Series solution of ordinary differential equation advance engineering mathematics The power series method is the standard method for solving linear ODEs with variable coefficients. It gives solutions in the form of power series. These series can be used for computing values, graphing curves, proving formulas, and exploring properties of solutions, as we shall see.

1. Series solution of ordinary differential
equation
PRANAV VEERANI 130280111117 311225
SHIKHA VERMA 130280111118 311226
DHWANI WAGHELA 130280111121 311227

2. Power series
• The power series method is the standard
method for solving linear ODEs with variable
coefficients. It gives solutions in the form of
power series. These series can be used for
computing values, graphing curves, proving
formulas, and exploring properties of
solutions, as we shall see.

3. Power Series

4. Ordinary and Singular Point
• Consider the linear differential equation
Y”+𝑃(𝑋)Y’+𝑄(𝑋)Y =0
where 𝑃(𝑥)and 𝑄(𝑥)are functions of x only.
• Ordinary Point
The point x=𝑥0 is called an ordinary point of the
equation Y”+𝑃(𝑋)Y’+𝑄(𝑋)Y =0 if both 𝑃(𝑋)and
𝑄(𝑋)are finite at x=𝑥0.

5. SINGULAR POINT
• Singular Point
The point x=𝑥0 is called a singular point of the
equation Y”+𝑃(𝑋)Y’+𝑄(𝑋)Y =0 if 𝑃(𝑋) or 𝑄(𝑋) or
both are infinite at x=𝑥0.
There are two types of singular points:
(1)Regular Singular Point:
The point x=𝑥0 is called regular singular point of
the equation Y”+𝑃(𝑋)Y’+𝑄(𝑋)Y =0 if both (x − 𝑥0)
𝑃(𝑋) and (𝑥 − 𝑥0)2
𝑄(𝑋) are finite at x=𝑥0.

6. SINGULAR POINT
(2)Irregular Singular Point
The point x=𝑥0 is called an irregular singular
point of the equation Y”+𝑃(𝑋)Y’+𝑄(𝑋)Y =0
if(x − 𝑥0) 𝑃(𝑋) or (𝑥 − 𝑥0)2
𝑄(𝑋) or both are
infinite at x=𝑥0.

7. EXAMPLE
(i)Determine if x=1 is a regular singular point of
(1− 𝑥2
)y" − 2xy’+n(n+1)y=0 where n is a
constant
Solution: Dividing both sides by (1− 𝑥2
),
y”---
2𝑥
1−𝑥2 y’+
𝑛(𝑛+1)
1−𝑥2 y=0
ComparingwithY”+𝑃(𝑋)Y’+𝑄(𝑋)Y =0,
𝑃(𝑋)= −
2𝑥
1−𝑥2,𝑄 𝑥 =
𝑛(𝑛+1)
1−𝑥2

8. EXAMPLE
• P(x) and Q(x) are infinite at x=1,
• X=1is a singular point.
• Now,
(x-1)P(x)=(x-1)(
−2𝑥
1−𝑥2)=
2𝑥
1+𝑥
(𝑥 − 1)2
Q(x)= (𝑥 − 1)2
(
𝑛(𝑛+1)
(1−𝑥)2 )=
1−𝑥 𝑛(𝑛+1)
1+𝑥
Both (x-1)P(x) and (𝑥 − 1)2
Q(x) are finite at x=1.
X=1 is regular singular point.

9. EXAMPLE
(ii)Determine the singular points of differential
equation 2x(𝑥 − 1)2
Y”+3xY’+(x-2)Y=0 and
classify them as regular or irregular.
• Dividing both sides by 2x(𝑥 − 1)2
,
• Y”+
3
2(𝑥−2)2Y’+
1
2𝑥(𝑥−2)
Y=0
• Comparing with Y”+P(x)Y’+Q(x)Y=0,
• P(x)=
3
2(𝑥−2)2 ,Q(x)=
1
2𝑥(𝑥−2)

10. EXAMPLE
• At x=2,P(x) is infinite.
• At x=0,2 ,Q(x) is infinite.
• X=0 and x=2 are singular points.
• For x=0
• (x-0)P(x)=x
3
2(𝑥−2)2
• (𝑥 − 0)2
Q(x)=𝑥2 1
2𝑥(𝑥−2)
=
𝑥
2(𝑥−2)
• Both (x-0)P(x) and (𝑥 − 0)2
Q(x) are finite at x=0.

11. EXAMPLE
• X=0 is a regular singular point.
• For x=2
• (x-2)P(x)=(x-2)
3
2(𝑥−2)2=
3
2(𝑥−2)
• (𝑥 − 2)2
Q(x)=(𝑥 − 2)2 1
2𝑥(𝑥−2)
=
𝑥−2
2𝑥
• (x-2)P(x) is infinite at x=2.
• x=2 is an irregular singular point.

