Initial Value Problem $dy/dx = (y+1)^{1/3}$ Consider the differential equation $$\frac{dy}{dx} = (y+1)^{1/3}$$
(a) State the region of the $xy$-plane in which the conditions of the existence and uniqueness theorem are satisfied (using any appropriate theorem).
(b) Let $S$ be the region of the $xy$-plane where the conditions of the existence and uniqueness theorem are NOT satisfied. State whether the given equation with the initial condition $y(x_0)=y_0$, where $(x_0, y_0)$ is an element of $S$, has a solution.
(c) solve this equation
(d) Using the result of (c), find whether the given equation with the initial condition $y(x_0) = y_0$, where $(x_0, y_0)$ is an element of $S$, has a unique solution.
Struggling massively with this question. Currently studying initial value problems but haven't been able to come across any text or vid explaining the technique to approach this. Would very much appreciate any help. 
I can only assume for part (a) the region of the xy plane is -infinity < x < infinity. And -1 < y < infinity. 
For (b) I don't know the technical method of approaching this question even after reading through a differential equations textbook because it was explained it riddles. 
For (c) i arrived at y = [[(2/3)(x+c)]^(3/2)]-1. Don't know if evaluating it further is necessary or not? 
For (d) again I used a book which didn't explain the steps to doing this

For part (a), you're on the right track;  note, however, that the equation is just as well-defined for $y < -1$ as it is for $y > -1$.  @crankk's suggestion to apply the Picard-Lindelöf Theorem would be the most rigorous way of showing this.
For part (b), I'm not sure that there's a "technical method" to address the question.  However, I will tell you that the answer is "yes", and that there's a fairly obvious solution to the above equation that goes through $(x_0, y_0) = (x, -1)$ for arbitrary $x$;  finding and exhibiting this solution should be sufficient, I would think, for a definitive answer to the question.
