$$(yy')^3 = 27x(y^2-2x^2)$$ I tried a lot, but one what i see is $yy'=(y^2)'$ and then we get $z'^3 = 216x(z-2x^2)$ I have no idea, please, hint a substitution.
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- $\begingroup$ you forgot the prime in the final equation. Also $yy' = \dfrac{1}{2}\left(y^2\right)'$ $\endgroup$Chinny84– Chinny842014-12-12 17:13:53 +00:00Commented Dec 12, 2014 at 17:13
- $\begingroup$ yes, thank you) $\endgroup$Simankov– Simankov2014-12-12 17:17:42 +00:00Commented Dec 12, 2014 at 17:17
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1 Answer
$\begingroup$ $\endgroup$
4 Let $y = \sqrt{2}x v$ $$ 2xv(xv)' = \left(27\cdot 2 x^3\left(v^2-1\right)\right)^{1/3} = 3(2)^{1/3}x\left(v^2-1\right)^{1/3} $$ or $$ 2xv^2 + 2x^2vv' = 3(2)^{1/3}x\left(v^2-1\right)^{1/3} $$ Can you see how to finish it? Let $u =v^2$ $$ x^2u' = 2x\left(3(2)^{-2/3}\left(u-1\right)^{1/3} -u\right) $$
- $\begingroup$ No.In my opinion, it became worse) $\endgroup$Simankov– Simankov2014-12-12 17:25:25 +00:00Commented Dec 12, 2014 at 17:25
- $\begingroup$ It becomes separable now I do admit the integral is tricky, $\endgroup$Chinny84– Chinny842014-12-12 17:29:31 +00:00Commented Dec 12, 2014 at 17:29
- $\begingroup$ yes, thank, got it $\endgroup$Simankov– Simankov2014-12-12 18:02:15 +00:00Commented Dec 12, 2014 at 18:02
- $\begingroup$ is it a good answer? $$ \frac{(\frac{y^2}{2x^2}-1)^{4/3}}{2^{7/3}} - ln{\frac{|y|}{\sqrt(2)}}= C$$ $\endgroup$Simankov– Simankov2014-12-15 19:12:24 +00:00Commented Dec 15, 2014 at 19:12