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CSIR UGC NET June 2019 Unit III Question 46
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Differential Equation
CSIR UGC NET June 2019 Unit III Question 46
+1
vote
238
views
Let
x
∗
(
t
)
be the curve which minimizes the functional
J
(
x
)
=
∫
1
0
[
x
2
(
t
)
+
˙
x
2
(
t
)
]
d
t
satisfying
x
(
0
)
=
0
,
x
(
1
)
=
1
. Then the value of
x
∗
(
1
2
)
is
1.
√
e
1
+
e
2.
2
√
e
1
+
e
3.
√
e
1
+
2
e
4.
2
√
e
1
+
2
e
net
2019
jun
asked
in
Differential Equation
by
Junky
Expert
|
238
views
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