TS EAMCET 5 May 2018 Shift 1 Solved Paper

The given equation is expressed as ,
(1+y2)dx=(tan−1y−x)dy

dx

dy

+

x

1+y2

=

tan−1y

1+y2

The equation is in the form of

dx

dy

+Px=Q
The integrating factor is calculated as,
IF=e∫

1

1+y2

dy
=etan−1y
The solution of linear differential equation is,
x.IF=∫Q×IF+C
x.etan−1y=∫

tan−1y

1+y2

.etan−1ydy+c
Assume
tan−1y=t

1

1+y2

dy=dt
=∫t.etdt+c =t‌∫et−∫(et.1)dt+c
=tet−et+c
=et(t−1)+c
Then,
x.etan−1y=etan−1y(tan−1y−1)+c