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I was going through the following question: Solve the following differential equation $(1+xy)ydx + (1-xy)xdy=0$.

I have been taught to solve differential equations of the following type:

1.Homogeneous

2.Variable separable

3.Linear differential equation

But, I'm unable to find to which category the given differential equation belongs.Please help. I would like some hints towards solving the question and please don't post the answer.

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  • $\begingroup$ Is your expression equal to something ? $\endgroup$ – Claude Leibovici Dec 4 '13 at 12:11
  • $\begingroup$ One of the $dy$ terms should be a $dx$. Then presumably this is an exact differential equation. $\endgroup$ – hunter Dec 4 '13 at 12:11
  • $\begingroup$ Sorry for the typo...I was just getting mad trying to solve this question. $\endgroup$ – Rajath Krishna R Dec 4 '13 at 12:12
  • $\begingroup$ Good luck ! It is just awful to me ! $\endgroup$ – Claude Leibovici Dec 4 '13 at 12:15
  • $\begingroup$ hmm, it's not quite exact after all. $\endgroup$ – hunter Dec 4 '13 at 12:18
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Set $\xi = xy$, $\eta = \frac{x}{y}$. Then $$\begin{align} \mathrm{d}\xi &= y\,\mathrm{d}x + x\,\mathrm{d}y & \mathrm{d}\eta &= \frac{y\,\mathrm{d}x - x\,\mathrm{d}y}{y^2} \end{align}$$ And therefore $$\mathrm{d}\xi + \frac{\xi^2}{\eta}\mathrm{d}\eta = 0$$ I suppose, you can take it from there.

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  • $\begingroup$ Thank you sir for your answer. While manipulating the given differential equation I didn't have the insight towards making it into this form that you have made. Does that insight come through more practice? $\endgroup$ – Rajath Krishna R Dec 4 '13 at 12:48
  • $\begingroup$ Of course, over time (with practice) one learns to recognize such patterns. Enjoy! $\endgroup$ – ccorn Dec 4 '13 at 12:54
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IF you have anything of form f(x.y)y dx + g(x.y)x dy = 0 ; then I.F. will be 1/(Mx-Ny) . Multiply this to make it exact differential equation.

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  • $\begingroup$ Your method would work, add more details. $\endgroup$ – Nosrati Aug 7 '18 at 15:30

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