Differential equation question

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This is not a homework question, but I am posting here so as not to run afoul of the "rules."

Homework Statement



(1/z) * dz/dx = a* \sqrt{dy/dx}

where x,y,z are variables and a is a constant.

Homework Equations



See above

The Attempt at a Solution



\left[ (1/z) * dz/dx = a*\sqrt{dy/dx} \right] *dx<br /> <br /> Thus,<br /> dz/z = a* \sqrt{dy * dx} <br /> <br /> \int dz/z = \int a* \sqrt{dy * dx}<br /> <br /> ln(z) = \int a \sqrt{dy * dx} <br /> <br /> ??
 
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And are you allowed to do something like:
<br /> (dz/z)^2 = a^2* dy * dx<br />
 
(1) I would say: first get rid of the square root, then continue. Those calculations with square roots and bare differentials are questionable.

(2) Since there are three variables (x,y,z), what are you supposed to do? For example: Let y be ANY function of x, plug it in and get a differential equation to solve for z... Would that be good for you?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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