(2)
Let
I =
1
1
tan
2
tan
1
y
y
e
dy
y
−
−
⎛
⎞
⎜
⎟
+
⎝
⎠
∫
Substituting tan–1 y = t so that 
112
dy
dt
y
⎛
⎞
=
⎜
⎟
⎝+
⎠
, we get
I =
t
∫t e dt
= t et – ∫1 et dt = t et – et = et (t – 1)
or
I =
tan1
y
e
−
(tan–1y –1)
Substituting the value of I in equation (2), we get
1
1
tan
tan
1 (tan
1)
C
y
y
x e
e
y
−
−
−
=
−
+
or
x =
1
1
tan
(tan
1)
C
y
y
e
−
−
−
−
+
which is the general solution of the given differential equation Miscellaneous Exercise on Chapter 9
1