This note solves in details the ODE
\[ x^{3}y^{\prime \prime }\left ( x\right ) =y\left ( x\right ) \]
Using asymptoes method using what is called the dominant balance submethod where it is assumed that \(y\left ( x\right ) =e^{S\left ( x\right ) }.\)
\(x=0\) is an irregular singular point. The solution is assumeed to be \(y\left ( x\right ) =e^{S\left ( x\right ) }\). Therefore \(y^{\prime }=S^{\prime }e^{S\left ( x\right ) }\) and \(y^{\prime \prime }=S^{\prime \prime }e^{S\left ( x\right ) }+\left ( S^{\prime }\right ) ^{2}e^{S\left ( x\right ) }\) and the given ODE becomes\begin{equation} x^{3}\left ( S^{\prime \prime }+\left ( S^{\prime }\right ) ^{2}\right ) =1\tag{1} \end{equation}
Assuming that\[ S^{\prime }\left ( x\right ) \thicksim cx^{\alpha }\] Hence \(S^{\prime \prime }\thicksim c\alpha x^{\alpha -1}\). and (1) becomes\begin{align*} x^{3}\left ( c\alpha x^{\alpha -1}+\left ( cx^{\alpha }\right ) ^{2}\right ) & \sim 1\\ c\alpha x^{\alpha +2}+c^{2}x^{2\alpha +3} & \sim 1 \end{align*}
Term \(c\alpha x^{\alpha +2}\ggg c^{2}x^{2\alpha +3}\), hence we set \(\alpha =\frac{-3}{2}\) to remove the subdominant term. Therefore the above becomes, after substituting for the found \(\alpha \)\begin{align*} \overset{x\rightarrow 0}{\overbrace{\frac{-3}{2}cx^{\frac{1}{2}}}}+c^{2} & \sim 1\\ c^{2} & =1 \end{align*}
Therefore \(c=\pm 1\). The result so far is \(S^{\prime }\left ( x\right ) \sim cx^{\frac{-3}{2}}\). Now another term is added. Let \[ S^{\prime }\left ( x\right ) \sim cx^{\frac{-3}{2}}+A\left ( x\right ) \] Now we will try to find \(A\left ( x\right ) \). Hence \(S^{^{\prime \prime }}\left ( x\right ) \sim \frac{-3}{2}cx^{\frac{-5}{2}}+A^{\prime }\) and \(x^{3}\left ( S^{\prime \prime }+\left ( S^{\prime }\right ) ^{2}\right ) =1\) now becomes\begin{align*} x^{3}\left ( \frac{-3}{2}cx^{\frac{-5}{2}}+A^{\prime }+\left ( cx^{\frac{-3}{2}}+A\right ) ^{2}\right ) & \sim 1\\ x^{3}\left ( \frac{-3}{2}cx^{\frac{-5}{2}}+A^{\prime }+c^{2}x^{-3}+A^{2}+2Acx^{\frac{-3}{2}}\right ) & \sim 1\\ \left ( \frac{-3}{2}cx^{\frac{1}{2}}+x^{3}A^{\prime }+c^{2}+x^{3}A^{2}+2Acx^{\frac{3}{2}}\right ) & \sim 1 \end{align*}
Since \(c^{2}=1\) from the above, then\[ \frac{-3}{2}cx^{\frac{1}{2}}+x^{3}A^{\prime }+x^{3}A^{2}+2Acx^{\frac{3}{2}}\sim 0 \] Dominant balance says to keep dominant term (but now looking at those terms in \(A\) only). From the above, since \(A\ggg A^{2}\) and \(A\) \(\ggg A^{\prime }\) then from the above, we can cross out \(A^{2}\) and \(A^{\prime }\) resulting in\[ \frac{-3}{2}cx^{\frac{1}{2}}+2Acx^{\frac{3}{2}}\sim 0 \] Hence we just need to find \(A\) to balance the above\begin{align*} \frac{-3}{2}cx^{\frac{1}{2}}+2Acx^{\frac{3}{2}} & \sim 0\\ 2Acx^{\frac{3}{2}} & \sim \frac{3}{2}cx^{\frac{1}{2}}\\ A & \sim \frac{3}{4x} \end{align*}
We found \(A\left ( x\right ) \) for the second term. Therefore, so far we have\[ S^{\prime }\left ( x\right ) =cx^{\frac{-3}{2}}+\frac{3}{4x}\] Or \[ S\left ( x\right ) =-2cx^{\frac{-1}{2}}+\frac{3}{4}\ln x+C_{0}\] But \(C_{0}\) can be dropped (subdominant to \(\ln x\) when \(x\rightarrow 0\)) and so far then we can write the solution as\begin{align*} y\left ( x\right ) & =e^{S\left ( x\right ) }W\left ( x\right ) \\ & =e^{S\left ( x\right ) }\sum _{n=0}^{\infty }a_{n}x^{nr}\\ & =\exp \left ( -2cx^{\frac{-1}{2}}+\frac{3}{4}\ln x\right ) \sum _{n=0}^{\infty }a_{n}x^{nr}\\ & =e^{-2cx^{\frac{-1}{2}}}x^{\frac{3}{4}}\sum _{n=0}^{\infty }a_{n}x^{nr}\\ & =e^{-\frac{2c}{\sqrt{x}}}\sum _{n=0}^{\infty }a_{n}x^{nr+\frac{3}{4}}\\ & =e^{\pm \frac{2}{\sqrt{x}}}\sum _{n=0}^{\infty }a_{n}x^{nr+\frac{3}{4}} \end{align*}
