
Yefim S. answered 07/21/22
Math Tutor with Experience
(2xcosy + 3x2y)dx + (x3 - x2siny - y)dy = 0;
∂/∂y(2xcosy + 3x2y) = (-2xsiny + 3x2) = ∂/∂x(x3 - x2siny - y). So, we have exact equation.
F(x,y) = ∫(2xcosy + 3x2y)dx = x2cosy + x3y + f(y);
dF/dy = - x2siny + x3 + f'(y) = x3 - 2xsiny - y; f'(y) = - y; f(y) = - y2/2 + C
F(x,y) = x2cosy + x3y - y2/2 + C