1.The solution of the differential equation (1+y2)+(x−etan−1y)dydx=0(1 + y^2) + (x - e^{\tan^{-1} y})\frac{dy}{dx} = 0(1+y2)+(x−etan−1y)dxdy=0 isa.xetan−1y=tan−1y+Cx e^{\tan^{-1} y} = \tan^{-1} y + Cxetan−1y=tan−1y+Cb.xe2tan−1y=e−tan−1y+Cx e^{2\tan^{-1} y} = e^{-\tan^{-1} y} + Cxe2tan−1y=e−tan−1y+Cc.2xetan−1y=e2tan−1y+C2x e^{\tan^{-1} y} = e^{2\tan^{-1} y} + C2xetan−1y=e2tan−1y+Cd.x2etan−1y=4e2tan−1y+Cx^2 e^{\tan^{-1} y} = 4e^{2\tan^{-1} y} + Cx2etan−1y=4e2tan−1y+CLogin to continueOnly logged in users canattempt or see the solution.
