Difficult Second-Order ODE solution?

Question

I'm looking at the functional

$$T[y] = \int_a^b\sqrt{\frac{1+y'^2}{2g(y-\mu x)}}\ \mathrm dx\tag{1}$$

and trying to minimize it via the Euler-Lagrange-equations, which I can do, however, it seems that the simplification

$$(1+y'^2)(1+\mu y')+2(y-\mu x)y'' =0\tag{2}$$

is correct, and exactly how the rather nasty output of the E-L-equation simplifies to this is eluding me. I'm also rather interested in how one goes about solving such differential equation... Any help is appreciated.

Answer 1

Hint.

You can get

$$(1+y'^2)(1+\mu y')+2(y-\mu x)y'' =0$$

keeping the numerator from the messy fraction. After that, making the transformation $Y = y-\mu x$ gives

$$ (1+(Y'+\mu)^2)(1+\mu(Y'+\mu))+YY''=0 $$

This nonlinear DE is solved with the contribution of MATHEMATICA

$$ y = \text{InverseFunction}\left[\frac{2 \left(\sqrt{\text{$\#$1} \left(e^{2 c_1} \left(\mu ^2+1\right)-\text{$\#$1}\right)}\pm\text{$\#$1} \mu \right)+e^{2 c_1} \left(\mu ^2+1\right) \tan ^{-1}\left(\frac{2 \sqrt{\text{$\#$1} \left(e^{2 c_1} \left(\mu ^2+1\right)-\text{$\#$1}\right)}}{2 \text{$\#$1}-e^{2 c_1} \left(\mu ^2+1\right)}\right)}{2 \left(\mu ^2+1\right)}\&\right]\left[c_2+x\right] $$

Answer 2

FWIW, it is possible to obtain a first integral via the Beltrami identity. The trick is to introduce Cesareo's variable $$Y~:=~y-\mu x$$ to remove explicit $x$-dependence in OP's Lagrangian $$ L~=~\sqrt{\frac{1+y^{\prime 2}}{y-\mu x}} ~=~ \sqrt{\frac{1+(Y^{\prime}+\mu)^2}{Y}}.$$ The momentum & energy read $$ P~=~\frac{\partial L}{\partial Y^{\prime}} ~=~\frac{y^{\prime}}{\sqrt{Y}\sqrt{1+y^{\prime 2}}}$$ and $$ E~=~PY^{\prime}-L~=~-\frac{1+\mu y^{\prime}}{\sqrt{Y}\sqrt{1+y^{\prime 2}}},$$ respectively. The energy is a first integral.