1. Higher Order Differential Equation
with Variable Coefficients
University Of Zakho
Faculty of Science
Mathematic department
Prepared by:
Halat H. Omer
Iman N. Abdulah
Supervised by:
Miss. Hakima Khudher

2. Outlines
• Introduction
• Cauchy–Euler equation
• Cauchy-Euler Equation General Form
• Recurrence Relations - Method of Summation Factors
• Methodology
• Series Solutions of Differential Equations
• Function Representations
• Differentiation of Power Series Functions
• Ordinary Points
• A Power Series Example

3. Introduction
Up to this point in the class, we have only specifically studied solution techniques for first-order
differential equations, i.e., equations whose highest derivative is the first. For example,
𝑦′ − 2𝑦 = 4 − 𝑥
is a first-order differential equation; we solved it in section 1.5. We do not yet have the tools to
solve an equation like
𝑦′′
− 2𝑦 = 4 − 𝑥;
this equation is second-order since its highest derivative is the second. In this chapter, we will
study solution techniques for differential equations of order 2, like
𝑦′′
− 2𝑦 = 4 − 𝑥,

4. Introduction
as well as higher-order equations. However, before we proceed, it would be
helpful to spend a few moments refreshing our memories on the information
we know about first-order differential equations; the information we will
gather about higher-order equations, while complementary to what we have
learned about first-order equations, can at first glance seem contradictory, so it
will be helpful to solidify our understanding of first-order equations before
proceeding.

5. Cauchy–Euler equation
In mathematics, an Euler–Cauchy equation, or Cauchy–Euler equation, or
simply Euler's equation is a linear homogeneous ordinary differential equation
with variable coefficients. It is sometimes referred to as an equidimensional
equation. Because of its particularly simple equidimensional structure, the
differential equation can be solved explicitly.

6. Cauchy-Euler Equation General Form
The differentiation equation gives the Cauchy-Euler differential equation of
order n as
𝑎𝑛𝑥𝑛𝑦 𝑛 + 𝑎𝑛−1𝑥𝑛−1𝑦 𝑛−1 𝑥 + ⋯ + 𝑎0𝑦 𝑥 = 0.
Here, ai; i = 1, 2, 3,…, n are constants and an ≠ 0.

7. Example 1:
Solve: 𝑥2𝑦′′ − 6𝑥𝑦′– 18𝑦 = 0
Solution:
Given second order Cauchy-Euler equation is:
𝑥2
𝑦′′
− 6𝑥𝑦′
− 18𝑦 = 0
Let 𝑦 = 𝑥𝑟
and substitute in the given differential equation.
𝑥2𝑦′′ − 6𝑥𝑦′ − 18𝑦 = 0
𝑥2 𝑥𝑟 ′′ − 6𝑥 𝑥𝑟 ′ − 18 𝑥𝑟 = 0
𝑥2
𝑟(𝑟 − 1)𝑥𝑟−2
− 6𝑥 𝑟𝑥𝑟−1
− 18𝑥𝑟
= 0
𝑟2
− 𝑟 𝑥𝑟
− 6𝑟𝑥𝑟
− 18𝑥𝑟
= 0
𝑟2 − 𝑟 − 6𝑟 − 18 𝑥𝑟 = 0

8. ⇒ 𝑟2 − 7𝑟 − 18 = 0
⇒ 𝑟2
+ 2𝑟 − 9𝑟 − 18 = 0
⇒ 𝑟(𝑟 + 2) − 9(𝑟 + 2) = 0
⇒ (𝑟 + 2)(𝑟 − 9) = 0
⇒ 𝑟 = −2, 𝑟 = 9
So, r = -2 and 9 are the two distinct possible values of r.
Therefore, the set of fundamental solution is {x-2, x9} and the general solution is
y = c1x-2 + c2x9.

9. Recurrence Relations - Method of
Summation Factors
There is another way of solving recurrence relations of the form 𝐴𝑎𝑛 =
𝐵𝑎𝑛−1 + 𝐶, where 𝐴, 𝐵 and 𝐶 are functions of 𝑛, which some references
call the method of summation factors. This method is straight forward
when A and B are linear functions of 𝑛, and it solves several recurrence
relations problems in recreational mathematics and computer science,
among others.