12. Power series solution about ordinary
point
• A power series solution of a differential
equation Y”+P(x)Y’+Q(x)Y=0 about ordinary
point x=𝑥0 can be obtained using the
following steps.
1 Assume that Y= 𝑘=0
∞
𝑎 𝑘 (𝑥 − 𝑥0) 𝑘
be the
solution of the given differential equation.
2 Find Y’= 𝑘=0
∞
𝑘 𝑎 𝑘(𝑥 − 𝑥0) 𝑘−1
and
Y”= 𝑘=0
∞
𝑘 (𝑘 − 1)𝑎 𝑘(𝑥 − 𝑥0) 𝑘−2

13. SOLUTION OF ORDINARY POINT
3 Substitute the value of Y , Y’ and Y” in given
differential equation.
4 Equate to zero the coefficients of various
powers of x and find 𝑎2,𝑎3,𝑎4 _ _ _etc.in terms
of 𝑎0and 𝑎1.
5 Substitute the values of the constants
𝑎2,𝑎3,𝑎4 _ _ _ in Y= 𝑘=0
∞
𝑎 𝑘 (𝑥 − 𝑥0) 𝑘
which
will be the required power series solution.

14. Mclaurin series
We see that Taylor series are power series.
From the last section we know that power
series represent analytic functions. And we
now show that every analytic function can be
represented by power series, namely, by
Taylor series (with various centers). A
Maclaurin series3 is a Taylor series with
center
00
z
00
z

15. Important maclurin series
𝑒 𝑥
= 1 +
𝑥
1!
+
𝑥2
2!
+
𝑥3
3!
+ ⋯ , −∞ < 𝑥 < ∞
cos 𝑥 = 1 −
𝑥2
2!
+
𝑥4
4!
−
𝑥6
6!
+ ⋯
sin 𝑥 = 𝑥 −
𝑥3
3!
+
𝑥5
5!
−
𝑥7
7!
+ ⋯
cos ℎ𝑥 = 1 +
𝑥2
2!
+
𝑥4
4!
+
𝑥6
6!
+ ⋯
sin ℎ𝑥 = 𝑥 +
𝑥3
3!
+
𝑥5
5!
+
𝑥7
7!
+ ⋯

16. SERIES FORMULAS
• Log(1+x)=x −
𝑥2
2
+
𝑥3
3
−
𝑥2
4
+----
• (1 + 𝑥)−1
= 1 − x + 𝑥2
− 𝑥3
+ 𝑥4
−------
• (1 − 𝑥)−1
= 1 + x + 𝑥2
+ 𝑥3
+ 𝑥4
+------

17. EXAMPLE

18. EXAMPLE
Y’= 𝑘=0
∞
𝑘𝑎 𝑘 𝑥 𝑘−1
Substituting above values in (1),we get
𝑘=0
∞
𝑘𝑎 𝑘 𝑥 𝑘−1- 2X 𝑘=0
∞
𝑎 𝑘=0
𝑘=0
∞
𝑘𝑎 𝑘 𝑥 𝑘−1- 𝑘=0
∞
2 𝑎 𝑘 𝑥 𝑘+1
=0
(kk-2)
𝑘=0
∞
𝑘𝑎 𝑘 𝑥 𝑘−1- 𝑘=2
∞
2 𝑎 𝑘−2 𝑥 𝑘−1
=0
𝑎1+ 𝑘=2
∞
𝑘𝑎 𝑘 𝑥 𝑘−1- 𝑘=2
∞
2 𝑎 𝑘−2 𝑥 𝑘−1
=0

19. EXAMPLE
𝑎1+ 𝑘=2
∞
[𝑘𝑎 𝑘 − 2𝑎 𝑘−2]𝑥 𝑘−1
=0
Equating the coefficients of
𝑥0
: 𝑎1=0
𝑥 𝑘−1
: k𝑎 𝑘-2𝑎 𝑘−2=0
𝑎 𝑘=
2
𝑘
𝑎 𝑘−2,k>=2
This is called the recurrence relation.
For k=2, 𝑎2=
2
2
𝑎0=𝑎0

20. EXAMPLE
For k=3, 𝑎3=
2
3
𝑎1=0
For k=4, 𝑎4=
2
4
𝑎2=
1
2
𝑎0
For k=5, 𝑎5=
2
5
𝑎3=0
For k=6, 𝑎6=
2
6
𝑎4=
1
3
(
𝑎0
2
)
Substituting above values in (2),we get
Y= 𝑘=0
∞
𝑎 𝑘 𝑥 𝑘

