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Question Number 180801 by Mastermind last updated on 17/Nov/22

Solve the Differential equation : (3xy+6y^2 )dx+(2x^2 +9xy)dy=0 Mastermind

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{Differential}\:\mathrm{equation}\:: \\ $$$$\left(\mathrm{3xy}+\mathrm{6y}^{\mathrm{2}} \right)\mathrm{dx}+\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{9xy}\right)\mathrm{dy}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Mastermind} \\ $$

Answered by qaz last updated on 17/Nov/22

(dy/dx)=−((3xy+6y^2 )/(2x^2 +9xy))=−((3(y/x)+6((y/x))^2 )/(2+9(y/x))) z=(y/x) (dy/dx)=((d(xz))/dx)=((zdx+xdz)/dx)=−((3z+6z^2 )/(2+9z)) ⇒((2+9z)/(5z+15z^2 ))dz=−(dx/x) ⇒(2/5)lnz+(1/5)ln(3z+1)+C=−lnx ⇒3x^2 y^3 +x^3 y^2 =C

$$\frac{{dy}}{{dx}}=−\frac{\mathrm{3}{xy}+\mathrm{6}{y}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}{xy}}=−\frac{\mathrm{3}\frac{{y}}{{x}}+\mathrm{6}\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} }{\mathrm{2}+\mathrm{9}\frac{{y}}{{x}}} \\ $$$${z}=\frac{{y}}{{x}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{d}\left({xz}\right)}{{dx}}=\frac{{zdx}+{xdz}}{{dx}}=−\frac{\mathrm{3}{z}+\mathrm{6}{z}^{\mathrm{2}} }{\mathrm{2}+\mathrm{9}{z}} \\ $$$$\Rightarrow\frac{\mathrm{2}+\mathrm{9}{z}}{\mathrm{5}{z}+\mathrm{15}{z}^{\mathrm{2}} }{dz}=−\frac{{dx}}{{x}} \\ $$$$\Rightarrow\frac{\mathrm{2}}{\mathrm{5}}{lnz}+\frac{\mathrm{1}}{\mathrm{5}}{ln}\left(\mathrm{3}{z}+\mathrm{1}\right)+{C}=−{lnx} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{3}} +{x}^{\mathrm{3}} {y}^{\mathrm{2}} ={C} \\ $$

Commented by Mastermind last updated on 17/Nov/22

Thanks

$$\mathrm{Thanks} \\ $$

Commented by SLVR last updated on 18/Nov/22

nicesir..

$${nicesir}.. \\ $$

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