Answer
the solution of initial value problem is given by
$$
y^{3}-4y- x^{3}+1=0
$$
the interval in which the solution is valid is $|x^{3}-1 | \lt \frac{16}{3\sqrt 3}$.
or
$ -1.276 \lt x \lt 1.597 $.
Work Step by Step
$$
y^{'}=\frac{dy}{dx}=\frac{3x^{2}}{(3y^{2}-4)} ,\quad \quad y(1)=0
$$
the differential equation can be written as
$$
(3y^{2}-4)dy=3x^{2}dx,\quad y(1)=0 \quad (*)
$$
integrating the left side with respect to $y$ and the right side with respect to $x$ gives
$$
y^{3}-4y=x^{3}+c \quad \quad (**)
$$
where c is an arbitrary constant.
To determine the solution satisfying the prescribed initial condition, we substitute $x=1$ and $y=0$ in eq. (**)
Hence the solution of initial value problem is given by
$$
y^{3}-4y- x^{3}+1=0 \quad \quad (**)
$$
The interval of validity of this solution extends
on either side of the initial point as long as the function remains differentiable, as we see in figure 1 which is attached.
We see that the interval ends when we reach points where the tangent line is vertical. It follows from the differential equation (*) that these are points where
$3y^{2}-4 =0$
i.e. when $y=\frac{2}{\sqrt 3} $ or $y=-\frac{2}{\sqrt 3} $
and from equation (**) the corresponding values of $x$ are respectively
$x^{3}-1=\frac{16}{3\sqrt 3} $ or $x^{3}-1=\frac{16}{3\sqrt 3} $ .
So the interval in which the solution is valid is $|x^{3}-1 | \lt \frac{16}{3\sqrt 3}$.
or
the interval in which the solution is valid is $ -1.276 \lt x \lt 1.597 $.
