How Do You Solve the Differential Equation 6yy' = x?

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mr_coffee
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hello everyone, this is probably simple but its in a weird form or maybe I'm just not seeing it:
6yy' = x; y(6) = 4;

Is x = g(x) and 6 = p(x)?
or is this not an integrating factor?
 
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It's separable: 6ydy = xdx
 
wow that was easy, thank you for the fast repsonce!, i don't know why i had a brain fart there:
sqrt(x^2/6+10)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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