Non-exact differential equation

Question

(y-x+xy cot(x))dx + xdy=0

Answer 1

(y-x+xy cot(x))dx + xdy=0;

M(x)(y - x + xycotx)dx +M(x)xdy = 0;

M(x)(1 + xcotx) = M'(x)x + M(x); M(x)xcotx = xM'(x); M'(x) = M(x)cotx; ln(M(x)) = ln(sinx); M(x) = sinx

We get integrated facor M(x) =sinx

sinx(y - x + xycotx)dx + xsinxdy = 0;

∂F(x, y)/∂y = xsinx; F(x, y) = ∫xsinx)dy = xysinx+ h(x);

∂F(x,y)/∂x = ysinx + xycosx + h'(x) = ysinx + xycosx - xsinx; h'(x) = - xsinx; h(x) = ∫-xsinxdx = xcosx - sinx .

So, we have solution: F(x, y) = xysinx + xcosx - sinx = C

Answer 2

Multiply both sides by sin(x). This will make the equation exact.