Example

Use a Maclaurin series to find the solution of $\begin{equation} y^{\prime \prime} =x^{2}y+e^{x}y^{\prime} \end{equation}$

given the initial conditions $y(0)=1$ and $y^{\prime} (0)=1$ .

Use the first five terms of the series to approximate values of

$y=y( x)$ for

$0\leq x\leq 1$ .

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 $y(x)=\sum_{k=0}^{\infty }\frac{y^{(k)}(0)}{k!}x^{n}=y(0)+y^{\prime} ( 0) x+\frac{y^{\prime \prime} (0)}{2!}x^{2}+\frac{y^{\prime \prime \prime} (0)}{3!}x^{3}+\cdots$

1093

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 $0\leq x\leq 1$ .

Maclaurin Series Approximation of

$y$ Using Five Terms of the Series

$x$	$0.0$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$	$0.9$	$1.0$
$y$	$1.0$	$1.1054$	$1.2231$	$1.3564$	$1.5088$	$1.6849$	$1.8898$	$2.1294$	$2.4101$	$2.7394$	$3.125$

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

Select Values of

$y$ Using a Numeric Equation Solver

$x$	$0.0$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$	$0.9$	$1.0$
$y$	$1.0$	$1.1054$	$1.2232$	$\ 1.3569$	$1.5114$	$1.6933$	$1.9127$	$2.1836$	$2.5271$	$2.9747$	$3.5752$

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 $0$ .

Problem 15.