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Question Number 40442 by rahul 19 last updated on 21/Jul/18

Solve : y^4 dx + 2xy^3 dy = ((ydx− xdy)/(x^3 y^3 )).

$$\mathrm{Solve}\:: \\ $$$$\mathrm{y}^{\mathrm{4}} \mathrm{d}{x}\:+\:\mathrm{2}{x}\mathrm{y}^{\mathrm{3}} \mathrm{dy}\:=\:\frac{\mathrm{yd}{x}−\:{x}\mathrm{dy}}{{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{3}} }. \\ $$

Answered by ajfour last updated on 21/Jul/18

y^2 (y^2 dx+2xydy)=((d(x/y))/(x^3 y)) ⇒ y^2 d(xy^2 )=((d(x/y))/(x^3 y)) ...(i) _____________________ ⇔ (xy^2 )^m d(xy^2 )=((x/y))^n d((x/y)) ⇒ m−n=3 and 2m+n=3 ⇒ m=2 and n=−1 _____________________ So (i) becomes ∫(xy^2 )^2 d(xy^2 )= ∫((d(x/y))/((x/y))) ⇒ (((xy^2 )^3 )/3)=ln (x/y)+c .

$${y}^{\mathrm{2}} \left({y}^{\mathrm{2}} {dx}+\mathrm{2}{xydy}\right)=\frac{{d}\left({x}/{y}\right)}{{x}^{\mathrm{3}} {y}} \\ $$$$\Rightarrow\:{y}^{\mathrm{2}} {d}\left({xy}^{\mathrm{2}} \right)=\frac{{d}\left({x}/{y}\right)}{{x}^{\mathrm{3}} {y}}\:\:\:\:\:\:...\left({i}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Leftrightarrow\:\left({xy}^{\mathrm{2}} \right)^{{m}} {d}\left({xy}^{\mathrm{2}} \right)=\left(\frac{{x}}{{y}}\right)^{{n}} {d}\left(\frac{{x}}{{y}}\right) \\ $$$$\Rightarrow\:\:{m}−{n}=\mathrm{3}\:\:\:{and} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{m}+{n}=\mathrm{3} \\ $$$$\Rightarrow\:\:\:{m}=\mathrm{2}\:\:{and}\:\:{n}=−\mathrm{1} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${So}\:\:\:\left({i}\right)\:{becomes} \\ $$$$\:\:\:\:\int\left({xy}^{\mathrm{2}} \right)^{\mathrm{2}} {d}\left({xy}^{\mathrm{2}} \right)=\:\int\frac{{d}\left({x}/{y}\right)}{\left({x}/{y}\right)} \\ $$$$\Rightarrow\:\frac{\left({xy}^{\mathrm{2}} \right)^{\mathrm{3}} }{\mathrm{3}}=\mathrm{ln}\:\frac{{x}}{{y}}+{c}\:. \\ $$

Commented by rahul 19 last updated on 22/Jul/18

Great! ����

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jul/18

y^2 dx+2xydy=(1/(x^3 y^3 )).d((x/y)) y^2 d(x)+xd(y^2 )=(1/(x^3 y^3 )).d((x/y)) d(xy^2 )=((d((x/y)))/(y^6 ((x/y))^3 )) y^6 d(xy^2 )=((d((x/y)))/(((x/y))^3 )) contd....wait pls...recheck the questionpls... (x^2 /y^2 ) .y^6 .d(xy^2 )=((d((x/y)))/(((x/y)))) (xy^2 )^2 d(xy^2 )=((d((x/y)))/(((x/y)))) (((xy^2 )^3 )/3)=ln((x/y))+lnc

$$\:{y}^{\mathrm{2}} {dx}+\mathrm{2}{xydy}=\frac{\mathrm{1}}{{x}^{\mathrm{3}} {y}^{\mathrm{3}} }.{d}\left(\frac{{x}}{{y}}\right) \\ $$$${y}^{\mathrm{2}} {d}\left({x}\right)+{xd}\left({y}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{{x}^{\mathrm{3}} {y}^{\mathrm{3}} }.{d}\left(\frac{{x}}{{y}}\right) \\ $$$${d}\left({xy}^{\mathrm{2}} \right)=\frac{{d}\left(\frac{{x}}{{y}}\right)}{{y}^{\mathrm{6}} \left(\frac{{x}}{{y}}\right)^{\mathrm{3}} } \\ $$$${y}^{\mathrm{6}} {d}\left({xy}^{\mathrm{2}} \right)=\frac{{d}\left(\frac{{x}}{{y}}\right)}{\left(\frac{{x}}{{y}}\right)^{\mathrm{3}} } \\ $$$${contd}....{wait}\:{pls}...{recheck}\:{the}\:{questionpls}... \\ $$$$\:\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\:.{y}^{\mathrm{6}} .{d}\left({xy}^{\mathrm{2}} \right)=\frac{{d}\left(\frac{{x}}{{y}}\right)}{\left(\frac{{x}}{{y}}\right)} \\ $$$$\left({xy}^{\mathrm{2}} \right)^{\mathrm{2}} {d}\left({xy}^{\mathrm{2}} \right)=\frac{{d}\left(\frac{{x}}{{y}}\right)}{\left(\frac{{x}}{{y}}\right)} \\ $$$$\frac{\left({xy}^{\mathrm{2}} \right)^{\mathrm{3}} }{\mathrm{3}}={ln}\left(\frac{{x}}{{y}}\right)+{lnc} \\ $$

Commented by rahul 19 last updated on 22/Jul/18

thank you sir ��

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