In this tutorial, we discuss about the solution of the differential equations in terms of power series. All the cases associated with this attachment has been discussed

1. Solution of Differential Equations in Power Series by Employing Frobenius
Method
Mehar Chand
Department of Mathematics
Baba Farid College, Bathind-151001, India
March 29, 2020
1 Frobenius Method
This method is applied to find the solution of the differential equation in terms of power series about x = α of
d2y
dx2
+ p(x)
dy
dx
+ q(x)y = 0 (1.1)
where x = α is its regular singular point.
If α is non-zero real, then we take x − α = z so that z = 0 becomes its regular singular point for the given
differential equation.
2 Key points to use this Method
We discuss here for x = 0 is a regular singular point point.
Step 1. let y(x) = xc(A0 + A1x + A2x2 + A3x3 + ....)
Step 2. Find the values of y (x) and y (x)
Step 3. Then put all the values of y(x), y (x), y (x) into the given differential equation (1.1).
Step 4. After putting these values in the given differential equation compare the coefficients of like power on both
sides
Step 5. The coefficient of xc or least power gives us c2 −c+cp0 +q0 = 0. which is know as INDICIAL EQUATION
and gives us two roots say c1, c2
And comparing the coefficient of xc+1, xc+2, ..., xc+k to find the values of A2, A3, A4, ...
There are four cases on the basis of roots of indicial equation
2.1 When the roots are differ by non-integer
Case 1. When the roots of the indicial equation are differ by a non-integer
Let c1, c2 where c1 = c2 be roots of indicial equation where c1−c2=non-integer. Then two independent solutions
will be given by
y(x)|c=c1
= Axc1
[power series in x] = Au(x)
1

2. y(x)|c=c2
= Bxc2
[power series in x] = Bv(x)
There for the general solution of given differential equation is given by y(x) = Au(x) + Bv(x), where A, B are
arbitrary constants and u(x), v(x) both are ascending power series in x.
Q-1: Solve
x2 d2y
dx2
+ x
dy
dx
+ (x2
− n2
)y = 0 (2.1)
where 2n is non-integral.
Solution: The given equation (2.1) can be written by dividing x2 on both side, as below
d2y
dx2
+
1
x
dy
dx
+
x2 − n2
x2
y = 0 (2.2)
Now, comparing the equation (2.2) with (1.1), we have
P(x) =
1
x
, Q(x) =
x2 − n2
x2
are not analytic at x = 0
and xP(x) = 1, x2
Q(x) = x2
− n2
both are now analytic at x = 0.
Therefore x = 0 is regular singular point, so the the solution of the given differential exists in terms of power series.
Step-1: Let y = xc[A0 + A1x + A2x2 + A3x3 + A4x4 + ... + Akxk + ....]
or
y = A0xc
+ A1xc+1
+ A2xc+2
+ A3xc+3
+ A4xc+4
+ ... + Akxc+k
+ .... (2.3)
Then
dy
dx
= A0cxc−1
+ A1(c + 1)xc
+ A2(c + 2)xc+1
+ A3(c + 3)xc+2
+ A4(c + 4)xc+3
+ ... + Ak(c + k)xc+k−1
+ ....
(2.4)
d2y
dx2
= A0c(c − 1)xc−2
+ A1(c + 1)cxc−1
+ A2(c + 2)(c + 1)xc
+ A3(c + 3)(c + 2)xc+1
+
A4(c + 4)(c + 3)xc+2
+ ... + Ak(c + k)(c + k − 1)xc+k−2
+ ....
(2.5)
Putting the values of y, y , y from equations (2.3), (2.4) and (2.5) in equation (2.1), then we have
x2
[A0c(c − 1)xc−2
+ A1(c + 1)cxc−1
+ A2(c + 2)(c + 1)xc
+ A3(c + 3)(c + 2)xc+1
+
A4(c + 4)(c + 3)xc+2
+ ... + Ak(c + k)(c + k − 1)xc+k−2
+ ....]
+ x[A0cxc−1
+ A1(c + 1)xc
+ A2(c + 2)xc+1
+ A3(c + 3)xc+2
+
A4(c + 4)xc+3
+ ... + Ak(c + k)xc+k−1
+ ....]
+ (x2
− n2
)xc
[A0 + A1x + A2x2
+ A3x3
+ A4x4
+ ... + Akxk
+ ....]
(2.6)
Step-2: Since, this equation satisfied the for all values of x, therefore co-efficient of different powers of x must
vanish separately. Equating the coefficient of like power, we have:
2

