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Question Number 181723 by Mastermind last updated on 29/Nov/22

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{xy}}^2+{2\mathrm{y}}^3}{{\mathrm{x}}^3+{\mathrm{x}}^2\mathrm{y}+{\mathrm{xy}}^2}$$$$$$$$\mathrm{Solve}$$$$$$

Answered by floor(10²Eta[1]) last updated on 29/Nov/22

$$\mathrm{let}\mathrm{y}=\mathrm{ux}\Rightarrow {\mathrm{y}}^{\prime }={\mathrm{u}}^{\prime }\mathrm{x}+\mathrm{u}$$$${\mathrm{y}}^{\prime }=\frac{{\mathrm{x}}^3{\mathrm{u}}^2+{2\mathrm{x}}^3{\mathrm{u}}^3}{{\mathrm{x}}^3+{\mathrm{x}}^3\mathrm{u}+{\mathrm{x}}^3{\mathrm{u}}^2}={\mathrm{u}}^{\prime }\mathrm{x}+\mathrm{u}$$$$\Rightarrow \frac{{\mathrm{u}}^2+{2\mathrm{u}}^3}{1+\mathrm{u}+{\mathrm{u}}^2}-\mathrm{u}={\mathrm{u}}^{\prime }\mathrm{x}$$$$\Rightarrow \frac{{\mathrm{u}}^3-\mathrm{u}}{1+\mathrm{u}+{\mathrm{u}}^2}=\frac{\mathrm{du}}{\mathrm{dx}}\mathrm{x}$$$$\Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\frac{1+\mathrm{u}+{\mathrm{u}}^2}{{\mathrm{u}}^3-\mathrm{u}}\mathrm{du}$$$$\ln \left(\mathrm{x}\right)+\mathrm{C}=\int \frac{1+\mathrm{u}+{\mathrm{u}}^2}{{\mathrm{u}}^3-\mathrm{u}}\mathrm{du}=\int \frac{1+\mathrm{u}+{\mathrm{u}}^2}{\mathrm{u}(\mathrm{u}-1)(\mathrm{u}+1)}\mathrm{du}$$$$\mathrm{use}\mathrm{partial}\mathrm{fractions}:$$$$\ln \left(\mathrm{x}\right)+\mathrm{C}=-\int \frac{\mathrm{du}}{\mathrm{u}}+\frac{3}{2}\int \frac{\mathrm{du}}{\mathrm{u}-1}+\frac{1}{2}\int \frac{\mathrm{du}}{\mathrm{u}+1}$$$$=-\ln \left(\mathrm{u}\right)+\frac{3}{2}\ln (\mathrm{u}-1)+\frac{1}{2}\ln (\mathrm{u}+1)$$$$=-\ln \left(\mathrm{u}\right)+\ln \sqrt{{(\mathrm{u}-1)}^3}+\ln \sqrt{\mathrm{u}+1}$$$$=\ln \left(\frac{\sqrt{{(\mathrm{u}-1)}^3(\mathrm{u}+1)}}{\mathrm{u}}\right)=\mathrm{lnx}+\mathrm{C}$$$$\Rightarrow \frac{\sqrt{{(\mathrm{u}-1)}^3(\mathrm{u}+1)}}{\mathrm{u}}=\mathrm{Cx}$$$$\frac{{(\mathrm{u}-1)}^3(\mathrm{u}+1)}{{\mathrm{u}}^2}={\mathrm{C}}^2{\mathrm{x}}^2$$$${\mathrm{u}}^4-{2\mathrm{u}}^3-{\mathrm{Cx}}^2{\mathrm{u}}^2+2\mathrm{u}-1=0$$$$\mathrm{solve}\mathrm{for}\mathrm{u}\therefore \mathrm{y}=\mathrm{ux}$$$$\mathrm{note}\mathrm{that}\mathrm{if}\mathrm{C}=0\Rightarrow \mathrm{u}=\pm 1$$$$\mathrm{y}=\pm \mathrm{x}\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{solution}$$

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