10. Methodology
First, define a summation factor 𝑠𝑛​, which, when multiplied to the recurrence relation, gives
𝐴𝑠𝑛𝑎𝑛 = 𝐵𝑠𝑛𝑎𝑛−1 + 𝐶𝑠𝑛.
This summation factor must be chosen such that the "new" sequence of numbers has an (𝑛 −
1)th term of 𝐵𝑠𝑛𝑎𝑛−1 and an 𝑛th term of 𝐴𝑠𝑛𝑎𝑛. This factor is chosen to be
𝑠𝑛 =
𝐴𝑛−1𝐴𝑛−2 ⋯ 𝐴2𝐴1
𝐵𝑛𝐵𝑛−1 … 𝐵3𝐵2
Thus, we can now let 𝑏𝑛 = 𝐴𝑠𝑛𝑎𝑛 and 𝑏𝑛−1 = 𝐵𝑠𝑛𝑎𝑛−1 . The recurrence relation now
becomes
𝑏𝑛−1 = 𝑏𝑛 + 𝐶𝑠𝑛

11. where 𝐶𝑠𝑛 can be thought of as the difference between the (𝑛 − 1)th and the 𝑛th
term. Thus, the closed form of 𝑏𝑛 is
𝑏𝑛 = 𝑏0 +
𝑘=1
𝑛
𝐶𝑠𝑘
where 𝐶 is a function of the index 𝑘. But 𝑏0 = 𝐴𝑛=0𝑠0𝑎0 and 𝑏𝑛 = 𝐴𝑠𝑛𝑎𝑛.
Therefore, the closed form of the original recurrence relation is
𝑎𝑛 =
1
𝐴𝑠𝑛
𝐴𝑛=0𝑠0𝑎0 +
𝑘=1
𝑛
𝐶𝑠𝑘 .

12. Example:
Find the closed form of the recurrence relation 𝑇0 = 0, 𝑇𝑛 = 2𝑇𝑛−1 + 1
Solution:
The summation factor 𝑠𝑛 with 𝐴(𝑛) = 1 and 𝐵(𝑛) = 2 is
𝑠𝑛 =
1 × 1 × ⋯ × 1
2 × 2 × ⋯ × 2
=
1
2𝑛−1
where the 1 's and 2's occur 𝑛 − 1 times each. Multiplying this to the recurrence
relation gives
1
2𝑛−1
𝑇𝑛 =
1
2𝑛−2
𝑇𝑛−1 +
1
2𝑛−1
Let 𝑏𝑛 =
1
2𝑛−1 𝑇𝑛. Then 𝑏𝑛−1 =
1
2𝑛−2 𝑇𝑛−1 and 𝑏0 =
1
20−1 𝑇0 = 0. So ,
𝑏𝑛 = 𝑏𝑛−1 +
1
2𝑛−1

13. It follows that the closed form of 𝑏𝑛 is
𝑏𝑛 = 0 +
𝑘=1
𝑛
1
2𝑘−1
=
1 ⋅
1
2
𝑛
− 1
1
2
− 1
= 2 −
1
2𝑛−1
.
Note that the summation above is a finite geometric series. But since 𝑏𝑛 =
1
2𝑛−1 𝑇𝑛, the closed form of 𝑇𝑛 is
𝑇𝑛 = 2𝑛−1 2 −
1
2𝑛−1
= 2𝑛 − 1.

14. Series Solutions of Differential Equations
Previously, we studied how functions can be represented as power series,
𝑦(𝑥) =
𝑛=0
∞
𝑎𝑛𝑥𝑛
We also saw that we can find series representations of the derivatives of such
functions by differentiating the power series term by term. This gives
𝑦′(𝑥) =
𝑛=1
∞
𝑛𝑎𝑛𝑥𝑛−1
And
𝑦″(𝑥) =
𝑛=2
∞
𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛−2.

15. In some cases, these power series representations can be used to find
solutions to differential equations.
The examples and exercises in this section were chosen for which
power solutions exist. However, it is not always the case that power
solutions exist.

16. Example:
𝑦″
− 𝑦 = 0
Solution:
Assume
𝑦 𝑥 =
𝑛=0
∞
𝑎𝑛𝑥𝑛
Then,
𝑦′(𝑥) =
𝑛=1
∞
𝑛𝑎𝑛𝑥𝑛−1
And
𝑦″(𝑥) =
𝑛=2
∞
𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛−2

17. We want to find values for the coefficients 𝑎𝑛 such that
𝑦″ − 𝑦 = 0
𝑛=2
∞
𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛−2 −
𝑛=0
∞
𝑎𝑛𝑥𝑛 = 0 .
𝑛=2
∞
𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛−2
=
𝑛=0
∞
(𝑛 + 2)(𝑛 + 1)𝑎𝑛+2𝑥𝑛
.
This gives
𝑛=0
∞
(𝑛 + 2)(𝑛 + 1)𝑎𝑛+2𝑥𝑛 −
𝑛=0
∞
𝑎𝑛𝑥𝑛 = 0
𝑛=0
∞
[(𝑛 + 2)(𝑛 + 1)𝑎𝑛+2 − 𝑎𝑛]𝑥𝑛 = 0.

