Leading Behaviour of a Third Order Linear ODE

Question

I'm trying to solve a problem taken from Advanced Mathematical Methods for Scientists and Engineers by Bender and Orszag.

3.33 b)

Find the leading behaviours of

$x^4y'''-3x^2y'+2y=0$ $(x\rightarrow0+)$

Here is my attempt at the solution:

We let

$y(x) = e^{S(x)}$

Then,

$y' = S'e^{S}$

$y''' = ((S')^3 + 3S'S''+ S''')e^{S}$

put these into the above ODE to obtain the equation:

$x^4((S')^3 + 3S'S''+ S''') -3x^2S' +2 = 0$

Since $x=0$ is an irregular singular point, we have that

$S''' \ll (S')^3 $ $\space$ $(x\rightarrow0+)$

$3S'S'' \ll 3(S')^3 $ $\space$ $(x\rightarrow0+)$

So we obtain the relation

$x^4(S')^3$ $\sim$ $3x^2S' -2$ $\space$ $(x\rightarrow0+)$

The reason why I'm stuck at this point is that I'm unsure if it is possible to make anymore approximation to the equation in $S'$. If not I'm still unsure how to solve this cubic relation in $S'$.

Any hints on how to proceed would be greatly appreciated. Thanks.

Answer 1

$$x^4(S')^3\sim 3x^2S' -2 \qquad (x\rightarrow0+)$$ Let $\quad S'\sim ax^b.$ We have to compare the order of magnitude of $\begin{cases}x^4(S')^3=a^3x^{3b+4}\\ 3x^2S'=3ax^{b+2}\\ 2x^0\end{cases}$

From the three possibilities $\begin{cases} 3b+4=b+2\quad\to\quad b=-1\quad\to\quad a^3x\sim 3ax-2\\ 3b+4=0\quad\to\quad b=-\frac{4}{3}\quad\to\quad a^3\sim 3ax^{2/3}-2\\ b+2=0\quad\to\quad b=-2\quad\to\quad a^3x^{-2}\sim 3a-2 \end{cases}$ ,

only $\quad b=-\frac{4}{3}\quad\to\quad a^3\sim -2\quad$ is valid $\quad\to\quad S'\sim -2^{1/3}x^{-4/3}\quad\to\quad S=2^{1/3}3\:x^{-1/3}$ $$y\sim e^{2^{1/3}3\:x^{-1/3}}\qquad (x\rightarrow0+)$$