(6x+y^2 )dx +y(2xy−3y^2 )dy = 0


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Question Number 124050 by john_santu last updated on 30/Nov/20

$(6 x + y^{2}) d x + y (2 x y - 3 y^{2}) d y = 0$
	Answered by mohammad17 last updated on 30/Nov/20

	$M = 6 x + y^{2} \to M_{y} = 2 y \to ()$ $N = 2 x y - 3 y^{2} \to N_{x} = 2 y \to ( )$ $f r o m () a n d (* *) t h e e q u a t i o n i s e x a c t$ $\int M d x + \int N d y = \int d (c)$ $\int (6 x + y^{2}) d x + \int (2 x y - 3 y^{2}) d y = \int d (c)$ $3 x^{2} + 2 {x y}^{2} + x y - y^{3} = C$ $b y : ⟨ M . O ⟩$
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