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Question Number 96012 by mhmd last updated on 29/May/20

Commented by john santu last updated on 29/May/20

$$(y-\sqrt{x^2-y^2})dx=xdy$$$$\frac{dy}{dx}=\frac{y-\sqrt{x^2-y^2}}{x}=\frac{y}{x}-\sqrt{1-{\left(\frac{y}{x}\right)}^2}$$$$sety=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$$$$\Rightarrow v+x\frac{dv}{dx}=v-\sqrt{1-v^2}$$$$x\frac{dv}{dx}=-\sqrt{1-v^2}$$$$\frac{dv}{\sqrt{1-v^2}}=-\frac{dx}{x}$$$$\int \frac{dv}{\sqrt{1-v^2}}=-\ln \left(x\right)+c$$$${\sin }^{-1}\left(v\right)=-\ln \left(x\right)+c$$$$\frac{y}{x}=\sin (c-\ln \left(x\right))$$$$y=x\sin (c-\ln \left(x\right))$$

Answered by Sourav mridha last updated on 29/May/20

$$(\mathit{y}-\sqrt{{\mathrm{x}}^2-{\mathrm{y}}^2})\mathrm{dx}-\mathrm{xdy}=0$$$$\Rightarrow \mathrm{ydx}-\mathrm{xdy}=\sqrt{{\mathrm{x}}^2-{\mathrm{y}}^2}\mathrm{dx}$$$$\Rightarrow -\left[\frac{\mathrm{xdy}-\mathrm{ydx}}{{\mathrm{x}}^2}\right]=\mathrm{x}\sqrt{1-{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^2}\frac{\mathrm{dx}}{{\mathrm{x}}^2}$$$$\mathrm{integrating}\mathrm{both}\mathrm{sides}--$$$$\Rightarrow -\int \left[\frac{\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}{\sqrt{1-{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^2}}\right]=\int \frac{\mathrm{dx}}{\mathrm{x}}$$$$\Rightarrow -{\sin }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\ln \left(\mathrm{cx}\right)$$$$\mathrm{so},\mathit{y}=-\mathit{x}\mathit{s}\mathit{i}\mathit{n}\left(\mathit{l}\mathit{n}\left(\mathit{c}\mathit{x}\right)\right).$$

Answered by Sourav mridha last updated on 29/May/20

$$\mathrm{if}\mathrm{this}\mathrm{is}\mathrm{like}\mathrm{that}$$$$(\mathrm{y}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2})\mathrm{dy}+\mathrm{xdx}=0$$$$\mathrm{then}----$$$$\mathrm{ydy}+\mathrm{xdx}=-\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\mathrm{dy}$$$$\Rightarrow \frac{1}{2}\mathrm{d}({\mathrm{x}}^2+{\mathrm{y}}^2)=-\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\mathrm{dy}$$$$\mathrm{integrating}\mathrm{both}\mathrm{sides}--$$$$\int \frac{\mathrm{d}({\mathrm{x}}^2+{\mathrm{y}}^2)}{2\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}=-\int \mathrm{dy}$$$$\Rightarrow \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}=-\mathrm{y}+\mathrm{c}$$$$\mathrm{or}\mathrm{you}\mathrm{get}\mathrm{the}\mathrm{ans}\mathrm{of}\mathrm{opption}\mathrm{A}$$$$\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}+\mathrm{y}=\mathrm{c}$$

Commented by mhmd last updated on 29/May/20

$$veryverythankyousir$$

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