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

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

Commented by bobhans last updated on 06/Jun/20

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}(\mathrm{y}-{\mathrm{y}}^2)}{{\mathrm{x}}^2+1}\iff \frac{\mathrm{dy}}{\mathrm{y}-{\mathrm{y}}^2}=\frac{\mathrm{x}\mathrm{dx}}{{\mathrm{x}}^2+1}$$$$\frac{\mathrm{dy}}{\mathrm{y}(1-\mathrm{y})}=\frac{\mathrm{d}({\mathrm{x}}^2+1)}{2({\mathrm{x}}^2+1)}\iff \frac{\mathrm{y}+1-\mathrm{y}}{\mathrm{y}(1-\mathrm{y})}\mathrm{dy}=\frac{1}{2}\ln ({\mathrm{x}}^2+1)+\mathrm{c}$$$$\int \frac{\mathrm{dy}}{\mathrm{y}}+\int \frac{\mathrm{dy}}{1-\mathrm{y}}=\frac{1}{2}\mathrm{lnC}({\mathrm{x}}^2+1)$$$$\ln \mid \frac{\mathrm{y}}{1-\mathrm{y}}\mid =\ln \sqrt{\mathrm{C}({\mathrm{x}}^2+1)}\iff \mid \frac{\mathrm{y}}{1-\mathrm{y}}\mid =\sqrt{\mathrm{C}({\mathrm{x}}^2+1)}$$$$$$

Commented by bemath last updated on 06/Jun/20

$$\mathrm{thanks}$$

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

$$\mathrm{e}\Rightarrow \frac{\mathrm{dy}}{\mathrm{xy}-{\mathrm{xy}}^2}=\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}-{\mathrm{y}}^2}=\frac{\mathrm{xdx}}{1+{\mathrm{x}}^2}\Rightarrow$$$$\int \frac{\mathrm{dy}}{\mathrm{y}-{\mathrm{y}}^2}=\int \frac{\mathrm{xdx}}{1+{\mathrm{x}}^2}=\frac{1}{2}\ln (1+{\mathrm{x}}^2)\mathrm{but}$$$$\int \frac{\mathrm{dy}}{\mathrm{y}-{\mathrm{y}}^2}=-\int \frac{\mathrm{dy}}{\mathrm{y}(\mathrm{y}-1)}=-\int (\frac{1}{\mathrm{y}-1}-\frac{1}{\mathrm{y}})\mathrm{dy}=\int (\frac{1}{\mathrm{y}}-\frac{1}{\mathrm{y}-1})\mathrm{dy}$$$$=\ln \mid \frac{\mathrm{y}}{\mathrm{y}-1}\mid \Rightarrow \ln \mid \frac{\mathrm{y}}{\mathrm{y}-1}\mid =\ln \sqrt{1+{\mathrm{x}}^2}+\mathrm{c}\Rightarrow \mid \frac{\mathrm{y}}{\mathrm{y}-1}\mid =\mathrm{k}\sqrt{1+{\mathrm{x}}^2}\Rightarrow$$$$\frac{\mathrm{y}}{\mathrm{y}-1}=\mathrm{k}\sqrt{1+{\mathrm{x}}^2}\Rightarrow \frac{\mathrm{y}-1+1}{\mathrm{y}-1}=\mathrm{k}\sqrt{1+{\mathrm{x}}^2}\Rightarrow 1+\frac{1}{\mathrm{y}-1}=\mathrm{k}\sqrt{1+{\mathrm{x}}^2}\Rightarrow$$$$\frac{1}{\mathrm{y}-1}=\mathrm{k}\sqrt{1+{\mathrm{x}}^2}-1\Rightarrow \mathrm{y}-1=\frac{1}{\mathrm{k}\sqrt{1+{\mathrm{x}}^2}-1}\Rightarrow \mathrm{y}=1+\frac{1}{\mathrm{k}\sqrt{1+{\mathrm{x}}^2}-1}$$

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