18. 𝑛=2
∞
𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛−2 =
𝑛=0
∞
(𝑛 + 2)(𝑛 + 1)𝑎𝑛+2𝑥𝑛 .
This gives
𝑛=0
∞
(𝑛 + 2)(𝑛 + 1)𝑎𝑛+2𝑥𝑛 −
𝑛=0
∞
𝑎𝑛𝑥𝑛 = 0
𝑛=0
∞
[(𝑛 + 2)(𝑛 + 1)𝑎𝑛+2 − 𝑎𝑛]𝑥𝑛 = 0.
𝑛 + 2 𝑛 + 1 𝑎𝑛+2 − 𝑎𝑛 = 0 for 𝑛 = 0,1,2, … .
𝑎2 =
𝑎0
2
𝑎4 =
𝑎2
4 ⋅ 3
=
𝑎0
4!
𝑎6 =
𝑎4
6 ⋅ 5
=
𝑎0
6!
⋮

19. Thus, in general, when 𝑛 is even,
𝑎𝑛 =
𝑎0
𝑛!
.
For the equations involving odd values of 𝑛, we see that
𝑎3 =
𝑎1
3 ⋅ 2
=
𝑎1
3!
𝑎5 =
𝑎3
5 ⋅ 4
=
𝑎1
5!
𝑎7 =
𝑎5
7 ⋅ 6
=
𝑎1
7!
⋮
Therefore, in general, when 𝑛 is odd,
𝑎𝑛 =
𝑎1
𝑛!
.

20. Putting this together, we have
𝑦 𝑥 =
𝑛=0
∞
𝑎𝑛𝑥𝑛
= 𝑎0 + 𝑎1𝑥 +
𝑎0
2
𝑥2
+
𝑎1
3!
𝑥3
+
𝑎0
4!
𝑥4
+
𝑎1
5!
𝑥5
+ ⋯ .
Re-indexing the sums to account for the even and odd values of 𝑛 separately, we obtain
𝑦 𝑥 = 𝑎0
𝑘=0
∞
1
2𝑘 !
𝑥2𝑘
+ 𝑎1
𝑘=0
∞
1
2𝑘 + 1 !
𝑥2𝑘+1
.

21. Function Representations
We can define a function f (x) using a power series as long as we restrict the
domain of f to the interval of convergence of the series. The function
𝑓 𝑥 =
𝑛=0
∞
𝑐𝑛(𝑥 − 𝑎)𝑛
is analytic at a point x if it can be defined at x by a power series.

22. Differentiation of Power Series
Functions
Functions defined by power series can be differentiated term-by-term. For
the function f (x) given above, this yields:
𝑓′ 𝑥 =
𝑛=1
∞
𝑐𝑛𝑛(𝑥 − 𝑎)𝑛−1 𝑎𝑛𝑑 𝑓′′ 𝑥
=
𝑛=2
∞
𝑐𝑛𝑛(𝑛 − 1)(𝑥 − 𝑎)𝑛−2

23. Ordinary Points
As we will see, there are two different types of power series solution methods.
In this section, we examine the first.
Ordinary Points
A point x0 is said to be an ordinary point of the differential equation 𝑦′′ +
𝑃(𝑥)𝑦′ + 𝑄(𝑥)𝑦 = 0 if both 𝑃(𝑥) and 𝑄(𝑥) are analytic at 𝑥0. A point is
singular if it is not ordinary.

24. Theorem
If 𝑥 = 𝑥0 is an ordinary point of a differential equation 𝑦′′ + 𝑃(𝑥)𝑦′ +
𝑄(𝑥)𝑦 = 0, we can always find two linearly independent solutions in the
form of power series centered at 𝑥0. A series solution converges at least on
some interval defined by |𝑥 − 𝑥0| < 𝑅 where R is the distance from x0 to
the closet singular point.
Note:
Such a solution is called a solution about an ordinary point.

25. A Power Series Example
Find two power series solutions to the equation 𝑦′′ − 2𝑥𝑦′ + 𝑦 = 0 about the
ordinary point 𝑥 = 0.
Assume the solution is of the form 𝑦(𝑥) = ∑𝑛=0
∞
𝑐𝑛𝑥𝑛. Since there are no finite
singular points, we have convergence on (−∞, ∞).
𝑛=0
∞
𝑐𝑛+2(𝑛 + 1)(𝑛 + 1) − 2𝑛𝑐𝑛 + 𝑐𝑛 𝑥𝑛 = 0
𝑦 = 𝑐0 1 −
1
2
𝑥2
−
3
4!
𝑥4
− ⋯ + 𝑐1 𝑥 +
1
3!
𝑥3
+
5
5!
𝑥5
+ ⋯

26. References
1. https://www.researchgate.net/publication/302204222_Higher-
Order_Differential_Equations_with_Variable_Coefficients
2. https://sites.lafayette.edu/thompsmc/files/2014/10/Section_3_11.pdf
3. https://en.m.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation
4. https://byjus.com/maths/cauchy-euler-equation/
5. https://brilliant.org/wiki/recurrence-relations-method-of-summation-factors/
6. https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/17%3A_Second-
Order_Differential_Equations/17.04%3A_Series_Solutions_of_Differential_Equations
7. https://users.math.msu.edu/users/gnagy/teaching/12-spring/mth235/L16-235.pdf
8. http://math.wallawalla.edu/~duncjo/courses/math312/spring07/notes/6-1_math312.pdf

27. THANK
YOU