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Eliminate arbitrary constant and obtain differential equation $x^2-y^2= a(x^2+y^2)^2$

$(a)\;y'= \frac{x}{2y} \\ (b)\;3y=2y'x \\ (c)\;y'= \frac{x}{y} \\ (d)\;y'=\frac{z}{y} $

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$2x-2yy'= a2(x^2+y^2(2x+2yy'))$
$x=yy'= 2a(x^2+y^2)(x+yy')$
$x-yy'= \large\frac{2(x^2-y^2)}{(x^2+y^2)^2}$$ (x^2+y^2) (x+yy')$
$(x-yy')e(x^2+y^2)=2(x^2-y^2)(x+yy')$
gives $y'=\large\frac{x}{y}$
Hence c is the correct answer.
answered Feb 7, 2014 by meena.p
 

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