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Question Number 121609 by benjo_mathlover last updated on 10/Nov/20

$$\mathrm{y}\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\mathrm{dx}-\mathrm{x}(\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2})\mathrm{dy}=0$$$$$$

Answered by liberty last updated on 10/Nov/20

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}{\mathrm{x}(\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2})}$$$$\mathrm{let}\mathrm{y}=\mathrm{vx}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}$$$$\Rightarrow \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{vx}\sqrt{{\mathrm{x}}^2+{\mathrm{v}}^2{\mathrm{x}}^2}}{{\mathrm{x}}^2+\mathrm{x}\sqrt{{\mathrm{x}}^2+{\mathrm{v}}^2{\mathrm{x}}^2}}$$$$\Rightarrow \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}\sqrt{1+{\mathrm{v}}^2}}{1+\sqrt{1+{\mathrm{v}}^2}}$$$$\Rightarrow \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}\sqrt{1+{\mathrm{v}}^2}-\mathrm{v}-\mathrm{v}\sqrt{1+{\mathrm{v}}^2}}{1+\sqrt{1+{\mathrm{v}}^2}}$$$$\Rightarrow \frac{(1+\sqrt{1+{\mathrm{v}}^2})}{\mathrm{v}}\mathrm{dv}=-\frac{\mathrm{dx}}{\mathrm{x}}$$$$\Rightarrow \ln \mid \mathrm{v}\mid +\int \frac{\sqrt{1+{\mathrm{v}}^2}\mathrm{dv}}{\mathrm{v}}=-\ln \mid \mathrm{x}\mid +\mathrm{c}$$$$\mathrm{let}\mathrm{I}=\int \frac{\sqrt{1+{\mathrm{v}}^2}}{\mathrm{v}}\mathrm{dv}=\int \frac{1+{\mathrm{v}}^2}{\mathrm{v}\sqrt{1+{\mathrm{v}}^2}}\mathrm{dv}$$$$\mathrm{I}=\int \frac{\mathrm{dv}}{\mathrm{v}\sqrt{1+{\mathrm{v}}^2}}+\int \frac{\mathrm{v}}{\sqrt{1+{\mathrm{v}}^2}}\mathrm{dv}$$$$\mathrm{I}=\frac{1}{2}\int \frac{\mathrm{d}(1+{\mathrm{v}}^2)}{\sqrt{1+{\mathrm{v}}^2}}+\int \frac{\sec {}^2\mathrm{u}\mathrm{du}}{\tan \mathrm{u}.\sec \mathrm{u}}[\mathrm{u}=\mathrm{arc}\tan \mathrm{v}]$$$$\mathrm{I}=\sqrt{1+{\mathrm{v}}^2}+\int \csc \mathrm{u}\mathrm{du}$$$$\mathrm{I}=\sqrt{1+{\mathrm{v}}^2}+\ln \mid \csc \mathrm{u}-\cot \mathrm{u}\mid$$$$$$$$\therefore \mathrm{the}\mathrm{solution}$$$$\ln \mid \mathrm{vx}\mid +\sqrt{1+{\mathrm{v}}^2}+\ln \mid \frac{\sqrt{1+{\mathrm{v}}^2}-1}{\mathrm{v}}\mid =\mathrm{C}$$$$\ln \mid \mathrm{x}\left(\frac{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}-\mathrm{x}}{\mathrm{x}}\right)\mid +\frac{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}{\mathrm{x}}=\mathrm{C}$$$$\ln \mid \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}-\mathrm{x}\mid +\frac{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}{\mathrm{x}}=\mathrm{C}$$$$$$

Answered by TANMAY PANACEA last updated on 10/Nov/20

$$y\sqrt{x^2+y^2}dx-x\sqrt{x^2+y^2}dy-x^{2}dy=0$$$$-\sqrt{x^2+y^2}(xdy-ydx)=x^{2}dy$$$$-x\times \sqrt{1+{\left(\frac{y}{x}\right)}^2}\times \left(\frac{xdy-ydx}{x^2}\right)=dy$$$$-\left(\frac{x}{y}\right)\sqrt{1+{\left(\frac{y}{x}\right)}^2}\times d\left(\frac{y}{x}\right)=\frac{dy}{y}$$$$\frac{dy}{y}+\left(\frac{x}{y}\right)\sqrt{1+{\left(\frac{y}{x}\right)}^2}\times d\left(\frac{y}{x}\right)=0=dC$$$$\frac{dy}{y}+\frac{\sqrt{1+p^2}dp}{p}=dC$$$$★intregation\int \frac{dy}{y}=lny+C_1$$$$◼\int \frac{\sqrt{1+p^2}}{p}dp\to k^2=1+p^2\to kdk=pdp$$$$\int \frac{\sqrt{1+p^2}}{p^2}\times pdp\to \int \frac{k\times kdk}{(k^2-1)}=\int \frac{k^2-1+1}{k^2-1}dk=\int (1+\frac{1}{k^2-1)})dk$$$$=k+\frac{1}{2}ln\left(\frac{k-1}{k+1}\right)+C_2$$$$=\sqrt{1+p^2}+\frac{1}{2}ln\left(\frac{\sqrt{1+p^2}-1}{\sqrt{1+p^2}+1}\right)+{\mathrm{C}}_2$$$$lny+\sqrt{1+p^2}+\frac{1}{2}ln\left(\frac{\sqrt{1+p^2}-1}{\sqrt{1+p^2}+1}\right)=C$$$$lny+\sqrt{1+{\left(\frac{y}{x}\right)}^2}+\frac{1}{2}ln\left(\frac{\sqrt{1+{\left(\frac{y}{x}\right)}^2-1}}{\sqrt{1+{\left(\frac{y}{x}\right)}^2}+1}\right)=C$$

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