Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value $y_0$, for $y_0 \neq 0$.
$y' + y^3 = 0$,
$y(0) = y_0$
A. $y(t) = \frac{y_0}{\sqrt{2y_0^2t + 1}}$, the solution exists when $t > -\frac{1}{2y_0^2}$
B. $y(t) = \frac{1}{\sqrt{2y_0^2t + 1}}$, the solution exists when $t > -\frac{1}{2y_0^2}$
C. $y(t) = \frac{y_0}{\sqrt{2y_0 + 1}}$, the solution exists when $t > -\frac{1}{2y_0^2}$
D. $y(t) = \frac{y_0}{\sqrt{2y_0^2t + 1}}$, the solution exists when $t > \frac{1}{2y_0^2}$
E. None
