I've worked my way through a variations problem and I've arrived to two integrals:
$$\int dx = \int {1\over \sqrt[]{{y^2\over c^2}-1}}dy $$
Solving for them, I find
$$x+c_2 = c_1\ln (c_1\sqrt[]{{y^2\over c_1^2}-1}+y)$$
And here I'm stuck. I'm looking for $y(x)$, but it doesn't seem clear to me how to approach this. I've looked at
$$e^{x+c_2\over c_1} =c_1 \sqrt[]{{y^2\over c_1^2}-1}+y$$
But that doesn't seem to get me anywhere. Can anyone point me in the right direction?
You have $$\left(\frac{e^{\frac{x+c_2}{c_1}}-y}{c_1}\right)^2=\frac{y^2}{c_1^2}-1.$$ Expanding, we get
$$\frac{y^2}{c_1^2}-2\frac{y}{c_1^2}e^{\frac{x+c_2}{c_1}}+\frac{e^{2\frac{x+c_2}{c_1}}}{c_1^2}=\frac{y^2}{c_1^2}-1.$$ Notice that $y^2/c_1^2$ cancels. You get $$-2\frac{y}{c_1^2}e^{\frac{x+c_2}{c_1}}+\frac{e^{2\frac{x+c_2}{c_1}}}{c_1^2}=-1.$$ Now just solve for $y$ in the obvious way: $$2\frac{y}{c_1^2}e^{\frac{x+c_2}{c_1}}=1+\frac{e^{2\frac{x+c_2}{c_1}}}{c_1^2},$$ $$y=\frac{c_1^2}{2}e^{-\frac{x+c_2}{c_1}}+\frac{1}{2}e^{\frac{x+c_2}{c_1}}.$$