Question

 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.

Answer 1

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