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In a problem making an integral stationary, I'm being told that this last implication is wrong, why? Is my integration wrong?

$$ \frac{d}{dx} y(x) = \pm\frac{c_1x^2}{\sqrt{x^2 - c_1^2}}$$ $$\implies y(x) = \pm c_1\sqrt{x^2-c^2} + c_2$$

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    $\begingroup$ Yes that integral is incorrect. Take the derivative and you will see that the equation is not returned to its original form $\endgroup$ Oct 5, 2014 at 5:31

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If the problem was $$\frac{d}{dx} y(x) = \pm\frac{c_1x}{\sqrt{x^2 - c_1^2}}$$ your result would be correct. But, for $$\frac{d}{dx} y(x) = \pm\frac{c_1x^2}{\sqrt{x^2 - c_1^2}}$$ it is different. Change variable $x=c \cosh(t)$ and use $\cosh(2t)=2\cosh^2(t)-1$. You should arrive to $$y(x)=\pm c_1 \left(\frac{1}{2} x \sqrt{x^2-c_1^2}+\frac{1}{2} c_1^2 \log \left(\sqrt{x^2-c_1^2}+x\right)\right)+c_2$$

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