Solve the initial value problem (ODE) and determine how the interval on which its solution exists depends on the initial value?

Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value $y_0$:

$$\frac {dy}{dx} + y^3 = 0,\;y(0)=y_0$$

So far I've solved the DE and got that $y=\frac {1}{\sqrt {2t+2C}}$... I tried plugging in the initial value and got that$C = \frac{1}{2y_0^2}$.

The answer to this problem is: $y=\frac{y0}{\sqrt{2ty_0^2+1}}$ if $y_0$ does not equal $0$. If $y_0$ does equal $0$, $\;y=0$.

I think I'm headed in the right direction but not sure because my constant value seems wrong - How did $y_0$ go to the numerator? Also, given the solution, how do you figure out how the interval depends on the initial value $y_0$?

$$\frac {dy}{dx}=-y^3$$
$$y^{-3} dy = -dx$$ $$\frac {-1}{2} y^{-2} = -x +c$$ $$y^{-2} =2x +c$$ $$y(0)=0 \implies c=y_0^{-2}$$
$$y= \frac {1}{\sqrt {2x+c}} = \frac {1}{\sqrt {2x+y_0^{-2}}} = \frac {y_0}{\sqrt {2xy_0^2 +1}}$$
The interval of existence is as far as you don't cross the vertical asymptote of $$\frac {y_0}{\sqrt {2xy_0^2 +1}}$$ which depends on the initial value.