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$$e^xy''+xy=0$$

How do I find the power series solution to this equation, or rather, how should I go about dealing with the $e^x$? Thanks!

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  • $\begingroup$ I edited your question in $\LaTeX$. Please check that I didn't alter the meaning of what's intended. $\endgroup$ – Git Gud Apr 15 '13 at 6:31
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When trying to find a series to represent something, it's important to decide what kind of a series you want. Even if you're dealing with power series, a small change in notation between $\displaystyle \sum a_n x^n$ and $\displaystyle\sum \frac{a_nx^n}{n!}$ can lead to substantial changes. In particular, we have that, if $f(x)=\sum a_n x^n$, then

$$e^x f(x) = \left(\sum \frac{x^n}{n!}\right) \left(\sum a_n x^n\right) = \sum \left(\sum_{i+j=n}\frac{a_i}{j!} \right) x^n, $$

which, while it will lead to a perfectly good recurrence yielding power series solution for a problem like this, is somewhat awkward, unwieldy, and likely not to lead to a recurrence that you can recognize or explicitly solve. However, if $f(x)=\sum \frac{a_n}{n!}$, then

$$e^x f(x) = \left(\sum \frac{x^n}{n!}\right) \left(\sum \frac{a_n x^n}{n!}\right) = \sum \left(\sum_{i+j=n}\frac{a_i n!}{i!j!} \right) \frac{x^n}{n!}=\sum \left(\sum_{k\leq n}a_k \binom{n}{k} \right) \frac{x^n}{n!}, $$

which is a nicer looking expression. Additionally, the power series expansion of $f'(x)$ has a nice form due to the cancellation between the $n$ in $n!$ and the $n$ in $D(x^n)=nx^{n-1}$

With this in mind, I suggest a slight change on the hint of @M.Strochyk

Hint: Expand $\displaystyle y(x)=\sum \frac{a_nx^n}{n!}$ and $\displaystyle e^x=\sum \frac{x^n}{n!}$ and substitute them into the equation.

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$e^xy''+xy=0$

$y''+xe^{-x}y=0$

Let $y=\sum\limits_{n=0}^\infty\dfrac{a_nx^n}{n!}$ ,

Then $y'=\sum\limits_{n=0}^\infty\dfrac{na_nx^{n-1}}{n!}=\sum\limits_{n=1}^\infty\dfrac{na_nx^{n-1}}{n!}=\sum\limits_{n=1}^\infty\dfrac{a_nx^{n-1}}{(n-1)!}$

$y''=\sum\limits_{n=1}^\infty\dfrac{(n-1)a_nx^{n-2}}{(n-1)!}=\sum\limits_{n=2}^\infty\dfrac{(n-1)a_nx^{n-2}}{(n-1)!}=\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}$

$\therefore\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+xe^{-x}\sum\limits_{n=0}^\infty\dfrac{a_nx^n}{n!}=0$

$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+x\left(\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^n}{n!}\right)\left(\sum\limits_{n=0}^\infty\dfrac{a_nx^n}{n!}\right)=0$

By the formula in http://en.wikipedia.org/wiki/Cauchy_product#Series,

$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a_kx^k(-1)^{n-k}x^{n-k}}{k!(n-k)!}=0$

$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}a_kx^n}{k!(n-k)!}=0$

$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+x\sum\limits_{n=0}^\infty\left(\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}n!a_k}{k!(n-k)!}\right)\dfrac{x^n}{n!}=0$

$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+\sum\limits_{n=0}^\infty\left(\sum\limits_{k=0}^n(-1)^{n-k}C_k^na_k\right)\dfrac{x^{n+1}}{n!}=0$

$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+\sum\limits_{n=3}^\infty\left(\sum\limits_{k=0}^{n-3}(-1)^{n-k-3}C_k^{n-3}a_k\right)\dfrac{x^{n-2}}{(n-3)!}=0$

$a_2+\sum\limits_{n=3}^\infty\left(\dfrac{a_n}{n-2}+\sum\limits_{k=0}^{n-3}(-1)^{n-k-3}C_k^{n-3}a_k\right)\dfrac{x^{n-2}}{(n-3)!}=0$

$\therefore\begin{cases}a_2=0\\\dfrac{a_n}{n-2}+\sum\limits_{k=0}^{n-3}(-1)^{n-k-3}C_k^{n-3}a_k=0\end{cases}$

But the recurrence relation is still very difficult to solve.

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Hint:
Expand $$y(x)=\sum\limits_{n=0}^{\infty}{a_n x^n}, \\ e^x=\sum\limits_{n=0}^{\infty}{\dfrac{x^n}{n!}}$$ and substitute them into equation.

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