OBJECTIVES

When you finish this section, you should be able to:

1. Use power series to solve a linear differential equation (p. 1089)

NEED TO REVIEW?

Power series are discussed in Section 8.8, pp. 600-609.

Use power series to solve the differential equation $y^{\prime} =y.$

Solution We assume that the solution of the differential equation can be expressed as the power series $$\begin{equation} y( x) =\sum\limits_{k=0}^{\infty }a_{k}x^{k} \tag{1} \end{equation}$$

Then $$\begin{equation*} y^{\prime} ( x) =\sum\limits_{k=1}^{\infty }ka_{k}x^{k-1} \end{equation*}$$

Since $y^{\prime} =y,$ this leads to $$\begin{equation*} \sum\limits_{k=1}^{\infty }ka_{k}x^{k-1}=\sum\limits_{k=0}^{\infty}a_{k}x^{k} \end{equation*}$$

To obtain relationships among the coefficients, we write out the terms. $$\begin{equation*} a_{1}x^{0}+2a_{2}x+3a_{3}x^{2}+4a_{4}x^{3}+\cdots =a_{0}x^{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \end{equation*}$$

Because the coefficients of corresponding powers of $x$ are equal, we have $$\begin{equation*} a_{1}=a_{0}\qquad 2a_{2}=a_{1}\qquad 3a_{3}=a_{2}\qquad 4a_{4}=a_{3}\ \ldots\ na_{n}=a_{n-1}\ldots \end{equation*}$$

We can express these relationships recursively. $$\begin{eqnarray*}{{\hspace{-5pc}}{\hspace{-5pc}}} & \enspace a_{1}=a_{0}\,\,\,\,\qquad a_{2}=\dfrac{1}{2}a_{1}=\dfrac{1}{2!}a_{0} &\hspace{-56pt}a_{3}=\dfrac{1}{3}a_{2}=\dfrac{1}{3\cdot 2}a_{0}=\dfrac{1}{3!}a_{0}\\[4pt] & \qquad \qquad \qquad \qquad a_{4} =\dfrac{1}{4}a_{3}=\dfrac{1}{4!}a_{0}\,\,\,\,\, \ldots\,\, \,\,\, a_{n}=\dfrac{1}{n}a_{n-1}=\dfrac{1}{n!}a_{0}\,\,\,\,\,\,\cdots \end{eqnarray*}$$

The power series (1) takes the form $$y( x) =\sum\limits_{k=0}^{\infty }a_{k}x^{k}=\sum\limits_{k=0}^{\infty }\dfrac{1}{k!}a_{0}x^{k}=a_{0}\sum \limits_{k=0}^{\infty }\dfrac{x^{k}}{k!}$$

which we recognize as $y( x) =a_{0}e^{x}$.

RECALL

As we learned in Section 8.9, $$e^{x}=\sum \limits_{k=0}^{\infty }\dfrac{x^{k}}{k!}$$

NOW WORK

Problem 1.

Use power series to solve the differential equation $y^{\prime} -2y=e^{-x}.$

Solution Assume that a solution $y=y( x)$ of the differential equation can be expressed as the power series $y(x)=\sum\limits_{k=0}^{\infty }a_{k}x^{k}.$ Then $y^{\prime}(x)=\sum\limits_{k=1}^{\infty }ka_{k}x^{k-1}.$ Since $e^{-x}=\sum \limits_{k=0}^{\infty }\dfrac{(-1)^{k}}{k!}x^{k},$ the differential equation can be written as $$\begin{equation} \underset{\color{#0066A7}{\hbox{\(y^{\prime}\)}}}{\underbrace{\sum\limits_{k=1}^{\infty }ka_{k}x^{k-1}}}-\underset{\color{#0066A7}{\hbox{\(y\)}}}{2\underbrace{\sum\limits_{k=0}^{\infty }a_{k}x^{k}}}=\underset{\color{#0066A7}{\hbox{\(e^{-x}\)}}}{\underbrace{\sum\limits_{k=0}^{\infty }\frac{(-1)^{k}}{k!}x^{k}}} \tag{2} \end{equation}$$

