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{\bf Question}

A solution in descending powers of $x$ is sought for the equation

$$y''+y=\df{1}{x},\ x\to +\infty$$

Find the first few terms by direct substitution of the ansatz

$$y(x) \sim \sum_{r=0}^\infty \df{a_r}{x^r}$$

\vspace{.25in}

{\bf Answer}

$y''+y=\df{1}{x}\ (1)\ x\to+\infty$

If $y \sim \sum_{r=0}^\infty \df{a_r}{x^r}$ then
$\begin{array}{l}y' \sim \sum_{r=0}^\infty -r\df{a_r}{x^{r+1}}\\
y'' \sim \sum_{r=0}^\infty \df{+r(r+1)a_r}{x^{r+2}}\end{array}$

Ignore Poincar\'{e} and differentiation of asymptotics. We're
doing formal maths!

Hence in (1):

$\sum_{r=0}^\infty \df{r(r+1)a_r}{x^{r+2}}+\sum_{r=0}^\infty
\df{a_r}{x^r}=\df{1}{x}$

We balance at \un{like} powers of $x$.

$\begin{array}{llll}O(x^0): & a_0=0\\ O(x^{-1}): & a_1=1\\
O(x^{-2}): & a_2=0\\ O(x^{-3}): & 1 \cdot 2\cdot a_1+a_3=0 &
\Rightarrow & a_3=-2a_1=-2\\ O(x^{-4}): & 2 \cdot 3 \cdot
a_2+a_4=0 & \Rightarrow & a_4=-6a_2=0\\ O(x^{-5}): & 3 \cdot 4
\cdot a_3+a_5=0 & \Rightarrow & a_5=-12a_3=+24 \end{array}$

Spot the pattern:

for $r=2n,\ a_r=0$

\ \ \ \ $r=2n+1,\ a_{2n+1}=-(2n-1)(2n)a_{2n-1}\ \ (n \geq 1)$

\ \ \ \ $a_1=1\ \ (n=0)$

This will, in principle, determine all coefficients.

The series diverges (by ratio test) for all $x$.


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