JEE Main 2003Annual, Paper 1MathematicsMediumDifferential Equations163× since 2002The solution of the differential equation (1+y2)+(x−etan−1y)dydx=0,\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,(1+y2)+(x−etan−1y)dxdy=0, is :AOption A. xe2tan−1y=etan−1y+kx{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + kxe2tan−1y=etan−1y+kBOption B. (x−2)=ke2tan−1y\left( {x - 2} \right) = k{e^{2{{\tan }^{ - 1}}y}}(x−2)=ke2tan−1yCOption C. 2xetan−1y=e2tan−1y+k2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k2xetan−1y=e2tan−1y+kDOption D. xetan−1y=tan−1y+kx{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + kxetan−1y=tan−1y+kCheckSkip · show solution
