# why this happens ? (dilation)

let be the dilation operator $x \frac{d}{dx}$ i know this operator is invariant under the change $y=ax$ for any positive 'a' real number

however let be the change $y= \frac{-1}{x}$ then the operator $x \frac{d}{dx}= y \frac{d}{dy}$ still remains invariant .. is there a mathematical reason to see why this happens ?? , given the group of dilations in one dimension, is there a bigger group containing it ??

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Why only positive reals? Also, it seems to me that $y\frac{d}{dy}=\color{Red}-x\frac{d}{dx}$. –  anon Apr 29 '12 at 9:56
BTW setting $x=g(y)$ looks like it leads to a simple DE. –  anon Apr 29 '12 at 10:02
To sum up, (i) $y=-1/x$ does NOT yield $xd/dx=yd/dy$, (ii) solving $xd/dx=yd/dy$ yields $y=ax$ for some nonzero $a$. –  Did May 12 '12 at 7:57