The solution of the differential equation (1+y^2 )+ (x-e^tan ^-1 y ) d y d x=0, is

$\left(1+y^2\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$(1+y2)+(x−etan−1y)dxdy​=0 $\left(1+y^2\right) \frac{d x}{d y}+x=e^{\tan ^{-1} y} \Rightarrow \frac{d x}{d y}+\frac{x}{\left(1+y^2\right)}=\frac{e^{\tan ^{-1} y}}{\left(1+y^2\right)}$(1+y2)dydx​+x=etan−1y⇒dydx​+(1+y2)x​=(1+y2)etan−1y​ I.F. $\displaystyle =e^{\int \frac{1}{\left(1+y^2\right)} d y}=e \tan ^{-1} y$=e∫(1+y2)1​dy=etan−1y $\displaystyle x\left(e^{\tan ^{-1} y}\right)=\int \frac{e^{\tan ^{-1} y}}{1+y} e^{\tan ^{-1} y d y}$x(etan−1y)=∫1+yetan−1y​etan−1ydy $x\left(e^{\tan ^{-1} y}\right)=\frac{e^{2 \tan ^{-1} y}}{2}+C \quad \therefore 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+k$x(etan−1y)=2e2tan−1y​+C∴2xetan−1y=e2tan−1y+k