solve-1-x-2-dy-dx-xy-xy-2-

Question Number 97001 by bemath last updated on 06/Jun/20

solve (1 + x^{2}) \frac{dy}{dx} = xy - {xy}^{2}

Commented by bobhans last updated on 06/Jun/20

\frac{dy}{dx} = \frac{x (y - y^{2})}{x^{2} + 1} \Leftrightarrow \frac{dy}{y - y^{2}} = \frac{x dx}{x^{2} + 1}

\frac{dy}{y (1 - y)} = \frac{d (x^{2} + 1)}{2 (x^{2} + 1)} \Leftrightarrow \frac{y + 1 - y}{y (1 - y)} dy = \frac{1}{2} \ln (x^{2} + 1) + c

\int \frac{dy}{y} + \int \frac{dy}{1 - y} = \frac{1}{2} lnC (x^{2} + 1)

\ln ∣ \frac{y}{1 - y} ∣ = \ln \sqrt{C (x^{2} + 1)} \Leftrightarrow ∣ \frac{y}{1 - y} ∣ = \sqrt{C (x^{2} + 1)}

Commented by bemath last updated on 06/Jun/20

thanks

Answered by mathmax by abdo last updated on 06/Jun/20

e \Rightarrow \frac{dy}{xy - {xy}^{2}} = \frac{dx}{1 + x^{2}} \Rightarrow \frac{dy}{y - y^{2}} = \frac{xdx}{1 + x^{2}} \Rightarrow

\int \frac{dy}{y - y^{2}} = \int \frac{xdx}{1 + x^{2}} = \frac{1}{2} \ln (1 + x^{2}) but

\int \frac{dy}{y - y^{2}} = - \int \frac{dy}{y (y - 1)} = - \int (\frac{1}{y - 1} - \frac{1}{y}) dy = \int (\frac{1}{y} - \frac{1}{y - 1}) dy

= \ln ∣ \frac{y}{y - 1} ∣ \Rightarrow \ln ∣ \frac{y}{y - 1} ∣ = \ln \sqrt{1 + x^{2}} + c \Rightarrow∣ \frac{y}{y - 1} ∣ = k \sqrt{1 + x^{2}} \Rightarrow

\frac{y}{y - 1} = k \sqrt{1 + x^{2}} \Rightarrow \frac{y - 1 + 1}{y - 1} = k \sqrt{1 + x^{2}} \Rightarrow 1 + \frac{1}{y - 1} = k \sqrt{1 + x^{2}} \Rightarrow

\frac{1}{y - 1} = k \sqrt{1 + x^{2}} - 1 \Rightarrow y - 1 = \frac{1}{k \sqrt{1 + x^{2}} - 1} \Rightarrow y = 1 + \frac{1}{k \sqrt{1 + x^{2}} - 1}