7.4 Series solutions of differential equations

Calculus volume 3 Page 1 / 2

Use power series to solve first-order and second-order differential equations.

In Introduction to Power Series , we studied how functions can be represented as power series, $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} .$ 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 $y^{'} (x) = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ and $y ″ (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} .$ In some cases, these power series representations can be used to find solutions to differential equations.

Be aware that this subject is given only a very brief treatment in this text. Most introductory differential equations textbooks include an entire chapter on power series solutions. This text has only a single section on the topic, so several important issues are not addressed here, particularly issues related to existence of solutions. 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. Those of you interested in a more rigorous treatment of this topic should consult a differential equations text.

Problem-solving strategy: finding power series solutions to differential equations

Assume the differential equation has a solution of the form $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} .$
Differentiate the power series term by term to get $y^{'} (x) = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ and $y ″ (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} .$
Substitute the power series expressions into the differential equation.
Re-index sums as necessary to combine terms and simplify the expression.
Equate coefficients of like powers of $x$ to determine values for the coefficients $a_{n}$ in the power series.
Substitute the coefficients back into the power series and write the solution.

Series solutions to differential equations

Find a power series solution for the following differential equations.

$y ″ - y = 0$
$(x^{2} - 1) y ″ + 6 x y^{'} + 4 y = −4$

Assume $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ (step 1). Then, $y^{'} (x) = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ and $y ″ (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$ (step 2). We want to find values for the coefficients $a_{n}$ such that

$\begin{matrix} y ″ - y & = & 0 \\ \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - \sum_{n = 0}^{\infty} a_{n} x^{n} & = & 0 (step 3). \end{matrix}$

We want the indices on our sums to match so that we can express them using a single summation. That is, we want to rewrite the first summation so that it starts with $n = 0 .$
To re-index the first term, replace n with $n + 2$ inside the sum, and change the lower summation limit to $n = 0 .$ We get

$\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} = \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} .$

This gives

$\begin{matrix} \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} - \sum_{n = 0}^{\infty} a_{n} x^{n} & = & 0 \\ \sum_{n = 0}^{\infty} [(n + 2) (n + 1) a_{n + 2} - a_{n}] x^{n} & = & 0 (step 4). \end{matrix}$

Because power series expansions of functions are unique, this equation can be true only if the coefficients of each power of x are zero. So we have

$(n + 2) (n + 1) a_{n + 2} - a_{n} = 0 for n = 0, 1, 2,….$

This recurrence relationship allows us to express each coefficient $a_{n}$ in terms of the coefficient two terms earlier. This yields one expression for even values of n and another expression for odd values of n . Looking first at the equations involving even values of n , we see that

$\begin{matrix} a_{2} & = & \frac{a_{0}}{2} \\ a_{4} & = & \frac{a_{2}}{4 \cdot 3} = \frac{a_{0}}{4!} \\ a_{6} & = & \frac{a_{4}}{6 \cdot 5} = \frac{a_{0}}{6!} \\ ⋮. \end{matrix}$

Thus, in general, when n is even, $a_{n} = \frac{a_{0}}{n!}$ (step 5).
For the equations involving odd values of n , we see that

$\begin{matrix} a_{3} & = & \frac{a_{1}}{3 \cdot 2} = \frac{a_{1}}{3!} \\ a_{5} & = & \frac{a_{3}}{5 \cdot 4} = \frac{a_{1}}{5!} \\ a_{7} & = & \frac{a_{5}}{7 \cdot 6} = \frac{a_{1}}{7!} \\ ⋮. \end{matrix}$

Therefore, in general, when n is odd, $a_{n} = \frac{a_{1}}{n!}$ (step 5 continued).
Putting this together, we have

$\begin{matrix} y (x) & = \sum_{n = 0}^{\infty} a_{n} x^{n} \\ = a_{0} + a_{1} x + \frac{a_{0}}{2} x^{2} + \frac{a_{1}}{3!} x^{3} + \frac{a_{0}}{4!} x^{4} + \frac{a_{1}}{5!} x^{5} + ⋯. \end{matrix}$

Re-indexing the sums to account for the even and odd values of n separately, we obtain

$y (x) = a_{0} \sum_{k = 0}^{\infty} \frac{1}{(2 k)!} x^{2 k} + a_{1} \sum_{k = 0}^{\infty} \frac{1}{(2 k + 1)!} x^{2 k + 1} (step 6).$

