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JEE Main Math Class 12 Differential Equations Part 2 Questions
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© examsnet.com
Question : 8
Total: 124
Let
y
=
y
(
x
)
be the solution of the differential equation
(
1
+
x
2
)
d
y
d
x
+
y
=
e
tan
−
1
x
,
y
(
1
)
=
0
. Then
y
(
0
)
is
[6 Apr 2024 Shift 1]
1
4
(
e
π
∕
2
−
1
)
1
2
(
1
−
e
π
∕
2
)
1
4
(
1
−
e
π
∕
2
)
1
2
(
e
π
∕
2
−
1
)
Validate
Solution:
👈: Video Solution
d
y
d
x
+
y
1
+
x
2
=
e
tan
−
1
x
1
+
x
2
I.F.
=
e
∫
1
1
+
x
2
dx
=
e
tan
−
1
x
y
⋅
e
tan
−
1
x
=
∫
(
e
tan
−
1
x
1
+
x
2
)
e
tan
−
1
x
⋅
dx
Let
tan
−
1
x
=
z
∴
dx
1
+
x
2
=
dz
∴
y
⋅
e
z
=
∫
e
2
z
d
z
=
e
2
z
2
+
C
y
⋅
e
tan
−
1
x
=
e
2
tan
−
1
x
2
+
C
⇒
y
=
e
tan
−
1
x
2
+
C
e
tan
−
1
x
∵
y
(
1
)
=
0
⇒
0
=
e
π
∕
4
2
+
C
e
π
∕
4
⇒
C
=
−
e
π
∕
2
2
∴
y
=
e
tan
−
1
x
2
−
e
π
∕
2
2
e
tan
−
1
x
∴
y
(
0
)
=
1
−
e
π
∕
2
2
© examsnet.com
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