To obtain relationships among the coefficients, we write out the terms of the power series. $$\begin{eqnarray*} &&\hspace{-.75pc}\big( a_{1}\,{+}\,2a_{2}x\,{+}\,3a_{3}x^{2}\,{+}\,4a_{4}x^{3}\,{+}\,5a_{5}x^{4}\,{+}\,\cdots \big)\!\,{-}\,2\big(a_{0}\,{+}\,a_{1}x\,{+}\,a_{2}x^{2}\,{+}\,a_{3}x^{3}\,{+}\,a_{4}x^{4}\,{+}\,\cdots \big)\\[4pt] && \enspace \quad =1-x+\frac{1}{2!}x^{2}-\frac{1}{3!}x^{3}+\dfrac{1}{4!}x^{4}+\cdots \end{eqnarray*}$$

Now we combine the coefficients of corresponding powers of $x$ to get $$\begin{eqnarray*} &&(a_{1}-2a_{0})+(2a_{2}-2a_{1})x+(3a_{3}-2a_{2})x^{2}+(4a_{4}-2a_{3})x^{3}+ \cdots\\[4pt] &&\qquad =1-x+\frac{1}{2!}x^{2}-\frac{1}{3!}x^{3}+\dfrac{1}{4!}x^{4}-\cdots \end{eqnarray*}$$

Since the coefficients of corresponding powers of $x$ are equal, we have $$\begin{equation*} \begin{array}{rcl@{\quad}l@{\quad}rcl@{\;}l} a_{1}-2a_{0}&=&1 & \hbox{or equivalently,} & a_{1}&=&2a_{0}+1 \\ 2a_{2}-2a_{1}&=&-1 & \hbox{or equivalently,} & a_{2}&=&\dfrac{2a_{1}-1}{2}=\dfrac{ 2\left( 2a_{0}+1\right) -1}{2}=2a_{0}+\dfrac{1}{2} \\ 3a_{3}-2a_{2}&=&\dfrac{1}{2!} & \hbox{or equivalently,} & a_{3}&=&\dfrac{2a_{2}+ \dfrac{1}{2!}}{3}=\dfrac{2}{3}\left( 2a_{0}+\dfrac{1}{2}\right) +\dfrac{1}{ 3\cdot 2!}\\ &&&&&=&\dfrac{4}{3}a_{0}+\dfrac{3}{3!} \\ 4a_{4}-2a_{3}&=&-\dfrac{1}{3!} & \hbox{or equivalently,} & a_{4}&=&\dfrac{2a_{3}- \dfrac{1}{3!}}{4}=\dfrac{1}{2}a_{3}-\dfrac{1}{4!}\\ &&&&& =&\dfrac{1}{2}\left( \dfrac{4 }{3}a_{0}+\dfrac{3}{3!}\right)-\dfrac{1}{4!}=\dfrac{2}{3}a_{0}+\dfrac{5}{4!}\\ 5a_{5}-2a_{4}&=&\dfrac{1}{4!} & \hbox{or equivalently,} & a_{5}&=&\dfrac{2a_{4}+ \dfrac{1}{4!}}{5}=\dfrac{2}{5}a_{4}+\dfrac{1}{5!}\\ &&&&&=&\dfrac{2}{5}\left( \dfrac{2 }{3}a_{0}+\dfrac{5}{4!}\right) +\dfrac{1}{5!}=\dfrac{4}{15}a_{0}+\dfrac{11}{ 5!} \end{array} \end{equation*}$$

This recursive formula $( n+1) a_{n+1}=2a_{n}+( -1) ^{n}\dfrac{1}{( n) !}$ can be used to find $a_{n}$ in terms of $a_{0,}$ as we did for $a_{1},$ $a_{2},$ $a_{3},$ $a_{4}$, and $a_{5}.$ The power series representation of the general solution of the equation is $$\begin{eqnarray*} y(x) =\sum\limits_{k=0}^{\infty }a_{k}x^{k}&=&a_{0}+(2a_{0}+1)x+\left( 2a_{0}+\frac{1}{2!}\right) x^{2}+\left( \frac{4}{3}a_{0}+\frac{3}{3!}\right) x^{3}\\[4pt] &&+\left( \frac{2}{3}a_{0}+\frac{5}{4!}\right)\! x^{4}+\!\left( \dfrac{4}{15} a_{0}+\dfrac{11}{5!}\right)\! x^{5}+\cdots \\[4pt] & =&a_{0}\!\left( 1+2x+2x^{2}+\frac{4}{3}x^{3}+\frac{2}{3}x^{4}+\dfrac{4}{15} x^{5}+\cdots \right) \\[4pt] &&+\,x+\frac{1}{2!}x^{2}+\frac{3}{3!}x^{3}+\frac{5}{4!} x^{4}+\dfrac{11}{5!}x^{5}+\cdots \end{eqnarray*}$$

where $a_{0}$ is arbitrary.

