4.4.1 Solution
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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 $$y\left ( x\right ) =e^{S\left ( x\right ) }.$$

4.4.1 Solution

4.8.0.1 Solution
4.8.0.2 References

$$x=0$$ 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$$x^{3}\left ( S^{\prime \prime }+\left ( S^{\prime }\right ) ^{2}\right ) =1\tag{1}$$

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

1.
Page 80-82 Bender and Orszag textbook.
2.
Lecture notes, Lecture 5, Tuesday Janurary 31, 2017. EP 548, University of Wisconsin, Madison by Professor Smith.
3.
Lecture notes from http://www.damtp.cam.ac.uk/