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 :

Given $\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$ <br><br>$\Rightarrow$ ${{dy} \over {dx}} = - {{\sqrt {1 - {y^2}} } \over {\sqrt {1 - {x^2}} }}$ <br><br>$\Rightarrow$ ${{dy} \over {\sqrt {1 - {y^2}} }} + {{dx} \over {\sqrt {1 - {x^2}} }} = 0$ <br><br>$\Rightarrow$ sin^-1 y + sin^-1 x = c <br><br>Given that $y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$ <br><br>So when x = ${1 \over 2}$ then y = ${{\sqrt 3 } \over 2}$ <br><br>$\therefore$ sin^-1 ${{\sqrt 3 } \over 2}$ + sin^-1 ${1 \over 2}$ = c <br><br>$\Rightarrow$ c = ${\pi \over 3}$ + ${\pi \over 6}$ = ${\pi \over 2}$ <br><br>So sin^-1 y + sin^-1 x = ${\pi \over 2}$ <br><br>$\Rightarrow$ sin^-1 y = ${\pi \over 2}$ - sin^-1 x <br><br>$\Rightarrow$ sin^-1 y = cos^-1 x <br><br>Now $y\left( { - {1 \over {\sqrt 2 }}} \right)$ means x = $- {1 \over {\sqrt 2 }}$ and find y. <br><br>Putting x = $- {1 \over {\sqrt 2 }}$ <br><br>sin^-1 y = cos^-1 $\left( { - {1 \over {\sqrt 2 }}} \right)$ = ${{3\pi } \over 4}$ <br/><br/>y (at x = $- {1 \over {\sqrt 2 }}$) = sin(${{3\pi } \over 4}$) = ${{1 \over {\sqrt 2 }}}$