Solve the Euler's equation for the functional:
$$J[y]=\int_1^2\frac{\sqrt{1+y'^2}}{x}dx$$
with $y(1)=0$ and $y(2)=1$
This would be the automatic approach:
$$\frac{d}{dx}\frac{\partial F}{\partial y'}-\frac{\partial F}{\partial y}=0$$
which gives
$$\frac{d}{dx}\frac{2y'}{2x\sqrt{1+y'^2}}=0$$
$$\frac{y'}{x\sqrt{1+y'^2}}=C$$
$$y'=Cx\sqrt{1+y'^2}$$
$$y'^2=Cx^2+cx^2y'^2$$
$$y'^2-Cx^2y'^2=Cx^2$$
$$y'^2=\frac{Cx^2}{1-Cx^2}$$
$$y'=\frac{\sqrt{C}x}{\sqrt{1-Cx^2}}$$
Integration gives
$$y(x)=-\frac{\sqrt{1-Cx^2}}{\sqrt{C}}+D$$
Then we rearrange:
$$y(x)=-\frac{\sqrt{1-Cx^2}}{\sqrt{C}}+D$$
$$y-D=-\frac{\sqrt{1-Cx^2}}{\sqrt{C}}$$
$$(y-D)^2=\frac{1}{C}-x^2$$
$$(y-D)^2+x^2=\frac{1}{C}$$
Insert for initial condition 1:
$$(D)^2+1^2=\frac{1}{C}\longrightarrow D=\sqrt{1/C-1}$$
Insert for initial condition 2:
$$(1-D)^2+2^2=\frac{1}{C}\longrightarrow D=1-\sqrt{1/C-4}$$
Any ideas how to solve this?
Thanks