<s>[INST] Solving non-linear ordinary differential equation: $(y'')^2-y'y'''=\left(\frac{y'}{x}\right)^2$ $$(y'')^2-y'y'''=\left(\frac{y'}{x}\right)^2$$
I have been struggling to solve this equation by doing following simplifications:
$$y'''y'=(y'')^2-\left(\frac{y'}{x}\right)^2$$
Dividing by $y'$:
$$y'''=\frac{(y'')^2}{y'}-\frac{y'}{x^2}$$
Dividing by $y''$:
$$\frac{y'''}{y''}=\frac{y''}{y'}-\frac{y'}{y''x^2}$$
I can't figure out what goes next since I have never encountered with the last fraction. I would really appreciate any help on the matter.
 [/INST] Solving : 
$$ \Bigg(\frac{d^2 y(x)}{dx^2}\Bigg)  - \frac{dy(x)}{dx}\frac{d^3 y(x)}{ dx^3} = \frac{\frac{dy(x)}{dx}^2}{x^2}$$
Let :
$$\frac{dy(x)}{dx} = v(x)$$
which yields :
$$\frac{d^2 y(x)}{dx^2} = \frac{dv(x)}{dx}$$ 
$$\frac{d^3 y(x)}{ dx^3}= \frac{d^2 v(x)}{dx^2}$$
You get :
$$\Bigg( \frac{dv(x)}{dx}\Bigg)^2 -v(x)\frac{d^2 v(x)}{dx^2} = \frac{v(x)^2}{x^2}$$
$$\Rightarrow$$
$$\Bigg[-x^2 \Bigg( \frac{dv(x)}{dx}\Bigg)^2 + v(x)^2 + x^2\frac{d^2 v(x)}{dx^2}v(x)\Bigg]\frac{1}{x^2} = 0 $$
$$\Rightarrow$$
$$\frac{1}{x^2} + \frac{\frac{d^2 v(x)}{dx^2}}{v(x)} - \frac{\Bigg(\frac{dv(x)}{dx}\Bigg)^2}{v(x)^2} = 0 $$
$$\Rightarrow$$
$$\int \Bigg[\frac{1}{x^2} + \frac{\frac{d^2 v(x)}{dx^2}}{v(x)} - \frac{\Bigg(\frac{dv(x)}{dx}\Bigg)^2}{v(x)^2}\Bigg]dx = \int0dx$$
$$\Rightarrow$$ 
$$- \frac{1}{x} + \frac{\frac{dv(x)}{dx}}{v(x)} = c_1$$
$$\Rightarrow$$ 
$$\frac{dv(x)}{dx} = \frac{(c_1x+1)v(x)}{x}$$
$$\Rightarrow$$ 
$$\int \frac{dv(x)}{dx} dx = \int \frac{(c_1x+1)v(x)}{x} dx$$
$$\Rightarrow$$ 
$$\ln{v(x)} = \ln x + c_1x + c_2$$
$$\Rightarrow$$ 
$$v(x) =  e^{c_1x + c_2}x$$
$$\Rightarrow$$
$$v(x) = c_2 c_1^x x$$
Substituting back for $\frac{dy(x)}{dx} = v(x)$, you get :
$$\frac{dy(x)}{dx} = c_2 c_1^x x $$
$$\Rightarrow$$
$$y(x) =  \int c_2 c_1^x x dx = \frac{c_2 (\ln{(c_1)}x-1)c_1^x}{\ln^2c_1} + c_3 $$
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