Answer
y=$-\frac{1}{4}\sqrt {1-4x^{2}}$+C
Work Step by Step
$\sqrt {1-4x^{2}}y'$=x
dy=$\frac{x}{\sqrt {1-4x^{2}}}$dx
$\int$dy=$\int$$\frac{x}{\sqrt {1-4x^{2}}}$dx
$\int$dy=$\frac{-1}{8}$$\int$(1-4$x^{2}$)$^{-\frac{1}{2}}$(-8xdx)
y=$-\frac{1}{4}\sqrt {1-4x^{2}}$+C
