Series Solutions: Introduction

Throughout these pages I will assume that you are familiar with power series and the concept of the radius of convergence of a power series.

Let us consider the example of the second order differential equation

\begin{displaymath}y''+4y=0,\quad y(0)=1,\ y'(0)=0.\end{displaymath}

You probably already know that the solution is given by $y(t)=\cos(2t)$; recall that this function has as its Taylor series

cos(2t) = 1 - $\displaystyle {\frac{{2^2 t^2}}{{2!}}}$ + $\displaystyle {\frac{{2^4 t^4}}{{4!}}}$ - $\displaystyle {\frac{{2^6 t^6}}{{6!}}}$ +...,


and that this representation is valid for all $t\in\Bbb R$.

Do you remember how to compute the Taylor series expansion for a given function?

If everything works out, then

\begin{displaymath}y(t)=a_0+a_1\cdot t+a_2 \cdot t^2 + a_3\cdot t^3\ldots,\end{displaymath}

where

\begin{displaymath}a_n=\frac{y^{(n)}(0)}{n!}.\end{displaymath}

Here y(n)(0) denotes the nth derivative of y at the point t=0.

What does this mean? To compute the Taylor series for the solution to our differential equation, we just have to compute its derivatives. Note that the initial conditions give us a head start: y(0)=1, so a0=1. y'(0)=0, so a1=0.

We can rewrite the differential equation as

y''=-4y,

so in particular

\begin{displaymath}y''(0)=-4y(0)=-4\cdot 1=-4.\end{displaymath}

Consequently a2=-4/2!=-2. By now, we have figured out that the solution to our differential equation has as its second degree Taylor polynomial the function 1-2t2.

Next we differentiate

y''=-4y

on both sides with respect to t, to obtain

y'''=-4y',

so in particular

y'''(0)=-4y'(0)=0,

yielding a3=0.

We can continue in this fashion as long as we like: Differentiating

y'''=-4y'

yields

y(4)=-4y'',

in particular

\begin{displaymath}y^{(4)}(0)=-4y''(0)=-4\cdot (-4)=16,\end{displaymath}

so a4=16/4!=2/3. At this point we have figured out that the solution to our differential equation has as its fourth degree Taylor polynomial the function

\begin{displaymath}1-2t^2+\frac{2}{3}t^4.\end{displaymath}

We expect that this Taylor polynomial is reasonably close to the solution y(t) of the differential equation, at least close to t=0. In the picture below, the solution $y(t)=\cos(2t)$ is drawn in red, while the power series approximation $f(t)=1-2t^2+\frac{2}{3}t^4$ is depicted in blue:


The method outlined works also in theory for non-linear differential equations, even though the computational effort usually becomes prohibitive after the first few steps. Let's consider the example

\begin{displaymath}y''+\sin(y)=0,\quad y(0)=0,\ y'(0)=1.\end{displaymath}

We try to find the Taylor polynomials for the solution, of the form

\begin{displaymath}a_0+a_1\cdot t+a_2 \cdot t^2 + a_3\cdot t^3\ldots,\end{displaymath}

where

\begin{displaymath}a_n=\frac{y^{(n)}(0)}{n!}.\end{displaymath}

The initial conditions yield a0=0 and a1=1. To find a2, we rewrite the differential equation as

\begin{displaymath}y''=-\sin(y),\end{displaymath}

and plug in t=0:

\begin{displaymath}y''(0)=-\sin(y(0))=-\sin(0)=0;\end{displaymath}

consequently a2=0.

Next we differentiate

\begin{displaymath}y''=-\sin(y)\end{displaymath}

on both sides with respect to t, to obtain

\begin{displaymath}y'''=-\cos(y)\cdot y',\end{displaymath}

so in particular

\begin{displaymath}y'''(0)=-\cos(y(0))\cdot y'(0)=-\cos(0)\cdot 1=-1,\end{displaymath}

yielding a3=-1/3!=-1/6. Note that since we were differentiating with respect to t, we had to use the chain rule to find the derivative of the term $\sin(y(t))$.

Let's continue a little bit longer: We differentiate

\begin{displaymath}y'''=-\cos(y)\cdot y',\end{displaymath}

to obtain

\begin{displaymath}y^{(4)}=\sin(y)\cdot (y')^2-\cos(y)\cdot y''.\end{displaymath}

Consequently y(4)(0)=0, and thus a4=0.

Differentiating

\begin{displaymath}y^{(4)}=\sin(y)\cdot (y')^2-\cos(y)\cdot y'' \end{displaymath}

yields

\begin{displaymath}y^{(5)}=\cos(y)\cdot (y')^3+\sin(y)\cdot 2(y')\cdot y''+\sin(y)\cdot y'\cdot y''-\cos(y)\cdot y'''.\end{displaymath}

You can check that y(5)=y'(0)3-y'''(0)=1-(-1)=2 and thus a5=2/5!.

The fifth degree Taylor polynomial approximation to the solution of our differential equation has the form

\begin{eqnarray*}f(t)&=&a_0+a_1\cdot t+a_2 \cdot t^2 + a_3\cdot t^3+a_4\cdot t^4 +a_5 \cdot t^5\\
&=&t-\frac{1}{6}t^3+\frac{2}{5!}t^5.
\end{eqnarray*}


We again expect that this Taylor polynomial is reasonably close to the solution y(t) of the differential equation, at least close to t=0. In the picture below, the solution, as computed by a numerical method, is drawn in red, while the power series approximation $f(t)=t-\frac{1}{6}t^3+\frac{2}{5!}t^5$ is depicted in blue:


The next sections will develop an organized method to find power series solutions for second order linear differential equations. Here are a couple of examples to practice what you have learned so far:

Exercise 1:

Find the fifth degree Taylor polynomial of the solution to the differential equation

\begin{displaymath}y''-3y=0,\quad y(0)=1,\ y'(0)=-1.\end{displaymath}

Answer.

Exercise 2:

Find the fourth degree Taylor polynomial of the solution to the differential equation

\begin{displaymath}y''+y^2=0,\quad y(0)=1,\ y'(0)=0.\end{displaymath}

Answer.

[Back] [Next]
[Algebra] [Trigonometry] [Complex Variables]
[Calculus] [Differential Equations] [Matrix Algebra]

S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Helmut Knaust
1998-06-29

Copyright 1999-2017 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour