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Question Number 68855 by ramirez105 last updated on 16/Sep/19

Commented by kaivan.ahmadi last updated on 16/Sep/19

$s e t M = 6 x + y^{2} a n d N = y (2 x - 3 y)$ $\frac{\partial M}{\partial y} = 2 y = \frac{\partial N}{\partial x} \Rightarrow t h e e q u a t i o n i s e x a c t$ $\frac{\partial u}{\partial x} = 6 x + y^{2} a n d \frac{\partial u}{\partial y} = y (2 x - 3 y) = 2 x y - 3 y^{2}$ $\Rightarrow u (x, y) = \int (6 x + y^{2}) d x = 3 x^{2} + {x y}^{2} + h (y)$ $\Rightarrow \frac{\partial u}{\partial y} = 2 x y - 3 y^{2} = 2 x y + \frac{d (h)}{d y} \Rightarrow \frac{d (h)}{d y} = - 3 y^{2} \Rightarrow$ $h = - y^{3} + c^{'}$ $u (x, y) = c \Rightarrow 3 x^{2} + {x y}^{2} - y^{3} + c^{'} = c \Rightarrow$ $u (x, y) = 3 x^{2} + {x y}^{2} - y^{3} = C$
Commented by ramirez105 last updated on 17/Sep/19

$thank you sir!$
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