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Mathematics
Problem
Let
y
=
y
(
x
)
y = y(x)
y
=
y
(
x
)
be a solution of the differential equation,
1
−
x
2
d
y
d
x
+
1
−
y
2
=
0
\sqrt{1-x^2} \frac{dy}{dx} + \sqrt{1-y^2} = 0
1
−
x
2
d
x
d
y
+
1
−
y
2
=
0
,
∣
x
∣
<
1
|x| < 1
∣
x
∣
<
1
. If
y
(
1
2
)
=
3
2
y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2}
y
(
2
1
)
=
2
3
, then
y
(
−
1
2
)
y\left(\frac{-1}{\sqrt{2}}\right)
y
(
2
−
1
)
is equal to (1)
−
3
2
-\frac{\sqrt{3}}{2}
−
2
3
(2)
1
2
\frac{1}{\sqrt{2}}
2
1
(3)
3
2
\frac{\sqrt{3}}{2}
2
3
(4)
−
1
2
-\frac{1}{\sqrt{2}}
−
2
1
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