3. Coeff. of xc) A0c(c − 1) + A0c − n2A0 = 0
⇒ A0(c2
− n2
) = 0
But A0 = 0 ⇒ c2 − n2 = 0 and hence we have c = ±n
Since 2n is not an integer therefore roots are different and differ by non integer.
( Since c1 − c2 = n + n = 2n = Non-integer)
coeff. of xc+1)
A1(c + 1)c + A1(c + 1) − n2
A1 = 0
A1[(c + 1)2
− n2
] = 0 ⇒ A1 = 0
Since, at c = n, −n the value of (c + 1)2 − n2 = 0
coeff. of xc+2)
A2(c + 2)(c + 1) + A2(c + 2) + A0 − n2
A2 = 0
A2[(c + 2)2
− n2
] + A0 = 0 ⇒ A2 = −
A0
(c + 2)2 − n2
coeff. of xc+k)
Ak[(c + k)2
− n2
] + Ak−2 ⇒ Ak = −
Ak−2
(c + k)2 − n2
(2.7)
putting the values of k = 3, 5, 7, 9, .... and using the the value of A1 in equation (2.7), we have A3 = A5 = A7 =
A9 = 0
Similarly if we choose k = 2, 4, 6, 8, ..., we have
A2 = −
A0
(c + 2)2 − n2
A4 =
A0
[(c + 2)2 − n2][(c + 4)2 − n2]
Step-3: At c = n, the values of A2, A4 reduced to the following the form:
A2 = −
A0
22(n + 1)
A4 =
A0
[(n + 2)2 − n2][(n + 4)2 − n2]
=
A0
242!(n + 1)(n + 2)
Similarly, we can obtained the value of A6 i.e.
A6 = −
A0
263!(n + 1)(n + 2)(n + 3)
Now putting the values of A2, A4, A6 in equation (2.3), we have first solution of the given equation
y(x)]c=n = A0xn
1 −
x2
22(n + 1)
+
x4
242!(n + 1)(n + 2)
−
x6
263!(n + 1)(n + 2)(n + 3)
+ ... = A0u(x) (2.8)
3

4. where
u(x) = xn
1 −
x2
22(n + 1)
+
x4
242!(n + 1)(n + 2)
−
x6
263!(n + 1)(n + 2)(n + 3)
+ ...
When c = −n, changing n to −n, the other solution of the given differential equation is
y(x)]c=−n = A0v(x) (2.9)
where
v(x) = x−n
1 −
x2
22(1 − n)
+
x4
242!(1 − n)(2 − n)
−
x6
263!(1 − n)(2 − n)(3 − n)
+ ...
Therefore the general solution of the given differential equation is given by
y(x) = c1A0u(x) + c2A0v(x)
⇒ y(x) = Au(x) + Bv(x)
(2.10)
where A = c1A0 and B = c2A0
2.2 When roots are equal
When roots of the indicial equal are equal i.e. c1 = c2
Let y(x) = u(x) be solution of the power series in terms of A0 and c, then the general is given as
y(x) = A[u(x)]c=c1 + B
∂u(x)
∂c c=c1
(2.11)
where A and B are arbitrary constants
Q-2: Solve
x
d2y
dx2
+
dy
dx
+ xy = 0 (2.12)
where 2n is non-integral.
Solution: First we compare the given differential equation (2.12) with (1.1), we have
P(x) =
1
x
, Q(x) =
1
x
are not analytic at x = 0
and xP(x) = 1, x2
Q(x) = x both are now analytic at x = 0.
Therefore x = 0 is regular singular point, so the the solution series of the given differential exists.
Putting the values of y, y , y from equations (2.3), (2.4) and (2.5) in equation (2.12), then we have
x[A0c(c − 1)xc−2
+ A1(c + 1)cxc−1
+ A2(c + 2)(c + 1)xc
+ A3(c + 3)(c + 2)xc+1
+
A4(c + 4)(c + 3)xc+2
+ ... + Ak(c + k)(c + k − 1)xc+k−2
+ ....]
+ [A0cxc−1
+ A1(c + 1)xc
+ A2(c + 2)xc+1
+ A3(c + 3)xc+2
+
A4(c + 4)xc+3
+ ... + Ak(c + k)xc+k−1
+ ....]
+ xc+1
[A0 + A1x + A2x2
+ A3x3
+ A4x4
+ ... + Akxk
+ ....]
(2.13)
Step-2: Since, this equation satisfied the for all values of x, therefore co-efficient of different powers of x must
vanish separately. Equating the coefficient of like power, we have:
4

