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,

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.