Determine if the differential equation (sin y − y sin x)dx + (cos x + x cos y − y)dy = 0 is exact. If it is exact, solve it.
"\\displaystyle\n\\text{Let $u = (\\sin y - y\\sin x)dx$ and $v= (\\cos x+x\\cos y - y)dy$}\\\\\n\\frac{\\partial (\\sin y - y \\sin x)}{\\partial y} = \\cos y - \\sin x\\\\\n\\frac{\\partial (\\cos x - x \\cos y-y)}{\\partial x} = \\cos y - \\sin x\\\\\n\\text{Next, we integrate u with respect to x}\\\\\n\\int (\\sin y- y\\sin x)dx = x\\sin y + y \\cos x + h(y)-(1)\\\\\n\\text{Next, we differentiate the expression above with respect to y}\\\\\nx\\cos y + \\cos x + h'(y)\\\\\n\\text{Comparing the expression above to v, we have}\\\\\nx\\cos y + \\cos x + h'(y) = \\cos x+x\\cos y - y\\\\\n\\implies h'(y) = -y \\therefore h(y) = -\\frac{y^2}{2}\\\\\n\\text{Substituting h(y) in (1), we have that}\\\\\nx\\sin y + y \\cos x-\\frac{y^2}{2}\\\\\n\\text{Hence the the solution of the differential equation is}\\\\\nx\\sin y + y \\cos x-\\frac{y^2}{2}=c"
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