Analysis for part a.
As expected for a second-order differential equation, this solution depends on two arbitrary constants. However, note that our differential equation is a constant-coefficient differential equation, yet the power series solution does not appear to have the familiar form (containing exponential functions) that we are used to seeing. Furthermore, since $y (x) = c_{1} e^{x} + c_{2} e^{− x}$ is the general solution to this equation, we must be able to write any solution in this form, and it is not clear whether the power series solution we just found can, in fact, be written in that form.
Fortunately, after writing the power series representations of $e^{x}$ and $e^{− x},$ and doing some algebra, we find that if we choose

$c_{0} = \frac{(a_{0} + a_{1})}{2}, c_{1} = \frac{(a_{0} - a_{1})}{2},$

we then have $a_{0} = c_{0} + c_{1}$ and $a_{1} = c_{0} - c_{1},$ and

$\begin{matrix} y (x) & = a_{0} + a_{1} x + \frac{a_{0}}{2} x^{2} + \frac{a_{1}}{3!} x^{3} + \frac{a_{0}}{4!} x^{4} + \frac{a_{1}}{5!} x^{5} + ⋯ \\ = (c_{0} + c_{1}) + (c_{0} - c_{1}) x + \frac{(c_{0} + c_{1})}{2} x^{2} + \frac{(c_{0} - c_{1})}{3!} x^{3} + \frac{(c_{0} + c_{1})}{4!} x^{4} + \frac{(c_{0} - c_{1})}{5!} x^{5} + ⋯ \\ = c_{0} \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} + c_{1} \sum_{n = 0}^{\infty} \frac{{(− x)}^{n}}{n!} \\ = c_{0} e^{x} + c_{1} e^{− x} . \end{matrix}$

So we have, in fact, found the same general solution. Note that this choice of $c_{1}$ and $c_{2}$ is not obvious. This is a case when we know what the answer should be, and have essentially “reverse-engineered” our choice of coefficients.
Assume $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ (step 1). Then, $y^{'} (x) = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ and $y ″ (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$ (step 2). We want to find values for the coefficients $a_{n}$ such that

$\begin{matrix} (x^{2} - 1) y ″ + 6 x y^{'} + 4 y & = & −4 \\ (x^{2} - 1) \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} + 6 x \sum_{n = 1}^{\infty} n a_{n} x^{n - 1} + 4 \sum_{n = 0}^{\infty} a_{n} x^{n} & = & −4 \\ x^{2} \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} + 6 x \sum_{n = 1}^{\infty} n a_{n} x^{n - 1} + 4 \sum_{n = 0}^{\infty} a_{n} x^{n} & = & −4 . \end{matrix}$

Taking the external factors inside the summations, we get

$\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} - \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} + \sum_{n = 1}^{\infty} 6 n a_{n} x^{n} + \sum_{n = 0}^{\infty} 4 a_{n} x^{n} = −4 (step 3).$

Now, in the first summation, we see that when $n = 0$ or $n = 1,$ the term evaluates to zero, so we can add these terms back into our sum to get

$\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} = \sum_{n = 0}^{\infty} n (n - 1) a_{n} x^{n} .$

Similarly, in the third term, we see that when $n = 0,$ the expression evaluates to zero, so we can add that term back in as well. We have

$\sum_{n = 1}^{\infty} 6 n a_{n} x^{n} = \sum_{n = 0}^{\infty} 6 n a_{n} x^{n} .$

Then, we need only shift the indices in our second term. We get

$\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} = \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} .$

Thus, we have

$\begin{matrix} \sum_{n = 0}^{\infty} n (n - 1) a_{n} x^{n} - \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} + \sum_{n = 0}^{\infty} 6 n a_{n} x^{n} + \sum_{n = 0}^{\infty} 4 a_{n} x^{n} & = & −4 (step 4). \\ \sum_{n = 0}^{\infty} [n (n - 1) a_{n} - (n + 2) (n + 1) a_{n + 2} + 6 n a_{n} + 4 a_{n}] x^{n} & = & −4 \\ \sum_{n = 0}^{\infty} [(n^{2} - n) a_{n} + 6 n a_{n} + 4 a_{n} - (n + 2) (n + 1) a_{n + 2}] x^{n} & = & −4 \\ \sum_{n = 0}^{\infty} [n^{2} a_{n} + 5 n a_{n} + 4 a_{n} - (n + 2) (n + 1) a_{n + 2}] x^{n} & = & −4 \\ \sum_{n = 0}^{\infty} [(n^{2} + 5 n + 4) a_{n} - (n + 2) (n + 1) a_{n + 2}] x^{n} & = & −4 \\ \sum_{n = 0}^{\infty} [(n + 4) (n + 1) a_{n} - (n + 2) (n + 1) a_{n + 2}] x^{n} & = & −4 \end{matrix}$

Looking at the coefficients of each power of x , we see that the constant term must be equal to $−4,$ and the coefficients of all other powers of x must be zero. Then, looking first at the constant term,

