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Question Number 89600 by jagoll last updated on 18/Apr/20

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}+\sqrt{{\mathrm{x}}^2-{\mathrm{y}}^2}}{\mathrm{x}}$$

Commented by mr W last updated on 18/Apr/20

$$\frac{dy}{dx}=\frac{y}{x}+\sqrt{1-{\left(\frac{y}{x}\right)}^2}$$$$y=xu$$$$\frac{dy}{dx}=u+x\frac{du}{dx}$$$$u+x\frac{du}{dx}=u+\sqrt{1-u^2}$$$$x\frac{du}{dx}=\sqrt{1-u^2}$$$$\frac{du}{\sqrt{1-u^2}}=\frac{dx}{x}$$$$\int \frac{du}{\sqrt{1-u^2}}=\int \frac{dx}{x}$$$${\sin }^{-1}u=\ln x+C$$$${\sin }^{-1}\frac{y}{x}=\ln x+C$$$$\Rightarrow y=x\sin (\ln x+C)$$

Commented by jagoll last updated on 18/Apr/20

$$\mathrm{thank}\mathrm{you}$$

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