Question Number 180801 by Mastermind last updated on 17/Nov/22
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{Differential}\:\mathrm{equation}\:: \\ $$$$\left(\mathrm{3xy}+\mathrm{6y}^{\mathrm{2}} \right)\mathrm{dx}+\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{9xy}\right)\mathrm{dy}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Mastermind} \\ $$
Answered by qaz last updated on 17/Nov/22
$$\frac{{dy}}{{dx}}=−\frac{\mathrm{3}{xy}+\mathrm{6}{y}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}{xy}}=−\frac{\mathrm{3}\frac{{y}}{{x}}+\mathrm{6}\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} }{\mathrm{2}+\mathrm{9}\frac{{y}}{{x}}} \\ $$$${z}=\frac{{y}}{{x}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{d}\left({xz}\right)}{{dx}}=\frac{{zdx}+{xdz}}{{dx}}=−\frac{\mathrm{3}{z}+\mathrm{6}{z}^{\mathrm{2}} }{\mathrm{2}+\mathrm{9}{z}} \\ $$$$\Rightarrow\frac{\mathrm{2}+\mathrm{9}{z}}{\mathrm{5}{z}+\mathrm{15}{z}^{\mathrm{2}} }{dz}=−\frac{{dx}}{{x}} \\ $$$$\Rightarrow\frac{\mathrm{2}}{\mathrm{5}}{lnz}+\frac{\mathrm{1}}{\mathrm{5}}{ln}\left(\mathrm{3}{z}+\mathrm{1}\right)+{C}=−{lnx} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{3}} +{x}^{\mathrm{3}} {y}^{\mathrm{2}} ={C} \\ $$
Commented by Mastermind last updated on 17/Nov/22
$$\mathrm{Thanks} \\ $$
Commented by SLVR last updated on 18/Nov/22
$${nicesir}.. \\ $$