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

$$...differentialequation...$$$$$$$$solve:$$$$$$$$\frac{dy}{dx}=\frac{1}{xy+2x^{2}y}$$$$generalsolution=???$$$$m.n.1970$$$$$$

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

$$ydy=\frac{dx}{x(1+2x)}$$$$ydy=(\frac{1}{x}-\frac{2}{2x+1})dx$$$$\int ydy=\int (\frac{1}{x}-\frac{2}{2x+1})dx$$$$\frac{y^2}{2}=\ln x-\ln (2x+1)+C$$$$\Rightarrow y^2=2\ln \mid \frac{Cx}{2x+1}\mid$$

Answered by Olaf last updated on 11/Oct/20

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

Commented by mnjuly1970 last updated on 11/Oct/20

$$grateful..$$

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

$$\frac{dy}{dx}=\frac{1}{y(x+2x^2)}$$$$\int ydy=\int \frac{dx}{x(1+2x)}$$$$\frac{y^2}{2}=\int \frac{1}{x}-\frac{2}{1+2x}dx$$$$\frac{y^2}{2}=log\left(x\right)-log(1+2x)+C$$$$\frac{y^2}{2}=log\left(\frac{C_{1}x}{1+2x}\right)$$$$y=\pm \sqrt{log\left(\frac{C_1^2{x}^2}{{(1+2x)}^2}\right)}$$

Commented by mnjuly1970 last updated on 11/Oct/20

$$thankyousomuch..$$

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