leading order behavior and finding the first two terms of the solutions of $t^3u''-u=0$ I am attempting to solve a problem regarding asymatotics. I would greatly appreciate if people would comment
Problem
Find the leading behavior of the ODE , ie the first two terms of each of the two solutions of the ode 
$t^3u''-u=0$
As $t \rightarrow 0^+$
My attempt 
If we let $u = e^\psi$, then that will imply $u'= \psi' e^\psi$ and $u''= \psi''e^\psi + (\psi ')^2e^\psi$. If we replace this into the ode, we get
$$t^3(\psi'' + (\psi ')^2)=1$$
From here I decided to guess that $\psi = At^p$. If we plug this back into the previous line we get
$$t^3(Ap(p-1)t^{p-2}+Apt^{p-2})=1$$
This observing the powers we get $3+p-2=0$ because on the left hand side we have a polynomial equation of power $3+p-2$ while on the right hand side we have a constant ie a polynomial of degree 0. Thus $p=-1$. I am not sure if this is correct, idea, please comment. I know some people do the method of dominant balance on $t^3(\psi'' + (\psi ')^2)=1$, but I am unsure how does that work and was thinking my idea has some merit. 
 A: Your method is a dominant balance :). If we already know that the solution is of the form $\psi = A t^p$, we can substitute this into the ODE and balance the highest powers of $t$. The exponents are $0, p + 1, 2 p + 1$. Taking $p + 1 = 0$ doesn't work because then $2 p + 1$ is the leading order. Taking $2 p + 1 = 0$ is consistent with $t^{p + 1}$ being a subdominant term. Equating the coefficients of the leading terms gives $A = \pm 2$.
If we do not know the form of $\psi$ in advance, we can apply the same reasoning to the ODE to conclude that the $\psi''$ term can be discarded, leaving $t^3 (\psi')^2 - 1 = 0$. The solutions of the simplified ODE are again $\pm 2/\sqrt t$.
Then we can look for the solution to the original ODE in the form $u = t^p e^{\pm 2/\sqrt t}$. Substitution gives the highest term $(2 p - 3/2) t^{p + 1/2}$. Therefore the solutions are $u \sim C t^{3/4} e^{\pm 2/\sqrt t}$.
To find the next term, take $u = t^{3/4} e^{\pm 2/\sqrt t} (1 + A t^p)$.
A: $$t^3u''-u=0$$
Let change $t=x$ and $u=y$ to have the notations according to the page referenced below :
$$x^3y''-y=0$$
$$y''-\frac{1}{x^3}y=0$$
One recognize the generalized form of Bessel equation, copy from : http://mathworld.wolfram.com/BesselDifferentialEquation.html

Thus $\quad2\alpha-1=0\quad;\quad 2\gamma-2=-3\quad;\quad \beta^2\gamma^2=-1\quad;\quad \alpha^2-n^2\gamma^2=0$.
$\alpha=\frac12\quad;\quad \gamma=-\frac12\quad;\quad \beta=\pm 2i\quad;\quad n=\pm 1$
$$y=c_1x^{1/2}J_1(2ix^{-1/2})+c_2x^{1/2}Y_1(2ix^{-1/2})$$
$$y=C_1x^{1/2}I_1(2x^{-1/2})+C_2x^{1/2}K_1(2x^{-1/2})$$
Bessel functions of the second kind.
For $x\to 0^+$  and $X=2x^{-1/2}\to\infty$
$I_1(X)\sim \frac{1}{\sqrt{2\pi X}}e^X \quad;\quad x^{1/2}I_1(2x^{-1/2})\sim \frac12\sqrt{\frac{x}{\pi}}\exp\left(\frac{2}{\sqrt{x}}\right)\to \infty$
$K_1(X)\sim \sqrt{\frac{\pi}{2X} }e^{-X}
\quad;\quad x^{1/2}K_1(2x^{-1/2})\sim \frac12\sqrt{\pi}e^{-2x^{-1/2}}x^{3/4}\to 0$
http://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/02/01/01/01/
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/06/02/01/02/
A: The two leading terms of the WKB approximation of $u''-q(t)u=0$, where $q(t)$ is large, give the two basis solutions
$$
u(t)=q(t)^{-1/4}\exp\left(\pm\int\sqrt{q(t)}dt\right).
$$
Inserting $q(t)=t^{-3}$ result in the same approximation as found in the other answers,
$$
u(t)=t^{3/4}\exp\left(\pm2t^{-1/2}\right).
$$
