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I'm solving the differential equation

$$y - yy'^2 = 6y'^2$$

I have a worked out solution but I got a much shorter approach. The solution is given in parametric equations, while mine isn't. Here is how I attempted to solve the problem:

We consider this as a differential equation where $x$ is a function of $y$. The differential equation becomes:

$$y - \frac{y}{x'^2} = \frac{6}{x'^2}$$

or equivalently:

$$yx'^2 - y = 6$$

or $$x' = \pm\sqrt{\frac{6+y}{y}}$$

the result follows by integrating with respect to $y$. Does this seem correct?

The solution my book suggests is to solve the initial equation for $y$ and then take the derivative of both sides where one makes the substitution $y' = p$.

Thanks in advance.

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    $\begingroup$ Why not solve directly $$\sqrt{1+\frac6y}y'=\pm1\ ?$$ $\endgroup$ – Did Jun 19 '17 at 12:01
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There isn't really any advantage to switching to $dx/dy$: you can rearrange the original equation to $$ y'^2 \frac{y+6}{y} = 1, $$ which is essentially the same thing. One can then take the square root and integrate by putting $u=(y+6)/y$ in the usual way.

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  • $\begingroup$ Thank you for your answer. Was my approach correct too? (I know it was overcomplicated) $\endgroup$ – user370967 Jun 19 '17 at 12:03
  • $\begingroup$ Yes, it's essentially equivalent. $\endgroup$ – Chappers Jun 19 '17 at 12:09

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