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Question Number 118962 by benjo_mathlover last updated on 21/Oct/20

$$xdy=(y+\sqrt{x^2+y^2})dx;y\left(\sqrt{3}\right)=1$$

Answered by TANMAY PANACEA last updated on 21/Oct/20

$$\frac{xdy-ydx}{x^2}=\frac{\sqrt{x^2+y^2}}{x^2}dx$$$$d\left(\frac{y}{x}\right)=\sqrt{1+{\left(\frac{y}{x}\right)}^2}\times \frac{dx}{x}$$$$\frac{d\left(\frac{y}{x}\right)}{\sqrt{1+{\left(\frac{y}{x}\right)}^2}}=\frac{dx}{x}$$$$ln(\frac{y}{x}+\sqrt{1+{\left(\frac{y}{x}\right)}^2})=lnx+lnc$$$$\frac{y+\sqrt{x^2+y^2}}{x}=xc$$$$\frac{1+\sqrt{3+1}}{\sqrt{3}}=\sqrt{3}\times c$$$$\frac{3}{3}=1=c$$$$y+\sqrt{x^2+y^2}=x^{2}answer$$

Answered by bramlexs22 last updated on 21/Oct/20

$$takey=mx\Rightarrow dy=xdm+mdx$$$$\Rightarrow x(xdm+mdx)=(mx+x\sqrt{1+m^2})dx$$$$\Rightarrow xdm+mdx=mdx+\sqrt{1+m^2}dx$$$$\Rightarrow xdm=\sqrt{1+m^2}dx$$$$\Rightarrow \frac{dm}{\sqrt{1+m^2}}=\frac{dx}{x}$$$$\Rightarrow \int \frac{dm}{\sqrt{1+m^2}}=\ln \left(x\right)+\lambda$$$$lettingm=\tan \psi$$$$\int \frac{\sec {}^2\psi }{\sqrt{1+\tan {}^2\psi }}d\psi =\ln \left(x\right)+\lambda$$$$\Rightarrow \psi =\ln \left(x\right)+c;{\tan }^{-1}\left(\frac{y}{x}\right)=\ln \left(Cx\right)$$$$\Rightarrow \frac{y}{x}=\tan \left(\ln \left(Cx\right)\right).$$$$put\{\begin{array}{l}x=\sqrt{3}\\ y=1\end{array}\Rightarrow \frac{1}{\sqrt{3}}=\tan \left(\ln C\sqrt{3}\right)$$$$\Rightarrow \frac{\pi }{6}=\ln \left(C\sqrt{3}\right);C=\frac{e^{\frac{\pi }{6}}}{\sqrt{3}}$$$$Hencethesolutionis$$$$y-x\tan \left(\ln \left(\frac{e^{\frac{\pi }{6}}.x}{\sqrt{3}}\right)\right)=0$$

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