Answer
$\displaystyle y = -\sqrt{2\sin(x) - 2x\cos(x) + 1}$
Work Step by Step
----$\displaystyle \frac{dy}{dx} = \frac{x\sin(x)}{y}$
Separation of Variables:
$\displaystyle \int y dy = \int x \sin(x) dx$
Integrate with respect to each variable (the right side requires integration by parts):
$\displaystyle \frac{y^2}{2} = x(-\cos(x))-\int(1)(-\cos(x))dx$
$\displaystyle \frac{y^2}{2} = -x\cos(x) + \int \cos(x)dx$
$\displaystyle \frac{y^2}{2} = -x\cos(x) + \sin(x) +C$
$\displaystyle y^2=2\sin(x)-2x\cos(x)+C$
--------------------------Plug in the initial condition to solve for C
$y(0) = -1$
$\displaystyle (-1)^2 = 2\sin(0)-2(0)\cos(0)+C$
$\displaystyle 1 = 0 - 0 + C$
$\displaystyle 1 = C$
--------------------------Plug in the initial condition and isolate y
$\displaystyle y^2 = 2\sin(x) - 2x\cos(0) + 1$
$\displaystyle y = \pm\sqrt{2\sin(x) - 2x\cos(x) + 1}$
Reject the positive solution since it doesn't fit the initial condition $y(0) = -1$
$\displaystyle y = -\sqrt{2\sin(x) - 2x\cos(x) + 1}$
