... (i) Now, eq. (i) is a linear differential equation of the form
dx
dy
+P1x=Q1 where P1=
1
1+y2
and Q1=
tan−1y
1+y2
Therefore, I.F =e∫
1
1+y2
‌dy=etan−1y Thus, the solution of the given differential equation is given by xetan−1y=∫(
tan−1y
1+y2
)etan−1y‌dy+C ...(ii) Let I=∫(
tan−1y
1+y2
)etan−1y‌dy On substituting tan−1y=t, so that (
1
1+y2
)‌dy=dt, we get I=∫tet‌dt=tet−∫1.et‌dt=tet−et=et(t−1) or I=etan−1y(tan−1y−1) On substituting the value of 1 in equation (i1), we get x.etan−1y=.etan−1y(tan−1y−1)+C or x=(tan−1y−1)+Cetan−1y which is the general solution of the given differntial equation.