(1_y2)+(x_etan-1y)dy/dx=0

Ans. \((1+y^2) + (x - e^{tan^{-1}y}) {dy\over dx} = 0\)

=> \((1+y^2){dx} + (x - e^{tan^{-1}y}) {dy} = 0\)

=> \((1+y^2){dx} = (e^{tan^{-1}y}-x) {dy} \)

\(=> \space \space (1+y^2){dx\over dy} = (e^{tan^{-1}y}-x)\)

\(=> \space \space (1+y^2){dx\over dy} +x = e^{tan^{-1}y}\)

\(=> \space \space {dx\over dy} +{x\over ({1+y^2})} = {e^{tan^{-1}y}\over ({1+y^2})} \space \space\space \space\space \space\space \space [Divide\space \space by\space \space ({1+y^2}) ]\)  

\(on \space \space Comparing \space \space with \space \space {dx\over dy} + P_1.x = Q_1, \space \space We\space \space get \)

\(P_1 = {1\over ({1+y^2})} \space and \space Q_1= {e^{tan^{-1}y}\over ({1+y^2})} \)

=> \(I.F. = {e ^{\int({1\over (1+y^2})dy}} = {e^{tan^{-1}y}}\)

Solution :

\(x . (I.F) = {\int { (Q_1 \times I.F})dy }+ C\)

=> \(x. {e^{tan^{-1}y}} = {\int} {{{e^{tan^{-1}y}\over (1+y^2)}}.{e^{tan^{-1}y}} dy } \space \space + C \)

\(let \space {e^{tan^{-1}y}} = t => ({1\over (1+y^2})dy = dt\)

=> \(\int e^{2t} dt = {e^{2t}\over 2} ={e^{2tan^{-1}y}\over 2}\)

So, 

\(=> x. {e^{tan^{-1}y}} = {{e^{2tan^{-1}y}}\over 2} \space \space + C \)

This is required Equation.