Since \(c=\pm 1\). We can now try adding one more term to \(S\left ( x\right ) \). Let \[ S^{\prime }\left ( x\right ) =cx^{\frac{-3}{2}}+\frac{3}{4x}+B\left ( x\right ) \] Hence \[ S^{\prime \prime }=\frac{-3}{2}cx^{\frac{-5}{2}}-\frac{3}{4x^{2}}+B^{\prime }\left ( x\right ) \] And \(x^{3}\left ( S^{\prime \prime }+\left ( S^{\prime }\right ) ^{2}\right ) \thicksim 1\) now becomes\begin{align*} x^{3}\left ( \left ( \frac{-3}{2}cx^{\frac{-5}{2}}-\frac{3}{4x^{2}}+B^{\prime }\left ( x\right ) \right ) +\left ( cx^{\frac{-3}{2}}+\frac{3}{4x}+B\left ( x\right ) \right ) ^{2}\right ) & \sim 1\\ x^{3}\left ( \frac{c^{2}}{x^{3}}-\frac{3}{16}x^{-2}+2cBx^{-\frac{3}{2}}+\frac{3}{2}Bx^{-1}+B^{2}+B^{\prime }\right ) & \sim 1\\ \left ( \overset{=1}{\overbrace{c^{2}}}-\frac{3}{16}x+2cBx^{\frac{3}{2}}+\frac{3}{2}Bx^{2}+x^{3}B^{2}+x^{3}B^{\prime }\right ) & \sim 1\\ -\frac{3}{16}x+2cBx^{\frac{3}{2}}+\frac{3}{2}Bx^{2}+x^{3}B^{2}+x^{3}B^{\prime } & \sim 0 \end{align*}
From the above, since \(B\left ( x\right ) \ggg B^{2}\left ( x\right ) \) and \(B\left ( x\right ) \ggg B^{\prime }\left ( x\right ) \) and for small \(x\), then we can cross out terms with \(B^{2}\) and \(B^{\prime }\) from above, and we are left with \[ -\frac{3}{16}x+2cBx^{\frac{3}{2}}+\frac{3}{2}Bx^{2}\backsim 0 \] Between \(2cBx^{\frac{3}{2}}\) and \(\frac{3}{2}Bx^{2}\), for small \(x\), then \(2cBx^{\frac{3}{2}}\ggg \frac{3}{2}Bx^{2}\), so we can cross out \(\frac{3}{2}Bx^{2}\) from above\begin{align*} -\frac{3}{16}x+2cBx^{\frac{3}{2}} & \backsim 0\\ 2cBx^{\frac{3}{2}} & \backsim \frac{3}{16}x\\ B & \backsim \frac{3}{32c}x^{-\frac{1}{2}} \end{align*}
We found \(B\left ( x\right ) \), Hence now we have \[ S^{\prime }\left ( x\right ) =cx^{\frac{-3}{2}}+\frac{3}{4x}+\frac{3}{32c}x^{-\frac{1}{2}}\] Or\[ S\left ( x\right ) =-2cx^{\frac{-1}{2}}+\frac{3}{4}\ln x+\frac{3}{16c}x^{\frac{1}{2}}+C_{1}\] But \(C_{1}\) can be dropped (subdominant to \(\ln x\) when \(x\rightarrow 0\)) and so far then we can write the solution as\begin{align*} y\left ( x\right ) & =e^{S\left ( x\right ) }W\left ( x\right ) \\ & =e^{S\left ( x\right ) }\sum _{n=0}^{\infty }a_{n}x^{nr}\\ & =\exp \left ( -2cx^{\frac{-1}{2}}+\frac{3}{4}\ln x+\frac{3}{16c}x^{\frac{1}{2}}\right ) \sum _{n=0}^{\infty }a_{n}x^{nr}\\ & =e^{-2cx^{\frac{-1}{2}}+\frac{3}{16c}x^{\frac{1}{2}}}x^{\frac{3}{4}}\sum _{n=0}^{\infty }a_{n}x^{nr}\\ & =e^{-2cx^{\frac{-1}{2}}+\frac{3}{16c}x^{\frac{1}{2}}}\sum _{n=0}^{\infty }a_{n}x^{nr+\frac{3}{4}} \end{align*}
For \(c=1\)\[ y_{1}\left ( x\right ) =e^{-2x^{\frac{-1}{2}}+\frac{3}{16}x^{\frac{1}{2}}}\sum _{n=0}^{\infty }a_{n}x^{nr+\frac{3}{4}}\] For \(c=-1\)\[ y_{2}\left ( x\right ) =e^{2x^{\frac{-1}{2}}-\frac{3}{16}x^{\frac{1}{2}}}\sum _{n=0}^{\infty }a_{n}x^{nr+\frac{3}{4}}\]
Hence \[ y\left ( x\right ) \thicksim Ay_{1}\left ( x\right ) +By_{2}\left ( x\right ) \]
Reference