(1)
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DIFFERENTIAL EQUATIONS
419
Now (1) is a linear differential equation of the form 
1P
dx
dy +
 x = Q1,
where,
P1 =
112
+y
 and 
1
1
2
tan
Q
1
y
y
−
=
+ Therefore,
I F = 
1
12
tan
1
dy
y
y
e
e
−
∫+
=
Thus, the solution of the given differential equation is
tan1
y
xe
−
 =
1
1
tan
2
tan
C
1
y
y
e
dy
y
−
−
⎛
⎞
+
⎜
⎟
+
⎝
⎠
∫ (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