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Solve y'=y/(y-2x) implicitly.
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@aaronq @abb0t @ParthKohli
substitute `y = ux`
y=ux u=y/x y'=u+u'x u'x+u=ux/(ux-2x)=u/(u-2) u'x=u/(u-2)-u=u/(u-2)-(u(u-2))/(u-2) =(u-u^2+2u)/(u-2)=(3u-u^2)/(u-2)=(u(3-u))/(u-2)
yes separate variables and do partial fractions
I don't know why the answer in the book is y^2(y-3x)=C when I got y^2(3x-y)=C as the answer. Am I right?
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I see why. I think it's because of the constant C. Since C can be both positive and negative because it's constant.
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