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Here's a question I'm struggling with and I'd like help on it. Please, this is not a homework problem.

I want to find the power series solutions about the origin of two linearly independent solutions of $$w''-zw=0.$$

Also, how do I show that these solutions are analytic?


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Look up the Airy function. – J. M. Feb 16 '12 at 6:47
For complex functions, once differentiable implies analytic. So you can assume power series... – Aryabhata Feb 16 '12 at 7:17
up vote 2 down vote accepted










$[n>2]$ $\frac{a_n}{(n-2)!}-\frac{a_{n-3}}{(n-3)!}=0$


$a_0=c_1$ $a_1=c_2$ $a_2=0$




if $n>=0$ then $a_{3n+2}=0$



$w(x)=c_1(1+\sum_{k=1}^\infty \frac{1.4.7...(3k-2)x^{3k}}{(3k)!})+c_2(x+\sum_{k=1}^\infty \frac{2.5.8...(3k-1)x^{3k+1}}{(3k+1)!})$

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Let's say that $w(z)$ is one such solution and suppose that it has a power series, convergent in some disk around the origin, given by $$ w(z) = \sum_n a_n z^n. $$ You just substitute that into the equation, and get a recurrence relation for the coefficients $a_0,a_1,a_2,\ldots$. But, in fact, you will get a linear recurrence relation that relates coefficients $a_2,a_3,\ldots$ to $a_0$ and $a_1$, so what you get is $$ w(z) = a_0 \sum_n a^{(1)}_n z^n + a_1 \sum_n a^{(2)}_n z^n, $$ where $a^{(1)}$ and $a^{(2)}$ are two distinct linearly independent solutions to the recurrence relation (for some equations it is a little trickier than this). The two power series define the two linearly independent solutions of the ODE.

Finally, a power series is analytic inside its radius of convergence. See here, for example.

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You seem to be going at it in reverse. If it is assumed to be differentiable (which it is!), then you can assume it is analytic, and therefore assume a power series representation. – Aryabhata Feb 16 '12 at 7:20

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