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JEE Main Math Class 12 Differential Equations Part 2 Questions
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© examsnet.com
Question : 43
Total: 124
Let
y
=
y
(
x
)
,
y
>
0
, be a solution curve of the differential equation
(
1
+
x
2
)
d
y
=
y
(
x
−
y
)
dx
. If
y
(
0
)
=
1
and
y
(
2
√
2
)
=
β
, then
[12-Apr-2023 shift 1]
e
3
β
−
1
=
e
(
5
+
√
2
)
e
3
β
−
1
=
e
(
3
+
2
√
2
)
e
β
−
1
=
e
−
2
(
3
+
2
√
2
)
e
β
−
1
=
e
−
2
(
5
+
√
2
)
Validate
Solution:
(
1
+
x
2
)
d
y
=
y
(
x
−
y
)
dx
Y
(
0
)
=
1
⋅
y
(
2
√
2
)
=
β
d
y
d
x
=
y
x
−
y
2
1
+
x
2
d
y
d
x
+
y
(
−
x
1
+
x
2
)
=
(
−
1
1
+
x
2
)
y
2
1
y
2
d
y
d
x
+
1
y
(
−
x
1
+
x
2
)
=
−
1
1
+
x
2
Put
1
y
=
t
then
−
1
y
2
d
y
d
x
=
d
t
d
x
d
t
d
x
+
t
x
1
+
x
2
=
1
1
+
x
2
If
=
e
∫
x
1
+
x
2
dx
=
e
1
2
ln
(
1
+
x
2
)
√
1
+
x
2
t
√
1
+
x
2
=
∫
1
√
1
+
x
2
dx
√
1
+
x
2
y
=
ln
(
x
+
√
x
2
+
1
)
+
c
y
(
0
)
=
1
⇒
c
=
1
⇒
√
1
+
x
2
=
y
ln
(
e
(
x
+
√
x
2
+
1
)
)
β
=
3
ln
(
e
(
3
+
2
√
2
)
)
⇒
3
β
=
ln
(
e
(
3
+
2
√
2
)
)
e
3
β
=
e
(
3
+
2
√
2
)
© examsnet.com
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