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 $$\begin{equation*} \begin{array}{rcl@{\quad}l@{\quad}rcl@{\;}l} a_{1}-2a_{0}&=&1 & \hbox{or equivalently,} & a_{1}&=&2a_{0}+1 \\ 2a_{2}-2a_{1}&=&-1 & \hbox{or equivalently,} & a_{2}&=&\dfrac{2a_{1}-1}{2}=\dfrac{ 2\left( 2a_{0}+1\right) -1}{2}=2a_{0}+\dfrac{1}{2} \\ 3a_{3}-2a_{2}&=&\dfrac{1}{2!} & \hbox{or equivalently,} & a_{3}&=&\dfrac{2a_{2}+ \dfrac{1}{2!}}{3}=\dfrac{2}{3}\left( 2a_{0}+\dfrac{1}{2}\right) +\dfrac{1}{ 3\cdot 2!}\\ &&&&&=&\dfrac{4}{3}a_{0}+\dfrac{3}{3!} \\ 4a_{4}-2a_{3}&=&-\dfrac{1}{3!} & \hbox{or equivalently,} & a_{4}&=&\dfrac{2a_{3}- \dfrac{1}{3!}}{4}=\dfrac{1}{2}a_{3}-\dfrac{1}{4!}\\ &&&&& =&\dfrac{1}{2}\left( \dfrac{4 }{3}a_{0}+\dfrac{3}{3!}\right)-\dfrac{1}{4!}=\dfrac{2}{3}a_{0}+\dfrac{5}{4!}\\ 5a_{5}-2a_{4}&=&\dfrac{1}{4!} & \hbox{or equivalently,} & a_{5}&=&\dfrac{2a_{4}+ \dfrac{1}{4!}}{5}=\dfrac{2}{5}a_{4}+\dfrac{1}{5!}\\ &&&&&=&\dfrac{2}{5}\left( \dfrac{2 }{3}a_{0}+\dfrac{5}{4!}\right) +\dfrac{1}{5!}=\dfrac{4}{15}a_{0}+\dfrac{11}{ 5!} \end{array} \end{equation*}$$

1091

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.