I'm currently trying to solve the differential equation: $$x^2y'^2-2(xy-4)y'+y^2=0,$$ but up to now I've had no succes. I rewrote it as


and substituted


hoping that the equation would become seperable. Unfortunately this only works well for linear equations. So I got: $$x^2v'^2-4vv'x+4v^2+8xv'-8v=0.$$ This seems pretty hard to solve. I don't know if there is a better substitution, or simply a better method to solve this one ?

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    $\begingroup$ The substitution $y/x$ may be more useful, since its derivative is $(x y'-y)/x^2$. $\endgroup$ – Semiclassical Oct 9 '14 at 11:51
  • $\begingroup$ @Nick probably you should use as mentioned above, $v= \frac{y}{x}$. This gets us $x^2v' = xy'-y$, and also that $y ' = \frac{x^2v'+y}{x} = xv' + v$ $\endgroup$ – Varun Iyer Oct 9 '14 at 11:54
  • $\begingroup$ @semiclassical, what might me the motivation for this substitution? Is it that by dividing the entire equation bij x^2 that you see this ? $\endgroup$ – Nick Oct 9 '14 at 12:03
  • $\begingroup$ That and a measure of intuition. It also amounts to writing $y=v x$, which based on your ODE seems like a good change of variables. $\endgroup$ – Semiclassical Oct 9 '14 at 12:25
  • $\begingroup$ @semiclassical, your substitution yields $x^4(v')^2+8xv'+8v=0$. It seems slightly better, but stil hard to solve :p $\endgroup$ – Nick Oct 9 '14 at 12:31

your rearrangement looks so handy i wondered what happens if we set: $$ v = xy'-y $$ this gives $$ v' =xy'' $$ the equation $$ v^2 +8y' =0 $$ may be differentiated to give $$ 2vv'+8y''=0 $$ multiplying by $x$ and factoring out $2$ gives $$ xvv' + 4v'=0 $$ i.e. $$ (xv+4)v'=0 $$

  • $\begingroup$ So then I could go with the cases $v'=0$ and $v=-4/x$ ? $\endgroup$ – Nick Oct 9 '14 at 12:40
  • $\begingroup$ i think that idea might be worth exploring. however the differentiation might introduce some complication. $v'=0$ gives the family of straight lines $y=ax+b$ subject to the constraint that $b^2=8a$. so these lines should define an envelope of the one-parameter family of solutions of $v=-\frac4{x}$ which are something like $y=x(c+\frac2{x^2})$. can you investigate this? $\endgroup$ – David Holden Oct 9 '14 at 14:00
  • $\begingroup$ I get the constraint that $b^2=-8a$. For the other equation I'm looking for a solution ... $\endgroup$ – Nick Oct 9 '14 at 14:16
  • $\begingroup$ you should find $\frac{d}{dx}\left(\frac{y}{x}\right) = -\frac4{x^3}$ $\endgroup$ – David Holden Oct 9 '14 at 14:33
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    $\begingroup$ my calculation agrees with that result, so well done! $\endgroup$ – David Holden Oct 9 '14 at 20:59

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