Consider the integral equation
$$y\left(x\right) = A + \int_{0}^{x}y^{2/3}\left(s\right) ds$$ where $A\in\mathbb{R}$. What does this limit of general solution when A goes to 0 imply ? Which is \begin{align*} \frac{dy}{dx} &= y^{2/3}\left(x\right) \\ \frac{dy}{y^{2/3}} &= dx \\ \int_{0}^{x}\frac{dy}{y^{2/3}} &= \int_{0}^{x} dx \\ 3y^{1/3}|_{0}^{x} &= x \\ \left(y\left(x\right)\right)^{1/3} - \left(y\left(0\right)\right)^{1/3} &= \frac{x}{3} \\ \left(y\left(x\right)\right)^{1/3} &= A^{1/3} + \frac{x}{3} \\ \phi\left(x,A\right) = y\left(x\right) &= \left(A^{1/3} + \frac{x}{3}\right)^{3} \end{align*} For what value(s) of $A$ can you be sure that the solution is unique ? What is the range of $\phi\left(x,A\right)$ for such value of $A$? Why ?