uniqueness problem with ODE?

Question

Consider the initial value problem $$ \frac{dy}{dt}=4t\sqrt{y}, \quad y(1)=1. $$ This is clearly separable, and using separation of variables we solve $$ \int y^{-1/2} dy=\int 4t dt $$ which gives $2\sqrt y=2t^2+C$. After replacing constants and squaring, this gives $$ y(t)=(t^2+C)^2 $$ as the general solution. Now, to solve the initial value problem, we consider $$ 1=y(1)=(1+C)^2=C^2+2C+1, $$ which has the two solutions $C=0,-2$. Therefore we should have two solutions $$ y_1(t)=t^4\qquad y_2(t)=(t^2-2)^2 $$ to the above initial value problem. My issue here is that, well, doesn't this violate the Uniqueness Theorem? (Recall, the Uniqueness Theorem says essentially that, for an initial value problem of the form $$ \frac{dy}{dt}=f(t,y),\quad y(t_0)=y_0, $$ if both $f$ and $\partial f/\partial y$ are continuous around $(t_0,y_0)$, then the IVP has a unique solution in a neighborhood of $(t_0,y_0)$.) What am I missing here?