Emden‐Fowler differential equation Good day.
I am trying to solve the following equation:
$$\ddot{y}(x)-\frac{A}{x}\dot{y}(x)+\frac{Bx^2}{2}y(x)=0.$$
WolframAlpha says it is an Emden‐Fowler equation, but I have no idea how to solve this. Can you give me some tips?
In case of $$A=B/2=1$$ WolframALpha gives an analytical solution $$y(x)=c_1 sin\frac{x^2}{2}+c_1 cos\frac{x^2}{2}.$$
If there is no analytical way to solve, can I do it numerical?
Thank you.
 A: $$\frac{d^2y}{dx^2}-\frac{A}{x}\frac{dy}{dx}+\frac{Bx^2}{2}y(x)=0.$$
Changing $x$ into $-x$ doesn't change the equation. This draw us to a change of variable 
$t=x^2 \quad;\quad dt=2x\:dx \quad;\quad \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=2x\frac{dy}{dt}\quad;\quad \frac{d^2y}{dx^2}=2\frac{dy}{dt}+2x\frac{d^2y}{dt^2}\frac{dt}{dx}=2\frac{dy}{dt}+4t\frac{d^2y}{dt^2}$
$$2\frac{dy}{dt}+4t\frac{d^2y}{dt^2}-2A\frac{dy}{dt} +\frac{B}{2}ty=0.$$
$$4t\frac{d^2y}{dt^2}+2(1-A)\frac{dy}{dt} +\frac{B}{2}ty=0.$$
This is an ODE of the Bessel kind. 
In the case of $A=1$ the ODE reduces to :
$$4\frac{d^2y}{dt^2} +\frac{B}{2}y=0.$$
The solution is wellknown :
$$y=c_1 \cos\left(\sqrt{\frac{B}{8}}t\right) + c_2 \sin\left(\sqrt{\frac{B}{8}}t\right)$$
$$y=c_1 \cos\left(\sqrt{\frac{B}{8}}x^2\right) + c_2 \sin\left(\sqrt{\frac{B}{8}}x^2\right)$$
In the case of $A\neq 1$ the solutions are expressed thanks to the Bessel functions :
$$y=C_1 x^{\frac{A+1}{2}}J_{\frac{A+1}{4}}\left(\sqrt{\frac{B}{8}}x^2 \right) + C_2 x^{\frac{A+1}{2}}Y_{\frac{A+1}{4}}\left(\sqrt{\frac{B}{8}}x^2 \right)$$
The functions $J$ and $Y$ are the Bessel functions of the first and second kind respectively.
A: For the example you give with the solution you have
$$
y''-\frac{1}{x}y'+x^2y = 0 \implies \frac{1}{x^2}y''-\frac{1}{x^3}y' +y = 0
$$
multiply by $y'$ we find
$$
\frac{1}{x^2}\dfrac{d}{dx}\frac{y'^2}{2}-\frac{1}{x^3}y'^2+yy'= \frac{1}{2x^2}\dfrac{d}{dx}y'^2-\frac{1}{x^3}y'^2+\dfrac{d}{dx}\frac{y^2}{2} = 0
$$
thus
$$\frac{1}{x^2}\dfrac{d}{dx}y'^2-\frac{2}{x^3}y'^2+\frac{d}{dx}y^2=0
$$
we can see
$$
\dfrac{d}{dx}\frac{y'^2}{x^2}=\frac{1}{x^2}\dfrac{d}{dx}y'^2-\frac{2}{x^3}y'^2
$$
thus you get
$$
\dfrac{d}{dx}\frac{y'^2}{x^2} +\dfrac{d}{dx}y^2 = 0
$$
or
$$
\frac{y'^2}{x^2}+y^2 = \lambda
$$
can you take it from here?
