Ideas/Methods for solving an non linear DE with non constant coefficients I have to solve the following equation.
$2 A x^2 u''(x)-2 A x u''(x)+2 A x u'(x)-A u'(x)+B u(x)+\frac{C u(x)}{x}$
I have been thinking in a way of solving this since it is pretty complicated.
I tried to solve it by using Laplace Transforms. But the Laplace transform of such equation seems to be another differential equation. What would be the meaning of that?
The only boundary condition I have right now is that $u(0)=0$
I would think to try to convert this "somehow" into the Hypergeometric differential equation in the form of the eq here:
https://mathworld.wolfram.com/HypergeometricDifferentialEquation.html
However the last term $Cu(x)/x$ is being something that is getting me into trouble.
If you have any idea or have you seen how to solve this. I would really appreaciate any idea/suggestion/recommendation.
Thanks
Jmtz
 A: Cool question. Basically we have an ODE
$$2\alpha~x(x-1)u''(x)+\alpha(2x-1)u'(x)+\left(\beta+\frac{\gamma}{x}\right)u(x)=0$$
We first divide through by $-2a$:
$$x(1-x)u''(x)+\left(\frac{1}{2}-x\right)u'(x)+\left(\mu+\frac{\nu}{x}\right)u(x)=0\tag{*}$$
Where $\mu=-\beta/(2\alpha),\nu=-\gamma/(2\alpha)$. This already looks quite close to the hypergeometric ODE, but there is that non-constant coefficient in front of the $u(x)$ that is mucking things up. This reminds me an awful lot of the transformation between the spherical Bessel equation and the ordinary Bessel equation, so we try a substitution of the form
$$u(x)=x^pv(x)$$
The derivatives are
$$
u'(x)=px^{p-1}v(x)+x^pv'(x)\\
u''(x)=(p-1)px^{p-2}v(x)+2px^{p-1}v'(x)+x^pv''(x)
$$
Let's plug this in and see if we can spot anything "nice". After some tedious algebra we get
$$(*)=\begin{matrix}\left(x^{p+1}-x^{p+2}\right)v''(x)\\+\left[\left(\frac{1}{2}+2p\right)x^p-\left(1+2p\right)x^{p+1}\right]v'(x)\\ +\left[\left(\nu -\frac{p}{2}+p^2\right)x^{p-1}+\left(\mu-p^2\right)x^p\right]v(x)\end{matrix}$$
Dividing through by $x^p$,
$$(*)=\begin{matrix}x(1-x)v''(x)\\+\left[\left(\frac{1}{2}+2p\right)-\left(1+2p\right)x\right]v'(x)\\ +\left[\left(\nu -\frac{p}{2}+p^2\right)x^{-1}+\left(\mu-p^2\right)\right]v(x)\end{matrix}$$
This is already looking quite close to what we want. All we need to do is choose a $p$ such that
$$\nu-\frac{p}{2}+p^2=0$$
We note that there are two such choices of $p$, being
$$p_{\pm}=\frac{1\pm\sqrt{1-16\nu}}{4}$$
It shouldn't matter which one we pick, so arbitrarily we choose $p=p_+$. Now,
$$(*)=x(1-x)v''(x)+\left[\left(\frac{1}{2}+2p\right)-(1+2p)x\right]v'(x)+(\mu-p^2)v(x)$$
In order for this to look exactly like the hypergeometric equation, we need to find $a,b,c$ such that
$$\left(\frac{1}{2}+2p\right)-(1+2p)x=c-(a+b+1)x \\ \mu-p^2=-ab$$
This clearly requires $c=\frac{1}{2}+2p$ (or, more precisely $c_{\pm}=\frac{1}{2}+2p_{\pm}$) so all that is left is the system of equations
\begin{align}
a+b &=2p \\ 
ab &=p^2-\mu
\end{align}
I'll let you check for yourself that this system of equations is solved by either of the following choices:
$$\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}p-\sqrt \mu \\ p+\sqrt \mu\end{bmatrix}~~\text{or}~~\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}p+\sqrt \mu \\ p-\sqrt \mu\end{bmatrix}$$
Again, it shouldn't matter which one we pick, so we arbitrarily choose
$$\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}p+\sqrt \mu \\ p-\sqrt \mu\end{bmatrix}$$
So, assuming that
$$c=\frac{1}{2}+2p=\frac{2+\sqrt{1-16\nu}}{2}=\frac{2+\sqrt{1+8\gamma/\alpha}}{2}\notin \{-1,-2,-3,\dots\}$$
Then the solutions are
$$v(x)=k_1~{}_2F_1\left(\begin{matrix}a,b\\c\end{matrix}~\bigg|~x\right)+k_2~x^{1-c}~{}_2F_1\left(\begin{matrix}a,b\\c\end{matrix}~\bigg|~x\right)$$
If, on the other hand $c$ is a negative integer, things get more complicated, see for example this reference. So, putting it all together,
$$u(x)=k_1~x^{p}~{}_2F_1\left(\begin{matrix}a,b\\c\end{matrix}~\bigg|~x\right)+k_2~x^{p+1-c}~{}_2F_1\left(\begin{matrix}a,b\\c\end{matrix}~\bigg|~x\right)$$
It is clear why it did not matter which solution for $a,b$ we picked, since the hypergeometric function is symmetric w.r.t $a,b$.
It is also a bit clearer why it did not matter which $p_{\pm}$ we chose, as you can check for yourself that $p_++1-c_+=p_-$ and likewise $p_-+1-c_-=p_+$. However, we need to rigorously confirm that the solution span is unchanged by this choice. I was.... not able to do this. But I hope you can.
A: Mathematica says:
$$
\left\{\left\{y(x)\to c_1(-1)^{\frac{\sqrt{A}-\sqrt{A+8C}}{4\sqrt{A}}}x^{\frac{\sqrt{A}-\sqrt{A+8C}}{4\sqrt{A}}}\,_2F_1\left(-\frac{i\sqrt{B}}{\sqrt{2}\sqrt{A}}-\frac{\sqrt{A+8C}}{4\sqrt{A}}+\frac{1}{4},\frac{i\sqrt{B}}{\sqrt{2}\sqrt{A}}-\frac{\sqrt{A+8C}}{4\sqrt{A}}+\frac{1}{4};1-\frac{\sqrt{A+8C}}{2\sqrt{A}};x\right)+c_2(-1)^{\frac{\sqrt{A+8C}+\sqrt{A}}{4\sqrt{A}}}x^{\frac{\sqrt{A+8C}+\sqrt{A}}{4\sqrt{A}}}\,_2F_1\left(-\frac{i\sqrt{B}}{\sqrt{2}\sqrt{A}}+\frac{\sqrt{A+8C}}{4\sqrt{A}}+\frac{1}{4},\frac{i\sqrt{B}}{\sqrt{2}\sqrt{A}}+\frac{\sqrt{A+8C}}{4\sqrt{A}}+\frac{1}{4};\frac{\sqrt{A+8C}}{2\sqrt{A}}+1;x\right)\right\}\right\}
$$