21. EXAMPLE
=𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥2
+ 𝑎3 𝑥3
+ 𝑎4 𝑥4
+ 𝑎5 𝑥5
+
𝑎6 𝑥6
+-------
=𝑎0 + 𝑎0 𝑥2
+
𝑎0 𝑥4
2
+
𝑎0 𝑥6
6
+-----
=𝑎0(1 + 𝑥2
+
𝑥4
2
+
𝑥6
6
+--)
=𝑎0 𝑒 𝑥2
basis ={𝑒 𝑥2
}

22. EXAMPLE
By power series method , solve
(1 − 𝑥2
)Y”-2xY’+2Y=0
Given equation
(1 − 𝑥2
)Y”-2xY’+2Y=0
Comparing with Y”+P(x)Y’+Q(x)Y=0,
P(x)=
−2𝑥
1−𝑥2 , Q(x)=
2𝑦
1−𝑥2
Both P(x) and Q(x) are finite at x=0.

23. EXAMPLE
• X=0 is an ordinary point.
Let the series solution of (1) be
Y= 𝑘=0
∞
𝑎 𝑘 𝑥 𝑘
Y’= 𝑘=0
∞
𝑘 𝑎 𝑘 𝑥 𝑘−1
Y”= 𝑘=0
∞
𝑘 (𝑘 − 1)𝑎 𝑘 𝑥 𝑘−2
Substituting above values in (1),we get
(1-𝑥2
) 𝑘=0
∞
𝑘 (𝑘 − 1)𝑎 𝑘 𝑥 𝑘−2
-
2x 𝑘=0
∞
𝑘 𝑎 𝑘 𝑥 𝑘−1
+2 𝑘=0
∞
𝑎 𝑘 𝑥 𝑘=0

24. EXAMPLE
• 𝑘=0
∞
𝑘 (𝑘 − 1)𝑎 𝑘 𝑥 𝑘−2
− 𝑘=0
∞
𝑘 (𝑘 − 1)𝑎 𝑘 𝑥 𝑘
−
𝑘=0
∞
2𝑘 𝑎 𝑘 𝑥 𝑘
+ 𝑘=0
∞
2 𝑎 𝑘 𝑥 𝑘
=0
• 𝑘=0
∞
𝑘 (𝑘 − 1)𝑎 𝑘 𝑥 𝑘−2
− 𝑘=0
∞
[𝑘 ( 𝑘 −

25. EXAMPLE
• 𝑘=0
∞
𝑘 (𝑘 − 1)𝑎 𝑘 𝑥 𝑘−2
−
𝑘=2
∞
[𝑘(𝑘 − 3)]𝑎 𝑘−2 𝑥 𝑘−2
=0
• 𝑘=2
∞
𝑘 (𝑘 − 1)𝑎 𝑘 𝑥 𝑘−2
−
𝑘=2
∞
[𝑘(𝑘 − 3)]𝑎 𝑘−2 𝑥 𝑘−2
=0
Equating the coefficient of
𝑥 𝑘−2
: k k − 1 𝑎 𝑘 − k k − 3 𝑎 𝑘−2 = 0
𝑎 𝑘=
𝑘(𝑘−3)
𝑘(𝑘−1)
𝑎 𝑘−2, k ≥ 2

26. EXAMPLE
• 𝑎 𝑘=
(𝑘−3)
(𝑘−1)
𝑎 𝑘−2, k ≥ 2
• For For k=2, 𝑎2 = −𝑎0
For k=3, 𝑎3 = 0
For k=4, 𝑎4 =
1
3
𝑎2 =
−1
3
𝑎0
For k=5, 𝑎5 =
2
4
𝑎3 = 0
For k=6, 𝑎6 =
3
5
𝑎4 =
3
5
−1
3
𝑎0 =
−1
5
𝑎0
Substituting above values in (2) ,we get

27. EXAMPLE
• Y= 𝑘=0
∞
𝑎 𝑘 𝑥 𝑘
= 𝑎0 +𝑎1 𝑥 + 𝑎2 𝑥2
+ 𝑎3 𝑥3
+ 𝑎4 𝑥4
+ 𝑎5 𝑥5
+
𝑎6 𝑥6
+-------
= 𝑎0 +𝑎1 𝑥 − 𝑎0 𝑥2
−
𝑎0 𝑥4
3
−
𝑎 𝑜 𝑥6
5
− −----
= 𝑎0 (1 − 𝑥2
−
𝑥4
3
−
𝑥6
5
− −----)+𝑎1x