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.

Using Power Series to Solve a Linear Differential Equation

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.

Using Power Series to Solve a Linear Differential Equation

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.

Using Power Series to Solve a Linear Differential Equation

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

Using a Maclaurin Series to Solve a Linear Differential Equation

1. Use a Maclaurin series to find the solution of
   \[ \begin{equation} y^{\prime \prime} =x^{2}y+e^{x}y^{\prime} \tag{3} \end{equation} \]
   given the initial conditions \(y(0)=1\) and \(y^{\prime} (0)=1\).
2. Use the first five terms of the series to approximate values of \(y=y( x)\) for \(0\leq x\leq 1\).
3. Use a numeric differential equation solver with \(10,000\) equally spaced numbers in the interval \([ 0,1]\) to solve the differential equation in (a). Then construct a table showing the values of \(y\) for \(x=0,0.1,0.2,\ldots,0.9,1.\)

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.