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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.

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

3 Answers 3

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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 $a \ge x_0$.

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$\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.

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

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