† ln x ‡ x r 2
X 1
mˆ0
a m x m :
…2:34†
…2:35†
While for a double root of Eq. (2.29) the second solution always contains a
logarithmic term, the coecient k p may be zero and so the logarithmic term may
be missing, as shown by the following example.
Example 2.15
Solve the di€erential equation
x 2 y 00 ‡ xy 0 ‡…x 2 1 4
†y ˆ 0:
Solution:
X 1
mˆ0
Substituting Eq. (2.28) and its derivatives into this equation, we obtain
‰…m ‡ r†…m ‡ r 1†‡…m ‡ r† 1 4 Ša mx m‡r ‡ P1
mˆ0
a m x m‡r‡2 ˆ 0:
By equating the coecient of x r to zero we get the indicial equation
r…r 1†‡r 1 4 ˆ 0 or r2 ˆ 1
4 :
The roots r 1 ˆ 1
2 and r 2 ˆ 1 2
di€er by an integer. By equating the sum of the
coecients of x s‡r to zero we ®nd
‰…r ‡ 1†r ‡…r 1† 1 4 Ša 1 ˆ 0 …s ˆ 1†: …2:36a†
‰…s ‡ r†…s ‡ r 1†‡s ‡ r 1 4 Ša s ‡ a s2 ˆ 0 …s ˆ 2; 3; ...†: …2:36b†
For r ˆ r 1 ˆ 1
2 , Eq. (2.36a) yields a 1 ˆ 0, and the indicial equation (2.36b)
becomes
…s ‡ 1†sa s ‡ a s2 ˆ 0:
From this and a 1 ˆ 0 we obtain a 3 ˆ 0, a 5 ˆ 0, etc. Solving the indicial equation
for a s and setting s ˆ 2p, we get
a 2p2
a 2p ˆ
2p…2p ‡ 1†
…p ˆ 1; 2; ...†:
92