Using the Existence and uniqueness theorem to find the interval of validity of a unique solution in a Linear ODE From the existness and uniqueness Theorem,the initial value problem

$$y'=3x(y-1)^{1/3}            ,          y(3)=-7$$

has a unique solution on some open interval that contains $x=3$. Find the solution and determine
the largest open interval on which it’s unique.

What i tried,
             First i tried to solve the equation by the seperable equation method to get,$$y=1+(x^2-5)^{1.5}$$. Then from here i used the existence and uniquness theorem to calculate $f(x,y)$ and $f_{y}$. From the calculations,when $y$ not equals to $1$, $f_{y}$ will be continuous, hence there will be a unique solution when $y$ not equals to $1$, according to the wxistness and uniquness theorem. However im stuck from here onwards as to finding the interval of validaty. Is my working correct. Could anyone explain. Thanks
once you have $y = 1 - (x^2 -5)^{3/2}$, you can see that $\sqrt 5 \le x &lt; \infty$ is the maximal interval of existence.
The answer is correct and it gives the technique for finding the interval of validity. By the way it is the first answer to be posted solving the issue of the interval of validity.

I'll advance based on your solution. To find the interval of validity we need the solution of the ode  


  $$ y = 1+ (x^2-5)^{3/2} $$ 


which implies imposing the condition


  $$ x^2-5 \geq 0 \implies x\in ( -\infty, -\sqrt{5} )\cup (\sqrt{5}, \infty ). $$


The above suggested that the interval of validity is $x\in (\sqrt{5}, \infty )$.