BITSAT 2021 Paper

The given differential equation can be written as

dx

dy

+

x

1+y2

=

tan−1y

1+y2

... (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.