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Question Number 42228 by rahul 19 last updated on 20/Aug/18

Solve :  2x^2 ydx −2y^4 dx+2x^3 dy+3xy^3 dy=0.

$$\mathrm{Solve}\:: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} {ydx}\:−\mathrm{2}{y}^{\mathrm{4}} {dx}+\mathrm{2}{x}^{\mathrm{3}} {dy}+\mathrm{3}{xy}^{\mathrm{3}} {dy}=\mathrm{0}. \\ $$

Answered by ajfour last updated on 20/Aug/18

2x^2 (ydx+xdy)=y^3 (2ydx−3xdy)  ⇒ 2x^2 d(xy)=y^3 {((d(x^2 /y^3 ))/(xy^2 ))}×y^6   ⇒ 2x^2 d(xy)=x^3 y((y^3 /x^2 ))^2 d((x^2 /y^3 ))  ⇒  2∫ ((d(xy))/(xy)) = ∫((d(x^2 /y^3 ))/((x^2 /y^3 )^2 ))  ⇒ 2ln (xy)=−(y^3 /x^2 ) +c .

$$\mathrm{2}{x}^{\mathrm{2}} \left({ydx}+{xdy}\right)={y}^{\mathrm{3}} \left(\mathrm{2}{ydx}−\mathrm{3}{xdy}\right) \\ $$$$\Rightarrow\:\mathrm{2}{x}^{\mathrm{2}} {d}\left({xy}\right)={y}^{\mathrm{3}} \left\{\frac{{d}\left({x}^{\mathrm{2}} /{y}^{\mathrm{3}} \right)}{{xy}^{\mathrm{2}} }\right\}×{y}^{\mathrm{6}} \\ $$$$\Rightarrow\:\mathrm{2}{x}^{\mathrm{2}} {d}\left({xy}\right)={x}^{\mathrm{3}} {y}\left(\frac{{y}^{\mathrm{3}} }{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} {d}\left(\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{3}} }\right) \\ $$$$\Rightarrow\:\:\mathrm{2}\int\:\frac{{d}\left({xy}\right)}{{xy}}\:=\:\int\frac{{d}\left({x}^{\mathrm{2}} /{y}^{\mathrm{3}} \right)}{\left({x}^{\mathrm{2}} /{y}^{\mathrm{3}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{2ln}\:\left({xy}\right)=−\frac{{y}^{\mathrm{3}} }{{x}^{\mathrm{2}} }\:+{c}\:. \\ $$

Commented by rahul 19 last updated on 20/Aug/18

Awesome ! ������

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Aug/18

excellent...

$${excellent}... \\ $$

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