solve (3x^5 y^4 +4y)dx+(2x^6 y^3 +3x)dy=0


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Question Number 88238 by M±th+et£s last updated on 09/Apr/20

$s o l v e$ $(3 x^{5} y^{4} + 4 y) d x + (2 x^{6} y^{3} + 3 x) d y = 0$
	Answered by ajfour last updated on 09/Apr/20

	$d i v i d i n g b y x^{3} y^{2}$ $(3 x^{2} y^{2} d x + 2 x^{3} y d y) + \frac{(4 y d x + 3 x d y)}{x^{3} y^{2}} = 0$ $x^{3} y^{2} d (x^{3} y^{2}) + (4 y d x + 3 x d y) = 0$ $m u l t i p l y i n g b y x^{3} y^{2}$ ${(x^{3} y^{2})}^{2} d (x^{3} y^{2}) + (4 x^{3} y^{3} d x + 3 x^{4} y^{2} d y) = 0$ ${(x^{3} y^{2})}^{2} d (x^{3} y^{2}) + d (x^{4} y^{3}) = 0$ $\Rightarrow \frac{{(x^{3} y^{2})}^{3}}{3} + x^{4} y^{3} = c .$
	Commented by M±th+et£s last updated on 09/Apr/20

	$n i c e s o l u t i o n . h o d b l e s s y o u$
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