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

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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 $0$.