Find critical points of the functional
$I[y] = c \int_0^L y(y')^3 dx$ with $y(0)=0$ and $y(L)=R$
Euler-Lagrange equation: I arrive at $(y')^3+3yy'y''=0$ and so solve $y'=0$ and $(y')^2+3yy''=0.$
First one is easy to solve: $y=c$.
Second one I have no idea to solve (could you explain the method please) ^ using wolfram I get $y=a(b+4x)^{\frac{3}{4}}$
Using the conditions I think $b=0, a=\frac{R}{(4L)^{\frac{3}{4}}}$
Is this correct?
$$(y')^2+3yy''=0$$ $$1-3y\frac{d(1/y')}{dx}=0$$ $$1-3y\frac{d(1/y')}{dy}y'=0$$ $$\frac{dy}{3y}=\frac{d(1/y')}{1/y'}$$ $$y^{1/3}=\frac{c}{y'}$$
You can take it from here