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Question Number 93818 by i jagooll last updated on 15/May/20

(y−xy^2 )dx +(x+x^2 y^2 )dy = 0

$$\left(\mathrm{y}−\mathrm{xy}^{\mathrm{2}} \right)\mathrm{dx}\:+\left(\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \right)\mathrm{dy}\:=\:\mathrm{0} \\ $$

Answered by i jagooll last updated on 15/May/20

ydx+xdy=xy^2 dx−x^2 y^2 dy d(xy) = x^2 y^2 ((dx/x)−dy) ((d(xy))/((xy)^2 )) = (dx/x)−dy ∫ ((d(xy))/((xy)^2 )) = ∫ (dx/x)−∫dy −(1/(xy)) = ln (x)−y+c (1/(xy)) = y+ln(x)+C ⇒ xy = (1/(y+ln(x)+C))

$$\mathrm{ydx}+\mathrm{xdy}=\mathrm{xy}^{\mathrm{2}} \mathrm{dx}−\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \mathrm{dy} \\ $$$$\mathrm{d}\left(\mathrm{xy}\right)\:=\:\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:\left(\frac{\mathrm{dx}}{\mathrm{x}}−\mathrm{dy}\right) \\ $$$$\frac{\mathrm{d}\left(\mathrm{xy}\right)}{\left(\mathrm{xy}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{dx}}{\mathrm{x}}−\mathrm{dy}\: \\ $$$$\int\:\frac{\mathrm{d}\left(\mathrm{xy}\right)}{\left(\mathrm{xy}\right)^{\mathrm{2}} }\:=\:\int\:\frac{\mathrm{dx}}{\mathrm{x}}−\int\mathrm{dy} \\ $$$$−\frac{\mathrm{1}}{\mathrm{xy}}\:=\:\mathrm{ln}\:\left(\mathrm{x}\right)−\mathrm{y}+\mathrm{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{xy}}\:=\:\mathrm{y}+\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{C}\:\Rightarrow\:\mathrm{xy}\:=\:\frac{\mathrm{1}}{\mathrm{y}+\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{C}}\: \\ $$

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