Use power series to find the general solution of the linear differential equation $$\begin{equation*} y^{\prime \prime} +xy^{\prime} +2y=0 \end{equation*}$$

Solution If $y(x)=\sum\limits_{k=0}^{\infty }a_{k}x^{k}$ is a solution to the differential equation, then $$y^{\prime} (x)=\sum\limits_{k=1}^{\infty }ka_{k}x^{k-1}\qquad \hbox{and}\qquad y^{\prime \prime} (x)=\sum\limits_{k=2}^{\infty }k(k-1)a_{k}x^{k-2}$$

Now we substitute these power series into the differential equation. $$\begin{eqnarray*} \sum\limits_{k=2}^{\infty }k(k-1)a_{k}x^{k-2}\,{+}\,x\!\sum\limits_{k=1}^{\infty}ka_{k}x^{k-1}+2\sum\limits_{k=0}^{\infty }a_{k}x^{k}\!&=&0\quad {\color{#0066A7}{y^{\prime \prime} +xy^{\prime} +2y=0}} \\[4pt] \sum\limits_{k=2}^{\infty }k(k-1)a_{k}x^{k-2}+\sum\limits_{k=1}^{\infty }ka_{k}x^{k}+\sum\limits_{k=0}^{\infty }2a_{k}x^{k}\!&=&0 \quad {\color{#0066A7}{\hbox{Move }x\hbox{ and } 2\hbox{ into the summation.}}} \end{eqnarray*}$$

Next we adjust the indexes of summation so that $x^{k}$ appears in each series. Here, only the first series needs modification. If we replace $k$ with $k+2$ in the first series, we obtain $$\begin{equation*} \underset{\color{#0066A7}{\hbox{Replace }k\hbox{ with }k+2}}{\underbrace{\sum\limits_{k=0}^{ \infty }( k+2) (k+1)a_{k+2}x^{k}}}+\sum\limits_{k=1}^{\infty }ka_{k}x^{k}+\sum\limits_{k=0}^{\infty }2a_{k}x^{k}=0 \end{equation*}$$

The index in the first and third series begins at $k=0$, and the index in the second series begins at $k=1$. We write out the $k=0$ term of the first and third series separately. Then each summation starts at $k=1.$ $$\begin{eqnarray*} \left( 2\right) \left( 1\right) a_{2}x^{0}+2a_{0}x^{0}+\sum\limits_{k=1}^{\infty }( k+2) (k+1)a_{k+2}x^{k}+\sum\limits_{k=1}^{\infty }ka_{k}x^{k}+\sum\limits_{k=1}^{\infty }2a_{k}x^{k}& =&0 \\[4pt] 2a_{2}+2a_{0}+\sum\limits_{k=1}^{\infty }\left[ (k+2)(k+1)a_{k+2}+(k+2) a_{k}\right] x^{k}& =&0 \end{eqnarray*}$$

Equating the coefficients of corresponding powers of $x$ (the coefficients on the right side are all $0)$, we have $$\begin{eqnarray*} \hspace{-5pt}2a_{2}+2a_{0}&=&0 \hspace{20pt}(n+2)(n+1)a_{n+2}+(n+2)a_{n}=0 & \\[2pt] a_{2}&=&-a_{0}\quad\hspace{140pt} a_{n+2}=-\dfrac{a_{n}}{(n+1)}\quad n=1,2,3,\ldots \end{eqnarray*}$$

