Prove uniqueness of solution in ODE I need to find function $f(x)$, such that $x > 0$ and $f(x) > 0$, so that the area limited by the function, the $y$ and $x$ axes and $x = a$ ($a > 0$) is $f^3(x)$ for every $a$, and prove that there is only one such function.
Let $y' = f(x)$ and $y = \int^{x}{f(t)dt}$ I get the ODE $(y')^3 = y \Longrightarrow y'=\sqrt[3]{y}$ which is separable thus I get $y = \sqrt[3]{(\frac{2}{3}(C+x))^2}$. But I'm not sure how I'm supposed to prove uniqueness. The existence and uniqueness theorem is defined for $D = \{(x, y)| \alpha < x <\beta, \gamma < y < \delta \}$ where $y$ is continuous and the $y$ partial derivative exists if $(x_0, y_0)\in D$ then there is a solution for the initial value problem with $(x_0, y_0)$ and that solution is unique, but can I use this here? and if so, how?
 A: $$
(y')^{3}=y,\Longrightarrow{y'}=\sqrt[3]{y},\implies\frac{dy}{dx}=y^{1/3},\implies\frac{dy}{y^{1/3}}=dx,\implies{y^{-1/3}}dy=dx,\implies
$$
$$
\implies\int{y^{-1/3}}dy=\int{dx},\implies{\frac{3}{2}y^{\frac{2}{3}}}=x+C.
$$
We see that $\bbox[lightgreen]{f(y, x)=\sqrt[3]{y}}$ is clearly continuous at all points $\bbox[lightgreen]{(y, x)}$ and $\bbox[lightblue]{\frac{\partial{f}}{\partial{y}}=\frac{1}{3}{y}^{-\frac{2}{3}}=\frac{1}{3}\cdot\frac{1}{y^{\frac{2}{3}}}}$ $\implies$ we have a critical coordinate at $\bbox[lightblue]{y=0}$ $\implies$ $\bbox[yellow]{\frac{\partial{f}}{\partial{y}}}$ is not continuous. Therefore, we shouldn't expect to have both existence and uniqueness if our an initial condition, for example, is $\bbox[pink]{y(0)=0}$ and in fact we don't have uniqueness, as we have seen.
Good luck!
A: From the question we get that $y = \sqrt{\frac{8}{27}}(x+C)^{3/2}$ so $f(x) = y'= \sqrt[3]{y}=\sqrt{\frac{2}{3}(x+C)}$ we know that $y(a) = \int_{0}^{a} f(x)dx = (y'(a))^3$ hence
$$
\int_{0}^{a} f(x)dx = \int_{0}^{a}  \sqrt{\frac{2}{3}(x+C)}dx = \sqrt{\frac{8}{27}}(x + C)^{3/2}|_0^a = \sqrt{\frac{8}{27}}(a + C)^{3/2} - \sqrt{\frac{8}{27}}(C)^{3/2} = \sqrt{\frac{8}{27}}(a + C)^{3/2} = \sqrt{\frac{2}{3}(a+C)}^3 = f^3(a) \Longrightarrow \sqrt{\frac{8}{27}}(C)^{3/2}=0 \Longrightarrow C=0
$$
Therefore there is only one unique f(x) in the first quadrant when $C=0$ and $f(x) =  \sqrt{\frac{2}{3}(x)}$
