# Differential Equation $dy/dx=y^{1/3}$ and condition $y(x_0)=y_0$ has infinite solutions

Prove that the differential equation $$\frac{dy}{dx}= y^{1/3}$$ with the initial value of $$y(x_0)=y_0$$ has infinite solutions.

I don't really understand the problem if I have to show that there are infinite solutions depending on the initial conditions or if it is something like if I proposed $$(x_0,y_0)=(0,0)$$ and prove that for that case the equation has infinite solutions. if you think is the first one could you explain how to do it.

• i have read somewhere that $\frac{dy}{dx}=y^{\alpha}$ conditon $y(x_0)=0$ has infinite solution when $1>\alpha$ Mar 29, 2020 at 5:28
• Wolfie Mar 29, 2020 at 5:31
• Define "has infinite solutions". Does it mean that it blows up? Or that it has infinitely many solutions? Mar 29, 2020 at 9:39

If $$y_0 > 0$$ then $$f(y) = y^{1/3}$$ is Lipschitz in a neighbourhood of $$y_0$$ (the derivative is bounded). So by Picard-Lindelöf there is a unique solution.

On the other hand, for $$y_0 = 0$$, $$f$$ is no longer Lipschitz and so we can no longer expect a unique solution. In fact two solutions are found readily:

$$y(x) = 0 \text{ and } y(x) = \left[\frac{2}{3}(x - x_0)\right]^{3/2}.$$

But as Wikipedia alludes to in its article on the Peano existence theorem, the transition between the two solutions can happen at any point (not just at $$x_0$$). So the general solution is

$$y_a(x) = \begin{cases} 0 & x \le a, \\ \left[\frac{2}{3}(x - a)\right]^{3/2} & x > a, \end{cases}$$

for any parameter $$a \ge x_0$$.

$$\frac{dy}{dx} = y^ \frac{1}{3}$$ , by using method of variable separable we will get $$y(x)= \left(\frac{2x}{3}+k\right)^\frac{3}{2}$$ ; where $$k$$ is an arbitrary constant now consider $$x_0 = 0$$ and $$y_0 = 0$$ as initial conditions then for every $$\alpha \ge0$$ we have $$f(x) = \begin{cases} 0, & x \le\alpha \\ \left(\frac{2x}{3}-\frac{2\alpha}{3}\right)^\frac{3}{2}, & x\gt\alpha \end{cases}$$ so uncountably infinite solution for this initial condition.

With $$\frac{dy}{dx} = \sqrt[3]{y} \quad y(x_{0}) = y_{0}$$ then \begin{align} y^{-1/3} \, dy &= dx \\ \frac{3}{2} \, y^{2/3} &= x + c_{0} \\ y(x) &= \left[ \frac{2}{3} \, (x + c_{0}) \right]^{3/2}. \end{align}

When $$y(x_{0}) = y_{0}$$ then $$c_{0} = \frac{3}{2} \, y_{0}^{2/3} - x_{0}$$ and $$y(x) = \left[ \frac{2}{3} \, (x - x_{0}) + \sqrt[3]{y_{0}} \right]^{3/2}.$$