$\begin{matrix} 4 a_{0} - 2 a_{2} & = & −4 \\ a_{2} & = & 2 a_{0} + 2 (step 3). \end{matrix}$

For $n \geq 1,$ we have

$\begin{matrix} (n + 4) (n + 1) a_{n} - (n + 2) (n + 1) a_{n + 2} & = & 0 \\ (n + 1) [(n + 4) a_{n} - (n + 2) a_{n + 2}] & = & 0 . \end{matrix}$

Since $n \geq 1,$ $n + 1 \neq 0,$ we see that

$(n + 4) a_{n} - (n + 2) a_{n + 2} = 0$

and thus

$a_{n + 2} = \frac{n + 4}{n + 2} a_{n} .$

For even values of n , we have

$\begin{matrix} a_{4} & = & \frac{6}{4} (2 a_{0} + 2) = 3 a_{0} + 3 \\ a_{6} & = & \frac{8}{6} (3 a_{0} + 3) = 4 a_{0} + 4 \\ ⋮. \end{matrix}$

In general, $a_{2 k} = (k + 1) (a_{0} + 1)$ (step 5).
For odd values of n , we have

$\begin{matrix} a_{3} & = & \frac{5}{3} a_{1} \\ a_{5} & = & \frac{7}{5} a_{3} = \frac{7}{3} a_{1} \\ a_{7} & = & \frac{9}{7} a_{5} = \frac{9}{3} a_{1} = 3 a_{1} \\ ⋮. \end{matrix}$

In general, $a_{2 k + 1} = \frac{2 k + 3}{3} a_{1}$ (step 5 continued).
Putting this together, we have

$y (x) = \sum_{k = 0}^{\infty} (k + 1) (a_{0} + 1) x^{2 k} + \sum_{k = 0}^{\infty} (\frac{2 k + 3}{3}) a_{1} x^{2 k + 1} (step 6).$

Got questions? Get instant answers now!

Questions & Answers

Biology is a branch of Natural science which deals/About living Organism.

Ahmedin Reply

what is phylogeny

Odigie Reply

evolutionary history and relationship of an organism or group of organisms

AI-Robot

Deng

what is biology

Hajah Reply

cell is the smallest unit of the humanity biologically

Abraham

what is biology

Victoria Reply

what is biology

Abraham

HOW CAN MAN ORGAN FUNCTION

Alfred Reply

the diagram of the digestive system

Assiatu Reply

allimentary cannel

Ogenrwot

How does twins formed

William Reply

They formed in two ways first when one sperm and one egg are splited by mitosis or two sperm and two eggs join together

Oluwatobi

what is genetics

Josephine Reply

Genetics is the study of heredity

Misack

how does twins formed?

Misack

What is manual

Hassan Reply

discuss biological phenomenon and provide pieces of evidence to show that it was responsible for the formation of eukaryotic organelles

Joseph Reply

what is biology

Yousuf Reply

the study of living organisms and their interactions with one another and their environment.

Wine

discuss the biological phenomenon and provide pieces of evidence to show that it was responsible for the formation of eukaryotic organelles in an essay form

Joseph Reply

what is the blood cells

Shaker Reply

list any five characteristics of the blood cells

Shaker

lack electricity and its more savely than electronic microscope because its naturally by using of light

Abdullahi Reply

advantage of electronic microscope is easily and clearly while disadvantage is dangerous because its electronic. advantage of light microscope is savely and naturally by sun while disadvantage is not easily,means its not sharp and not clear

Abdullahi

cell theory state that every organisms composed of one or more cell,cell is the basic unit of life

Abdullahi

is like gone fail us

DENG

cells is the basic structure and functions of all living things

Ramadan

What is classification

ISCONT Reply

is organisms that are similar into groups called tara

Yamosa

Got questions? Join the online conversation and get instant answers!

Jobilize.com Reply

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 3' conversation and receive update notifications?

Ask

	How much do you love him? By Zarina Chocolate Start Quiz
	Unit 1 Geography Test By Qqq Qqq Start Flashcards
	1 CDL Quiz - Air Brakes Endorsement Part 1 By Jazzycazz Jackson Start Quiz
	OOP with Java - Quiz 1 By Vongkol HENG Start Quiz
	Principles of economics By OpenStax Read Online Course
	English Proficiency Test By Anindyo Mukhopadhyay Start Quiz
	Managerial Psychology Exam By Dan Ariely Start Exam
	6 AP 06 Skeletal System MCQ By OpenStax Start Quiz
	3 Week 3 Social Psych By Yacoub Jayoghli Start Quiz
	34 Biology 34 Animal Nutrition Digestive System By OpenStax Start Quiz