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Question Number 93818 by i jagooll last updated on 15/May/20 | ||
$$\left(\mathrm{y}−\mathrm{xy}^{\mathrm{2}} \right)\mathrm{dx}\:+\left(\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \right)\mathrm{dy}\:=\:\mathrm{0} \\ $$ | ||
Answered by i jagooll last updated on 15/May/20 | ||
$$\mathrm{ydx}+\mathrm{xdy}=\mathrm{xy}^{\mathrm{2}} \mathrm{dx}−\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \mathrm{dy} \\ $$$$\mathrm{d}\left(\mathrm{xy}\right)\:=\:\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:\left(\frac{\mathrm{dx}}{\mathrm{x}}−\mathrm{dy}\right) \\ $$$$\frac{\mathrm{d}\left(\mathrm{xy}\right)}{\left(\mathrm{xy}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{dx}}{\mathrm{x}}−\mathrm{dy}\: \\ $$$$\int\:\frac{\mathrm{d}\left(\mathrm{xy}\right)}{\left(\mathrm{xy}\right)^{\mathrm{2}} }\:=\:\int\:\frac{\mathrm{dx}}{\mathrm{x}}−\int\mathrm{dy} \\ $$$$−\frac{\mathrm{1}}{\mathrm{xy}}\:=\:\mathrm{ln}\:\left(\mathrm{x}\right)−\mathrm{y}+\mathrm{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{xy}}\:=\:\mathrm{y}+\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{C}\:\Rightarrow\:\mathrm{xy}\:=\:\frac{\mathrm{1}}{\mathrm{y}+\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{C}}\: \\ $$ | ||