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Question Number 113333 by mohammad17 last updated on 12/Sep/20

Answered by mr W last updated on 13/Sep/20

$$letu=xy$$$$y=\frac{u}{x}$$$$y^{\prime }=-\frac{u}{x^2}+\frac{u^{\prime }}{x}$$$$-\frac{u}{x^2}+\frac{u^{\prime }}{x}=\frac{u^2-2}{x^2}$$$${xu}^{{}^{\prime }}=u^2+u-2$$$$\frac{du}{u^2+u-2}=\frac{dx}{x}$$$$\int \frac{du}{u^2+u-2}=\int \frac{dx}{x}$$$$\int \frac{du}{(u-1)(u+2)}=\int \frac{dx}{x}$$$$\frac{1}{3}\int (\frac{1}{u-1}-\frac{1}{u+2})du=\int \frac{dx}{x}$$$$\frac{1}{3}\ln \mid \frac{u-1}{u+2}\mid =\ln x+C$$$$\frac{u-1}{u+2}={cx}^3$$$$1-\frac{3}{u+2}={cx}^3$$$$u=\frac{3}{1-{cx}^3}-2$$$$xy=\frac{3}{1-{cx}^3}-2$$$$\Rightarrow y=\frac{1}{x}(\frac{3}{1-{cx}^3}-2)$$

Answered by bobhans last updated on 13/Sep/20

$$\mathrm{set}\mathrm{y}=\frac{2\mathrm{u}}{\mathrm{x}}={2\mathrm{ux}}^{-1}\to \frac{\mathrm{dy}}{\mathrm{dx}}=-{2\mathrm{ux}}^{-2}+{2\mathrm{x}}^{-1}\frac{\mathrm{du}}{\mathrm{dx}}$$$$\iff -\frac{2\mathrm{u}}{{\mathrm{x}}^2}+\frac{2}{\mathrm{x}}\frac{\mathrm{du}}{\mathrm{dx}}=\frac{{4\mathrm{u}}^2-2}{{\mathrm{x}}^2}$$$$\iff -2\mathrm{u}+2\mathrm{x}\frac{\mathrm{du}}{\mathrm{dx}}={4\mathrm{u}}^2-2$$$$\Rightarrow \mathrm{x}\frac{\mathrm{du}}{\mathrm{dx}}={2\mathrm{u}}^2+\mathrm{u}-1$$$$\Rightarrow \frac{\mathrm{du}}{{2\mathrm{u}}^2+\mathrm{u}-1}=\frac{\mathrm{dx}}{\mathrm{x}}$$$$\Rightarrow \frac{\mathrm{du}}{(2\mathrm{u}-1)(\mathrm{u}+1)}=\frac{\mathrm{dx}}{\mathrm{x}}$$$$\int (\frac{2}{2\mathrm{u}-1}-\frac{1}{\mathrm{u}+1})\mathrm{du}=\int \frac{3\mathrm{dx}}{\mathrm{x}}$$$$\int \frac{\mathrm{d}(2\mathrm{u}-1)}{2\mathrm{u}-1}-\int \frac{\mathrm{d}(\mathrm{u}+1)}{\mathrm{u}+1}=\ln \left({\mathrm{Cx}}^3\right)$$$$\ln \left(\frac{2\mathrm{u}-1}{\mathrm{u}+1}\right)=\ln \left({\mathrm{Cx}}^3\right)$$$$\frac{\mathrm{xy}-1}{\frac{\mathrm{xy}}{2}+1}={\mathrm{Cx}}^3\Rightarrow \frac{2\mathrm{xy}-2}{\mathrm{xy}+2}={\mathrm{Cx}}^3$$

Commented by bobhans last updated on 13/Sep/20

$$\mathrm{hahaha}..\mathrm{yes}.\mathrm{you}\mathrm{are}\mathrm{right}$$

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