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I found this separable differential equation $y(y')^3=a$ at Dr. Chris Tisdell's video on separable equations (moment 33:01) , since I haven't worked an example that looks like this I don't really see how the separation works,

First by separation and integrating both sides w.r.t $x$ each time then cancelling : $\int y (\frac{dy}{dx})^3 dx = \int a dx $ , what I get is $y = [4ax^3 +12c_1 x^2 +24c_2x+c_3]^{1/4} $ which is not the correct answer shown down the video,

Second thing, if we keep integrating then we want to verify for the particular solution, $y(c)=d$ I don't see why the constants are solved in one equation of three variables,

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  • $\begingroup$ Before you can integrate, you need to take the cube root on both sides, I’m pretty sure you can’t integrate $y(y’)^3$ $\endgroup$
    – Robo300
    Feb 28 '20 at 22:58
  • $\begingroup$ Realise that you if you have a variable $x$, then you can integrate it if and only if you have a $dx$ present, which I remind is an operator. I don't see how you have integrated. Please revise your integration again. $\endgroup$
    – PCeltide
    Feb 28 '20 at 23:03
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See that $y(y')^3=a$ so, $y'=(\frac{a}{y})^{\frac{1}{3}}$. This leads to $$\int y^{\frac{1}{3}}dy=\int (a)^{\frac{1}{3}}dx$$ which solves to $$\frac{3}{4}y^{\frac{4}{3}}=x\cdot a^{\frac{1}{3}} +c$$ for some constant c.

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