Let us solve the equation "\\sin xdx + ydy = 0; y(0)=-2":
"\\sin xdx =- ydy"
"\\int \\sin xdx = -\\int ydy"
"-\\cos x=-\\frac{y^2}{2}+C"
Since "y(0)=-2", we have the equation "-\\cos 0=-\\frac{(-2)^2}{2}+C" which is equivalent to "-1=-2+C". So, "C=1."
Therefore,
"-\\cos x=-\\frac{y^2}{2}+1" or
"y^2=2\\cos x+2"
Since "y(0)=-2", the solution is
"y=-\\sqrt{2\\cos x+2}"
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