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General solution of $(2x^3-xy^2)dx+(2y^3-x^2y)dy=0$ is

$(a)\;x^4+x^2y^2-y^4=c \\ (b)\;x^4-x^2y^2+y^4=c \\ (c)\;x^4-x^2y^2-y^4=c \\ (d)\;x^2+x^2y^2+y^4=c $

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$2x^3dx+2y^3dy-(xy^2dx+x^2ydx)=0$
$d \bigg( \large\frac{x^4}{2}\bigg) +d \bigg( \large\frac{y^4}{2}\bigg) -d \bigg(\large\frac{x^2y^2}{2}\bigg)=0$
$d( x^4+y^2-x^2y^2)=0$
$x^2+y^4-x^2y^2=c$
Hence a is the correct answer
answered Feb 3, 2014 by meena.p
 

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