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How to approach the Differential equation $$2(x-y \, \sin(2x)) \, dx + (3y^{2} + \cos(2x)) \, dy=0$$ I cannot see whether the equation is in any standard form. Initial steps and/or hints appreciated.

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  • $\begingroup$ Have you heard of "exact differential equation"? $\endgroup$ – bof Sep 17 '18 at 6:15
  • $\begingroup$ The DE is in the form $M(x,y)dx+N(x,y)dy=0.$ What is the partial derivative of $M$ with rspect to $y$? What is the partial derivative of $N$ with respect to $x$? $\endgroup$ – bof Sep 17 '18 at 6:17
  • $\begingroup$ Is x and y supposed to be some function of time or something? Your notation looks a lot like a differential form. $\endgroup$ – user2662833 Sep 17 '18 at 6:36
  • $\begingroup$ Thank you. I had already done it. It was very easy infact. $\endgroup$ – Arka Seth Sep 17 '18 at 8:43
  • $\begingroup$ @user2662833. Divide the equation formally with $dx$ and — voilà — a differential equation appears. $\endgroup$ – md2perpe Sep 17 '18 at 17:50
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$$2(x-y \, \sin(2x)) \, dx + (3y^{2} + \cos(2x)) \, dy=0$$ $$2(x-y \, \sin(2x)) + (3y^{2} + \cos(2x)) y'=0$$ $$2x +(y\cos(2x))'+(y^3)'=0$$ Integrate $$x^2 +y\cos(2x)+y^3=K$$

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    $\begingroup$ Thank you... I'd already solved it by the same method. $\endgroup$ – Arka Seth Sep 17 '18 at 13:05
  • $\begingroup$ yw @ArkaSeth... $\endgroup$ – LostInSpace Sep 17 '18 at 13:23

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