If $n=1$, we find $a_3=-\frac{a_1}{2}$. Then we use the recursion formula on the left to obtain all of the coefficients in terms of $a_{0}$ and $a_{1}$. That is, $$\begin{equation*} \begin{array}{@{}l@{\quad\ \ \ }l@{\quad\ \ \ }l@{\quad\ \ \ }l@{}} a_{2}= -a_{0} & a_{4}=-\dfrac{a_{2}}{3}=\dfrac{a_{0}}{3} & a_{6}=-\dfrac{a_{4}}{5}=-\dfrac{a_{0}}{3\cdot 5} & a_{8}=-\dfrac{a_{6}}{7}=\dfrac{a_{0}}{3\cdot 5\cdot 7}\\ a_{3}=-\dfrac{a_{1}}{2} & a_{5}=-\dfrac{a_{3}}{4}=\dfrac{a_{1}}{2\cdot 4} & a_{7}=-\dfrac{a_{5}}{6}=-\dfrac{a_{1}}{2\cdot 4\cdot 6} \end{array} \end{equation*}$$

and so on. Since $a_{0}$ and $a_{1}$ can be chosen arbitrarily, the power series representation for the general solution is $$\begin{eqnarray*} y(x)&=&a_{0}\!\left( 1-x^{2}+\frac{x^{4}}{3}-\frac{x^{6}}{3\cdot 5}+\frac{x^{8}}{3\cdot 5\cdot 7}-\cdots \right)\\[5pt] &&+\,a_{1}\!\left( x-\frac{x^{3}}{2}+\frac{x^{5}}{2\cdot 4}-\frac{x^{7}}{2\cdot 4\cdot 6}+\cdots \right) \end{eqnarray*}$$

NOW WORK

Problem 7.

NEED TO REVIEW?

Maclaurin series are discussed in Section 8.9, pp. 611-614

Solution (a) We assume that the solution of the differential equation is given by the Maclaurin series $$y(x)=\sum_{k=0}^{\infty }\frac{y^{(k)}(0)}{k!}x^{n}=y(0)+y^{\prime} ( 0) x+\frac{y^{\prime \prime} (0)}{2!}x^{2}+\frac{y^{\prime \prime \prime} (0)}{3!}x^{3}+\cdots$$

We substitute the initial conditions, $y(0)=1$ and $y^{\prime} (0)=1$, into (3). Then $$\begin{equation*} y^{\prime \prime} ( 0) =0^{2}\cdot 1+e^{0}\cdot 1=1\qquad {\color{#0066A7}{\hbox{\(x=0; y(0) =1; y^{\prime} (0) =1; y^{\prime \prime}(x)= x^{2}y + e^{x}y^{\prime}\)}}} \end{equation*}$$

Now we differentiate $y^{\prime \prime} =x^{2}y+e^{x}y^{\prime}$ with respect to $x$ to find $y^{\prime \prime \prime} ( 0) .$ $$\begin{eqnarray*} y^{\prime \prime \prime} ( x) &=&( x^{2}y^{\prime} +2xy) +( e^{x}y^{\prime \prime} +e^{x}y^{\prime}) \\[3pt] y^{\prime \prime \prime} ( 0) &=& ( 0^{2}\cdot 1+2\cdot 0\cdot 1) +( e^{0}\cdot 1+e^{0}\cdot 1) =2 \end{eqnarray*}$$

We continue differentiating and evaluating the derivative at $x=0.$ $$\begin{eqnarray*} y^{(4)}( x) & =&x^{2}y^{\prime \prime} +4xy^{\prime} +2y+e^{x}y^{\prime \prime \prime} +2e^{x}y^{\prime \prime} +e^{x}y^{\prime} \\[3pt] y^{(4)}(0)& =&0^{2}\cdot 1+4\cdot 0\cdot 1+2\cdot 1+e^{0}\cdot 2+2\cdot e^{0}\cdot 1+e^{0}\cdot 1=7 \end{eqnarray*}$$

and so on. The Maclaurin series then becomes $$\begin{equation*} y(x)=1+x+\frac{1}{2!}x^{2}+\frac{2}{3!}x^{3}+\frac{7}{4!}x^{4}+\cdots \end{equation*}$$

(b) We construct Table 1 that uses the first five terms of the series to approximate select values of $y$ in the interval $0\leq x\leq 1$.

(c) Using every thousandth term from the numeric solution, we construct Table 2.

From the results of Tables 1 and 2, we can see how the five-term approximation to the series solution of the differential equation deteriorates as we move away from the center of convergence $0$.

NOW WORK

Problem 15.