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Question Number 117396 by mnjuly1970 last updated on 11/Oct/20

...differential equation... solve : (dy/dx)=(1/(xy+2x^2 y)) general solution =??? m.n.1970

$$\:\:\:\:\:\:\:\:...{differential}\:\:{equation}...\: \\ $$$$ \\ $$$$\:\:\:\:{solve}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{xy}+\mathrm{2}{x}^{\mathrm{2}} {y}} \\ $$$$\:\:\:\:\:\:\:\:\:{general}\:\:{solution}\:=??? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$$$\: \\ $$

Answered by mr W last updated on 11/Oct/20

ydy=(dx/(x(1+2x))) ydy=((1/x)−(2/(2x+1)))dx ∫ydy=∫((1/x)−(2/(2x+1)))dx (y^2 /2)=ln x−ln (2x+1)+C ⇒y^2 =2 ln ∣((Cx)/(2x+1))∣

$${ydy}=\frac{{dx}}{{x}\left(\mathrm{1}+\mathrm{2}{x}\right)} \\ $$$${ydy}=\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}\right){dx} \\ $$$$\int{ydy}=\int\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}\right){dx} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{ln}\:{x}−\mathrm{ln}\:\left(\mathrm{2}{x}+\mathrm{1}\right)+{C} \\ $$$$\Rightarrow{y}^{\mathrm{2}} =\mathrm{2}\:\mathrm{ln}\:\mid\frac{{Cx}}{\mathrm{2}{x}+\mathrm{1}}\mid \\ $$

Answered by Olaf last updated on 11/Oct/20

(dy/dx) = (1/(xy(2x+1))) (dy/dx)y = (1/(x(2x+1))) = (1/x)−(2/(2x+1)) (1/2)y^2 = ln∣x∣−ln∣2x+1∣+C y = ±(√(2ln∣(x/(2x+1))∣+K))

$$\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{{xy}\left(\mathrm{2}{x}+\mathrm{1}\right)} \\ $$$$\:\frac{{dy}}{{dx}}{y}\:=\:\frac{\mathrm{1}}{{x}\left(\mathrm{2}{x}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} \:=\:\mathrm{ln}\mid{x}\mid−\mathrm{ln}\mid\mathrm{2}{x}+\mathrm{1}\mid+\mathrm{C} \\ $$$${y}\:=\:\pm\sqrt{\mathrm{2ln}\mid\frac{{x}}{\mathrm{2}{x}+\mathrm{1}}\mid+\mathrm{K}} \\ $$

Commented by mnjuly1970 last updated on 11/Oct/20

grateful..

$${grateful}.. \\ $$

Answered by Dwaipayan Shikari last updated on 11/Oct/20

(dy/dx)=(1/(y(x+2x^2 ))) ∫ydy=∫(dx/(x(1+2x))) (y^2 /2)=∫(1/x)−(2/(1+2x))dx (y^2 /2)=log(x)−log(1+2x)+C (y^2 /2)=log(((C_1 x)/(1+2x))) y=±(√(log(((C_1 ^2 x^2 )/((1+2x)^2 )))))

$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{y}\left({x}+\mathrm{2}{x}^{\mathrm{2}} \right)} \\ $$$$\int{ydy}=\int\frac{{dx}}{{x}\left(\mathrm{1}+\mathrm{2}{x}\right)} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{2}}=\int\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}{x}}{dx} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{2}}={log}\left({x}\right)−{log}\left(\mathrm{1}+\mathrm{2}{x}\right)+{C} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{2}}={log}\left(\frac{{C}_{\mathrm{1}} {x}}{\mathrm{1}+\mathrm{2}{x}}\right) \\ $$$${y}=\pm\sqrt{{log}\left(\frac{{C}_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }\right)} \\ $$

Commented by mnjuly1970 last updated on 11/Oct/20

thank you so much..

$${thank}\:{you}\:{so}\:{much}.. \\ $$

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