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Mathematics
Problem
Consider the differential equation,
y
2
d
x
+
(
x
−
1
y
)
d
y
=
0
y^2 dx + \left(x - \frac{1}{y}\right) dy = 0
y
2
d
x
+
(
x
−
y
1
)
d
y
=
0
. If value of
y
y
y
is 1 when
x
=
1
x = 1
x
=
1
, the the value of
x
x
x
for which
y
=
2
y = 2
y
=
2
, is : [JEE-MAIN ONLINE-2019] (1)
1
2
+
1
e
\frac{1}{2} + \frac{1}{\sqrt{e}}
2
1
+
e
1
(2)
3
2
−
e
\frac{3}{2} - \sqrt{e}
2
3
−
e
(3)
5
2
+
1
e
\frac{5}{2} + \frac{1}{\sqrt{e}}
2
5
+
e
1
(4)
3
2
−
1
e
\frac{3}{2} - \frac{1}{\sqrt{e}}
2
3
−
e
1
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