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

1090

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 $y( x) =a_{0}e^{x}$.