Existences and uniqueness theorem, finding a unique solution of a first order ODE Consider the initial value problem
$$y′=3x(y−1)^{1/3},y(x_{0})=y_{0}$$
Using the existness and uniqueness theorem, for what points ($x_{0}$,$y_{0}$) imply that the above IVP  has a unique solution on some open interval that contains $x_{0}$?
What i tried
Using the seperable equation method, 
I got  a solution of $$y=1+(x^{2}+c)^{1.5}$$
Im unsure of how to find the value of $c$ given the inital conditions and whether is it necessary to find the value of $c$ for this problem.
While using the  existence and uniqueness theorem, $y_{0}$ must not be equal to $1$ in order for a unique solution to exist. But from here im unsure of how to find the points  ($x_{0}$,$y_{0}$), I was thinking that since i already know that $y_{0}$ does not equals to $1$, i must combine this result with the above solution that i got  in order to get $x_{0}$ and hence the point ($x_{0}$,$y_{0}$) but im unsure how to do so, especially when there is a $c$ involved. Could anyone explain. Thanks
 A: Since you have been given an initial condition then you need to use it to find $c$ as

$$ y_0 = 1+(x_0^2+c)^{3/2}. $$

Solve the above equation for $c$ and then substitute back the value of $c$ you get in the solution of the differential equation 

$$ y = 1+(x^2+c)^{3/2}. $$

A: The theory tells you about the existence of solution if the function $F(x,y)$ is continuous and existence and uniqueness if the function is $C^1$ ( in fact locally Lipschitz in the second variable is OK). Your function $3x(y-1)^{1/3}$ has not second partial derivative at points with $y=1$. So this will be the problem points.
By separation of variable we get
$$\frac{dy}{dx} = 3 x (y-1)^{\frac{1}{3}}$$ or
$$\frac{2/3}{(y-1)^{1/3}}dy  = 2 x dx$$
or 
$d((y-1)^{2/3} = d x^2$ or $(y-1)^{2/3} = x^2 + c$ . For $(x_0, y_0)$ get 
$$(y_0-1)^{2/3} = x_0^2 + c$$ so $c = (y_0-1)^{2/3} - x_0^2$. So we get the solutions
$$y(x) = 1 \pm \sqrt{(x^2 - x_0^2 + (y_0-1)^{2/3})^3}$$
If $y_0\ne 1$ then we have to choose the square root that makes $\pm \sqrt{(y_0-1)^2}= y_0-1$. Therefore, for $y_0 <1$ the solution is
$$y(x) = 1 - \sqrt{(x^2 - x_0^2 + (y_0-1)^{2/3})^3}$$
while for $y_0 >1$ the solution is 
$$y(x) = 1 + \sqrt{(x^2 - x_0^2 + (y_0-1)^{2/3})^3}$$
However, at points $(x_0, 1)$ , both choices of signs work, however, we have to move on the side on which $x^2 \ge x_0^2$, unless $x_0$ is also $0$, when we can move both ways. We also have the solution $y(x) \equiv 1$. Therefore, at points $(x_0, 1)$ we do not have uniqueness. Quite interesting...
