Example: Frobenius Series Solution
Solving a differential equation near a regular singular point
Problem
Find a series solution to
3ty ′′ + y ′ − y = 0,
t > 0.
Step 1: Put the equation in standard form
Dividing by 3t, we obtain
y ′′ +
1 ′
1
y − y = 0.
3t
3t
Here
1
1
,
Q(t) = − .
3t
3t
The point t = 0 is not an ordinary point because P (t) and Q(t) are not analytic at t = 0.
P (t) =
However,
tP (t) = t
1
3t
1
t
t Q(t) = t −
=− ,
3t
3
1
= ,
3
2
2
and both of these are analytic at t = 0. Therefore, t = 0 is a regular singular point.
Step 2: Use a Frobenius trial solution
Since t = 0 is a regular singular point, we look for a solution of the form
y=t
λ
∞
X
n
an t =
n=0
∞
X
an tn+λ ,
a0 ̸= 0.
n=0
Then
′
y =
∞
X
(n + λ)an tn+λ−1 ,
n=0
and
y ′′ =
∞
X
(n + λ)(n + λ − 1)an tn+λ−2 .
n=0
Step 3: Substitute into the differential equation
The differential equation is
3ty ′′ + y ′ − y = 0.
1
Substituting the series expressions gives
′′
3ty =
∞
X
3(n + λ)(n + λ − 1)an tn+λ−1 ,
n=0
y′ =
∞
X
(n + λ)an tn+λ−1 ,
n=0
and
−y =
∞
X
(−an )tn+λ .
n=0
Therefore,
∞ h
∞
i
X
X
3(n + λ)(n + λ − 1) + (n + λ) an tn+λ−1 −
an tn+λ = 0.
n=0
n=0
The first sum begins with the power tλ−1 , while the second begins with tλ . To combine them, we
separate the first term of the first sum and shift indices.
For the first sum, the n = 0 term is
3λ(λ − 1) + λ a0 tλ−1 .
For the remaining terms, we shift the index so that all powers are written as tn+λ . Equivalently,
after collecting powers, we obtain
∞
λ−1 X
3λ(λ − 1) + λ a0 t
+
3(n + 1 + λ)(n + λ) + (n + 1 + λ) an+1 − an tn+λ = 0.
n=0
Since this series is identically zero, each coefficient must be zero.
Step 4: Find the indicial equation
The coefficient of the lowest power tλ−1 must be zero:
3λ(λ − 1) + λ a0 = 0.
Since a0 ̸= 0, we must have
3λ(λ − 1) + λ = 0.
Simplifying,
3λ2 − 3λ + λ = 0,
so
3λ2 − 2λ = 0.
Thus,
λ(3λ − 2) = 0.
Therefore, the roots of the indicial equation are
λ=0
or
2
2
λ= .
3
Step 5: Find the recurrence relation
For each n ≥ 0, the coefficient of tn+λ must also be zero:
3(n + 1 + λ)(n + λ) + (n + 1 + λ) an+1 − an = 0.
Factor out (n + 1 + λ):
(n + 1 + λ) 3(n + λ) + 1 an+1 − an = 0.
Thus,
an+1 =
an
,
(n + 1 + λ)(3n + 3λ + 1)
n = 0, 1, 2, . . . .
Step 6: First solution, using λ = 0
For λ = 0, the recurrence relation becomes
an+1 =
Starting with a0 , we get
an
.
(n + 1)(3n + 1)
a0
= a0 ,
(1)(1)
a1
a0
a2 =
= ,
(2)(4)
8
a0
a0
a2
=
=
.
a3 =
(3)(7)
8 · 21
168
a1 =
Therefore,
1 2
1 3
y1 (t) = a0 1 + t + t +
t + ··· .
8
168
Since multiplying a solution by a nonzero constant gives another solution, we usually take a0 = 1.
Thus,
1 3
1
y1 (t) = 1 + t + t2 +
t + ··· .
8
168
A compact form for the coefficients is
y1 (t) = 1 +
∞
X
n=1
1
tn .
n! 1 · 4 · 7 · · · (3n − 2)
Step 7: Second solution, using λ = 23
For λ = 23 , the recurrence relation becomes
an+1 =
n + 1 + 23
3
an
.
3n + 3 · 23 + 1
Since
n+1+
2
5
=n+ ,
3
3
3n + 3 ·
we have
an+1 =
n + 53
2
+ 1 = 3n + 3,
3
a
n
.
(3n + 3)
Now compute the first few coefficients:
a1 =
a2 =
5
3
a0
a0
= ,
5
(3)
8
3
a1
=
(6)
raca1 16 =
raca0 80,
a3 =
a2
=
(9)
11
3
raca2 33 =
raca0 2640. Therefore,
a0
a0
a0 3
y2 (t) = t2/3 a0 + t + t2 +
t + ··· .
5
80
2640
Again, taking a0 = 1, we obtain
1
1
1 3
y2 (t) = t2/3 1 + t + t2 +
t + ··· .
5
80
2640
A compact form for the coefficients is
y2 (t) = t
2/3
1+
∞
X
n=1
1
tn
n! 5 · 8 · 11 · · · (3n + 2)
!
.
Step 8: General solution
The two roots of the indicial equation are
2
λ2 = .
3
λ1 = 0,
They are distinct, and the corresponding Frobenius solutions are linearly independent. Therefore,
the general solution for t > 0 is
y(t) = C1 y1 (t) + C2 y2 (t) .
That is,
1 2
1 3
1
1 2
1 3
2/3
y(t) = C1 1 + t + t +
t + · · · + C2 t
1+ t+ t +
t + ··· .
8
168
5
80
2640
4