Given "\\frac{x}{2}\\cos(y)y'=-1" which can be expressed as
"\\frac{x}{2}\\cos(y)\\frac{dy}{dx}=-1"
Equivalently we have
"\\frac{x}{2}\\cos(y)dy + dx=0"
The ODE is not exact since "\\frac{\\partial(\\frac{x}{2}\\cos (y))}{\\partial x}=\\frac{1}{2}\\cos(y) \\ne \\frac{\\partial(1)}{\\partial y}=0"
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