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Differential Equations question

2020 · 8 Jan · Shift 1 · Q31

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JEE Main
Mathematics
Differential Equations
2020 · 8 Jan · Shift 1 · Q31

Differential Equations question

2020 · 8 Jan · Shift 1 · Q31

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

Let y = y(x) be a solution of the differential equation,

\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0

, |x| < 1. If

y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}

, then

y\left( { - {1 \over {\sqrt 2 }}} \right)

is equal to :

A
$- {{\sqrt 3 } \over 2}$
B
None of those
C
${{1 \over {\sqrt 2 }}}$
D
$-{{1 \over {\sqrt 2 }}}$

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