5. Coeff. of xc−1) A0c(c − 1) + A0c = 0
⇒ A0c2
= 0
But A0 = 0 ⇒ c2 = 0 and hence we have c = 0, 0
⇒ Both the roots of the indicial equation are equal.
coeff. of xc)
A1(c + 1)c + A1(c + 1) = 0
A1(c + 1)2
= 0 ⇒ A1 = 0
Since, at c = 0 the value of (c + 1)2 = 0
coeff. of xc+1)
A2(c + 2)(c + 1) + A2(c + 2) + A0 = 0
A2(c + 2)2
+ A0 = 0 ⇒ A2 = −
A0
(c + 2)2
coeff. of xc+k)
Ak(c + k)2
+ Ak−2 ⇒ Ak = −
Ak−2
(c + k)2
(2.14)
putting the values of k = 3, 5, 7, 9, .... and using the the value of A1 in equation (2.7), we have A3 = A5 = A7 =
A9 = 0
Similarly if we choose k = 2, 4, 6, 8, ..., we have
A2 = −
A0
(c + 2)2
A4 =
A0
(c + 2)2(c + 4)2
A6 = −
A0
(c + 2)2(c + 4)2(c + 6)2
Step-3: Now putting the values of A0, A1, A2, A3, A4, A5, A6, ... in the equation (2.3), we have
y(x) = A0xc
1 −
x2
(c + 2)2
+
x4
(c + 2)2(c + 4)2
−
x6
(c + 2)2(c + 4)2(c + 6)2
+ ... (2.15)
∂y
∂c
= A0xc
log x 1 −
x2
(c + 2)2
+
x4
(c + 2)2(c + 4)2
−
x6
(c + 2)2(c + 4)2(c + 6)2
+ ...
+ A0xc 2x2
(c + 2)3
−
2x4
(c + 2)3(c + 4)2
−
2x4
(c + 2)2(c + 4)3
+ ....
(2.16)
At c = 0, the above equations (2.15) and (2.16) reduced to the following the form:
y(x)]c=0 = A0 1 −
x2
(2)2
+
x4
(2)2(4)2
−
x6
(2)2(4)2(6)2
+ ... = A0u(x) (2.17)
5

6. ∂y
∂c c=0
= A0 log(x) 1 −
x2
(2)2
+
x4
(2)2(4)2
−
x6
(2)2(4)2(6)2
+ ...
+ A0
2x2
(2)3
−
2x4
(2)3(4)2
−
2x4
(2)2(4)3
+ ....
= A0 log(x)u(x) + A0v(x)
(2.18)
where
u(x) = A0 1 −
x2
(2)2
+
x4
(2)2(4)2
−
x6
(2)2(4)2(6)2
+ ...
v(x) =
2x2
(2)3
−
2x4
(2)3(4)2
−
2x4
(2)2(4)3
+ ....
(2.19)
Therefore the general solution of the given differential equation is given by
y(x) = c1A0u(x) + c2[A0 log(x)u(x) + A0v(x)]
⇒ y(x) = c1A0u(x) + c2A0u(x) log(x) + c2A0v(x)
⇒ y(x) = Au(x) + Bu(x) log(x) + Bv(x)
⇒ y(x) = [A + B log(x)]u(x) + Bv(x)
(2.20)
where A = c1A0 and B = c2A0
2.3 Roots the indicial equation are different and differ by an integer, making a coefficient
infinite
Let c1 − c2 = λ a positive constant i.e. c1 > c2 and η = c1 − c2 + λ and let y(x) = u(x) be solution in power series
form in terms of A0 and c. Then general solution is given as follow:
y(x) = Au(x) + Bv(x) (2.21)
where u(x) = η]c=c2 and v(x) =
∂η
∂c c=c2
Q-3: Solve the following differential equations in power series.
x2 d2y
dx2
+ x
dy
dx
+ (x2
− 1)y = 0 (2.22)
Solution: First we compare the given differential equation (2.22) with (1.1), we have
P(x) =
1
x
, Q(x) =
x2 − 1
x
are not analytic at x = 0
and xP(x) = 1, x2
Q(x) = x2
− 1 both are now analytic at x = 0.
Therefore x = 0 is regular singular point, so the the solution series of the given differential exists.
Putting the values of y, y , y from equations (2.3), (2.4) and (2.5) in equation (2.22), then we have
x2
[A0c(c − 1)xc−2
+ A1(c + 1)cxc−1
+ A2(c + 2)(c + 1)xc
+ A3(c + 3)(c + 2)xc+1
+
A4(c + 4)(c + 3)xc+2
+ ... + Ak(c + k)(c + k − 1)xc+k−2
+ ....]
+ x[A0cxc−1
+ A1(c + 1)xc
+ A2(c + 2)xc+1
+ A3(c + 3)xc+2
+
A4(c + 4)xc+3
+ ... + Ak(c + k)xc+k−1
+ ....]
+ (x2
− 1)[A0xc
+ A1xc+1
+ A2xc+2
+ A3xc+3
+ A4xc+4
+ ... + Akxc+k
+ ....]
(2.23)
6

