(1+y2)dx=(tan−1y−x)dx
∴dxdy=tan−1y−x1+y2
∴dxdy=tan−1y1+y2−x1+y2
∴dxdy+x1+y2=tan−1y1+y2 (1)
dxdy+P(y)x=Q(y) which is linear differential equation of the type
∴P=1+y2 Q(A)=tan−1y1+y2
∴ Integrating factor =e∫P(y)dy
=e∫11+y2dy
∴ Multiplying equation (1) by etan−1y
∴etan−1ydxdy+etan−1y1+y−−=etan−1y(tan−1y1+y2)
∴ddx(etan−1y⋅x)=etan−1y⋅(tan−1y1+y2)
∴ Integrating both sides pairwise
∴xetan−1y=∫etan−1y⋅(tan−1y1+y2)dy .(2)
∴ Now, finding ∫etan−1y⋅(tan−1y1+y2)dy
Taking tan−1y=t
∴11+y2=dy=dt
∴xetan−1y=∫etdt
∴∫et[(t−1)+1]dt
∴ Here taking f(t)=1
∴xetan−1y=∫et(f(t)+f′(t))dt
etf(t)+c
∴xetan−1y=et(t−1)+c
∴ Result (2) becames as follows
∴xetan−1y=etan−1y(tan−1y−1)+c
∴x=(tan−1y−1)+c⋅e−tan−1y
∴ Which is required solution of given differential equation.