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Question Number 129656 by Ahmed1hamouda last updated on 17/Jan/21

Commented by Ahmed1hamouda last updated on 17/Jan/21

$$solvethedifferen\mathrm{tial}\mathrm{equation}$$

Answered by liberty last updated on 17/Jan/21

$$\mathrm{let}\mathrm{u}={\tan }^{-1}\left(\mathrm{y}\right)\Rightarrow \mathrm{y}=\tan \left(\mathrm{u}\right)$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\sec {}^2\left(\mathrm{u}\right)\frac{\mathrm{du}}{\mathrm{dx}}\Rightarrow \mathrm{dy}=\sec {}^2\left(\mathrm{u}\right)\mathrm{du}$$$$(\ast )(1+\tan {}^2\mathrm{u})\mathrm{dx}=(\mathrm{u}-\mathrm{x})\left(\sec {}^2\mathrm{u}\right)\mathrm{du}$$$$\left(\sec {}^2\mathrm{u}\right)\mathrm{dx}=(\mathrm{u}-\mathrm{x})\sec {}^2\mathrm{u}\mathrm{du}$$$$\mathrm{dx}+\mathrm{x}\mathrm{du}=\mathrm{u}\mathrm{du}$$$$\mathrm{d}\left(\mathrm{xu}\right)=\mathrm{u}\mathrm{du}$$$$\int \mathrm{d}\left(\mathrm{xu}\right)=\int \mathrm{u}\mathrm{du}$$$$\mathrm{xu}=\frac{1}{2}{\mathrm{u}}^2+\mathrm{C}$$$$\mathrm{x}=\frac{1}{2}\mathrm{u}+\frac{\mathrm{C}}{\mathrm{u}}$$$$\mathrm{x}=\frac{1}{2}{\tan }^{-1}\left(\mathrm{y}\right)+\frac{\mathrm{C}}{{\tan }^{-1}\left(\mathrm{y}\right)}$$$$$$

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