7. Step-2: Since, this equation satisfied the for all values of x, therefore co-efficient of different powers of x must
vanish separately. Equating the coefficient of like power, we have:
Coeff. of xc) A0c(c − 1) + A0c + A0 = 0
⇒ A0(c2
− 1) = 0
But A0 = 0 ⇒ c2 − 1 = 0 and hence we have c = 1, −1
⇒ c1 = 1, c2 = −1, therefore λ = c1 − c2 = 2 = An integer.
coeff. of xc+1)
A1c(c + 1) + A1(c + 1) − A1 = 0
A1c(c + 2) = 0 ⇒ A1 = 0
[Since, c = 0, c = −2 therefore c(c + 2) = 0]
coeff. of xc+2)
A2[(c + 1)(c + 2) + (c + 2) − 1] + A0 = 0
A2[(c + 1)(c + 3)] + A0 = 0 ⇒ A2 = −
A0
(c + 1)(c + 3)
coeff. of xc+k)
Ak[(c + k)2
− 1] + Ak−2 ⇒ Ak = −
Ak−2
(c + k)2 − 1
(2.24)
putting the values of k = 3, 5, 7, 9, .... and using the the value of A1 in equation (2.7), we have A3 = A5 = A7 =
A9 = 0
Similarly if we choose k = 2, 4, 6, 8, ..., we have
A2 = −
A0
(c + 1)(c + 3)
A4 =
A0
(c + 1)(c + 3)2(c + 5)
A6 = −
A0
(c + 1)(c + 3)2(c + 5)2(c + 7)
Step-3: Now putting the values of A0, A1, A2, A3, A4, A5, A6, ... in the equation (2.3), we have
y(x) = A0xc
1 −
x2
(c + 1)(c + 3)
+
x4
(c + 1)(c + 3)2(c + 5)
−
x6
(c + 1)(c + 3)2(c + 5)2(c + 7)
+ ... (2.25)
At c = −1, the second coefficient and all other following coefficients becomes infinite.
therefore η = (c − c1 + λ)y = (c + 1)y, by using the value of η from above equation (2.25), we have:
7

8. η = (c + 1)A0xc
1 −
x2
(c + 1)(c + 3)
+
x4
(c + 1)(c + 3)2(c + 5)
−
x6
(c + 1)(c + 3)2(c + 5)2(c + 7)
+ ...
= A0xc
(c + 1) −
x2
(c + 3)
+
x4
(c + 3)2(c + 5)
−
x6
(c + 3)2(c + 5)2(c + 7)
+ ...
(2.26)
∂η
∂c
= A0xc
log(x) (c + 1) −
x2
(c + 3)
+
x4
(c + 3)2(c + 5)
−
x6
(c + 3)2(c + 5)2(c + 7)
+ ...
+ A0xc
1 +
x2
(c + 3)2
−
x4
(c + 3)2(c + 5)2
−
2x4
(c + 3)3(c + 5)
+ ...
(2.27)
∂η
∂c
= η log(x) + A0xc
1 +
x2
(c + 3)2
−
x4
(c + 3)2(c + 5)2
−
2x4
(c + 3)3(c + 5)
+ ... (2.28)
Now put c = −1 in equation (2.26) and (2.28), we have
η]c=−1 = A0x−1
−
x2
2
+
x4
22.4
−
x6
22.42.6
+ ... = A0u(x) (2.29)
∂η
∂c c=−1
= A0u(x) log(x) + A0x−1
1 +
x2
22
−
1
22.42
−
2
23.4
x4
+ ... = A0v(x) (2.30)
where
u(x) = x−1
−
x2
2
+
x4
22.4
−
x6
22.42.6
+ ...
v(x) = u(x) log(x) + x−1
1 +
x2
22
−
1
22.42
−
2
23.4
x4
+ ...
Therefore the general solution of the given differential equation is given by
y(x) = c1A0u(x) + c2A0v(x) ⇒ y(x) = Au(x) + Bv(x) (2.31)
where A = c1A0 and B = c2A0
2.4 Roots the indicial equation are differ by an integer, making a coefficient indeterminate
Then general solution is given as follow:
y(x) = Au(x) + Bv(x) (2.32)
As all coefficients will become in terms of A0 and A1
Q-4: Solve the following differential equations in power series.
d2y
dx2
− 2x
dy
dx
+ 2ny = 0 (2.33)
Solution: First we compare the given differential equation (2.33) with (1.1), we have
8

