Series solution to an ODE always giving me 0

Question

I have the ODE

$$y''-9x^4y=0,$$ and I want a series solution of the form $$y(x)=\frac{1}{\sqrt{3}x}e^{x^3}\sum_{n=0}^\infty a_n x^{n\alpha}$$ for some constant $\alpha$.

Through all my attempts, I get $\alpha=3$ and $$\sum_{n=0}^\infty 18 na_nx^{3n}+\sum_{n=-1}^\infty (3n+2)(3n+1)a_{n+1}x^{3n}=0.$$ However, this immediately tells me $a_0=0$ and $$a_{n+1}=\frac{18n}{(3n+2)(3n+1)}a_n$$ so the whole series is 0. I know this is wrong because I know the solution is supposed to be $$\sqrt{2\pi x}I_\frac{1}{6}(x^3),$$ where $I_\nu(x)$ is the modified Bessel function of the first kind. Can anyone tell me where the hole in my thinking is? Thank you in advance.

Answer 1

$$y''=\sum_{n=0}^\infty a_n\frac{x^{3n-3}\left(9x^6+18nx^3+9n^2-9n+2\right)\mathrm{e}^{x^3}}{\sqrt{3}}$$

Answer 2

While the factor $x^{-1}$ is part of the asymptotic formula for $y''-Qy=0$, $Q>0$, at large $x$ $$ y(x)\sim Q(x)^{-\frac14}\exp\left(\pm\int\sqrt{Q(x)}dx\right), $$ the power series expansion is valid for $x\approx 0$. As the equation is not singular, it makes no sense to include it in the factorization. Thus set $$ y(x)=e^{x^3}u(x)\implies y'(x)=e^{x^3}\left(u'(x)+3x^2u(x)\right)\\ \implies y''(x)=e^{x^3}\left(u''(x)+6x^2u'(x)+(6x+9x^2)u(x)\right) $$ which gives the differential equation for the factor $u$ $$ u''(x)+6x^2u'(x)+6xu(x)=0. $$ With the power series expansion $u(x)=\sum_{n=0}^\infty a_nx^n$ we get the coefficient equations for $x^n$ (with the understanding that $a_n=0$ for $n<0$) $$ (n+2)(n+1)a_{n+2}+6(n-1)a_{n-1}+6a_{n-1}=0 $$ Thus only the coefficients for $n=3m$ are not automatically zero and $$ a_{3m+3}=-\frac{6(3m+1)}{(3m+3)(3m+2)}a_{3m}. $$

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In short, your problem was that you only had the coefficients for $n=3m-1$ in your series, which are automatically zero. You would have needed to include a "phase shift" $r$ in the exponent, $$ y(x)=\frac{e^{x^3}}{\sqrt3x}\sum a_nx^{n\alpha+r}. $$