Is this true for the Initial Value Problem

$xy'=2y$, $y(a)=b$

The function $y=x^2$ for $y\leq 0$ and $y=cx^2$ for $x\geq 0$ is one of the solutions.

The answer is given to be correct.

I find it rather amusing that the first solution is only satisfied by (0,0), which already is in the second solution. 

Drawing the graph in mathematica, the solution seems to be

$y=x^2$ for ***$x\leq 0$***, and $y=cx^2$ for $x\geq 0$

So, according to me the statement is incorrect..

So am i right . Or did I miss something?