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J-S-y-x-x-2-y-2-x-2-dx-dy-




Question Number 108053 by john santu last updated on 14/Aug/20
           ((⋏J S⋏)/^⇉ )       ((y/x)+(√((x^2 +y^2 )/x^2 )))dx=dy
$$\:\:\:\:\:\:\:\:\:\:\:\frac{\curlywedge\mathcal{J}\:\mathbb{S}\curlywedge}{\:^{\rightrightarrows} } \\ $$$$\:\:\:\:\:\left(\frac{{y}}{{x}}+\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\right){dx}={dy} \\ $$
Answered by bemath last updated on 14/Aug/20
Answered by Dwaipayan Shikari last updated on 14/Aug/20
(v+(√(1+v^2 ))  )=(dy/dx)          v=(y/x)  ,(dy/dx)=v+(dv/dx)x  v+(dv/dx)x=v+(√(1+v^2 ))  ∫(dv/( (√(1+v^2 ))))=∫(dx/x)  log(v+(√(1+v^2 )))=logx+logC  (y/x)+(√(1+(y^2 /x^2 ) )) =Cx  y+(√(x^2 +y^2 )) =Cx^2
$$\left({v}+\sqrt{\mathrm{1}+{v}^{\mathrm{2}} }\:\:\right)=\frac{{dy}}{{dx}}\:\:\:\:\:\:\:\:\:\:{v}=\frac{{y}}{{x}}\:\:,\frac{{dy}}{{dx}}={v}+\frac{{dv}}{{dx}}{x} \\ $$$${v}+\frac{{dv}}{{dx}}{x}={v}+\sqrt{\mathrm{1}+{v}^{\mathrm{2}} } \\ $$$$\int\frac{{dv}}{\:\sqrt{\mathrm{1}+{v}^{\mathrm{2}} }}=\int\frac{{dx}}{{x}} \\ $$$${log}\left({v}+\sqrt{\mathrm{1}+{v}^{\mathrm{2}} }\right)={logx}+{logC} \\ $$$$\frac{{y}}{{x}}+\sqrt{\mathrm{1}+\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:}\:={Cx} \\ $$$${y}+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:={Cx}^{\mathrm{2}} \\ $$

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