9. P(x) = x, Q(x) = 2n are analytic at x = 0
Therefore x = 0 is ordinary point, so the the solution series of the given differential exists.
Putting the values of y, y , y from equations (2.3), (2.4) and (2.5) in equation (2.33), then we have
[A0c(c − 1)xc−2
+ A1(c + 1)cxc−1
+ A2(c + 2)(c + 1)xc
+ A3(c + 3)(c + 2)xc+1
+
A4(c + 4)(c + 3)xc+2
+ ... + Ak(c + k)(c + k − 1)xc+k−2
+ ....]
2x[A0cxc−1
+ A1(c + 1)xc
+ A2(c + 2)xc+1
+ A3(c + 3)xc+2
+
A4(c + 4)xc+3
+ ... + Ak(c + k)xc+k−1
+ ....]
+ 2n[A0xc
+ A1xc+1
+ A2xc+2
+ A3xc+3
+ A4xc+4
+ ... + Akxc+k
+ ....]
(2.34)
Step-2: Since, this equation satisfied the for all values of x, therefore co-efficient of different powers of x must
vanish separately. Equating the coefficient of like power, we have:
Coeff. of xc−2) A0c(c − 1) + A0c + A0 = 0
⇒ A0c(c − 1) = 0
But A0 = 0 ⇒ c(c − 1) = 0 and hence we have c = 0, 1
⇒ c1 = 0, c2 = 1
coeff. of xc−1)
A1c(c + 1) = 0 ⇒ A1 =
0
c(c − 1)
therefore c = 0 makes it indeterminate
coeff. of xc)
A2[(c + 1)(c + 2)] − 2A0c + 2nA0 = 0
A2[(c + 1)(c + 2)] = 2(c − n)A0 ⇒ A2 =
2(c − n)A0
(c + 1)(c + 2)
coeff. of xc+1)
A3[(c + 2)(c + 3)] − 2A1(c + 1) + 2nA1 = 0
A3[(c + 2)(c + 3)] = 2(c + 1 − n)A1 ⇒ A3 =
2(c + 1 − n)A0
(c + 2)(c + 3)
coeff. of xc+2)
A4[(c + 3)(c + 4)] − 2A2(c + 2) + 2nA2 = 0
A4[(c + 3)(c + 34)] = 2(c + 2 − n)A2 ⇒ A4 =
2(c + 2 − n)A0
(c + 3)(c + 4)
similarly, we have
A5 =
2(c + 3 − n)A0
(c + 4)(c + 5)
and so on Step-3: At c = 0 the values of A2, A3, A4, A5, ... reduces to the following form
A2 =
−2n
2!
A0
9

10. A3 =
−2(n − 1)
3!
A1
A4 =
22(n − 2)
4!
A0
A5 =
22(n − 1)(n − 3)
5!
A1
A6 =
−23(n − 2)(n − 4)
6!
A0
and so on ...
Now putting the values of A2, A3, A4, A5, A6, ... in the equation (2.3), after simplification we have the following
y(x) = A0 1 −
2n
2!
x2
+
22n(n − 2)
4!
x4
... + A1x 1 −
2(n − 1)
3!
x3
+
22(n − 1)(n − 3)
4!
x5
... (2.35)
Which is the required solution of the given differential equation.
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