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2xy-tan-y-dx-x-2-x-sec-2-y-dy-0-




Question Number 122163 by benjo_mathlover last updated on 14/Nov/20
 (2xy−tan y) dx + (x^2 −x sec^2 y) dy = 0
$$\:\left(\mathrm{2}{xy}−\mathrm{tan}\:{y}\right)\:{dx}\:+\:\left({x}^{\mathrm{2}} −{x}\:\mathrm{sec}\:^{\mathrm{2}} {y}\right)\:{dy}\:=\:\mathrm{0}\: \\ $$
Answered by liberty last updated on 14/Nov/20
⇒d(x^2 y)−d(xtan y) = 0  ⇒ d(x^2 y) = d(xtan y)   ⇒∫ d(x^2 y) = ∫ d(x tan y)  ⇒ x^2 y = x tan y + C  ⇒x^2 y−x tan y = C . ▲
$$\Rightarrow\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} \mathrm{y}\right)−\mathrm{d}\left(\mathrm{xtan}\:\mathrm{y}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} \mathrm{y}\right)\:=\:\mathrm{d}\left(\mathrm{xtan}\:\mathrm{y}\right)\: \\ $$$$\Rightarrow\int\:\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} \mathrm{y}\right)\:=\:\int\:\mathrm{d}\left(\mathrm{x}\:\mathrm{tan}\:\mathrm{y}\right) \\ $$$$\Rightarrow\:\mathrm{x}^{\mathrm{2}} \mathrm{y}\:=\:\mathrm{x}\:\mathrm{tan}\:\mathrm{y}\:+\:\mathrm{C} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} \mathrm{y}−\mathrm{x}\:\mathrm{tan}\:\mathrm{y}\:=\:\mathrm{C}\:.\:\blacktriangle\: \\ $$

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