What's the general solution of the differential equation?

Question

$\frac{dy}{dx} = 3(y^2)^{\frac{1}3}$

(This is the same as saying the gradient is $3$ times $y$ to the power of $\frac{2}3$, defined to avoid confusion)

Initially, there doesn't seem to be too much trouble.

Integral of $\frac{1}3(y^2)^{-\frac{2}3}$ with respect to $y =$ integral of $1$ with respect to $x$

$y = (x+c)^3$ where c is a constant

But, according to my lecturer the general equation apparently has to be defined piecewise, and involves multiple constants and an indeterminately long period where the $y$ value is $0$ in the middle?

There is the special case where $y = 0$ and therefore cannot be divided through, but I still don't understand where multiple constants could come from. Surely $y$ and $\frac{dy}{dx}$ can only equal $0$ when $x = -c$?

Answer 1

Short answer. The method of separation

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = 3|y|^{\frac{2}{3}} \qquad\Rightarrow\qquad 3 |y|^{-\frac{2}{3}} \, \mathrm{d}y = \mathrm{d} x \qquad \Rightarrow \qquad \int 3 |y|^{-\frac{2}{3}} \, \mathrm{d}y = x + c $$

works only on an interval $I$ over which $y$ is non-zero. Moreover, the constant of integration $c$ may depend on that interval. There is no guarantee that this interval $I$ extends to all of $\mathbb{R}$.

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Long answer. Let's try to solve the equation rigorously.

Suppose that $y = y(x)$ is a differentiable function on $\mathbb{R}$ such that

$$ y'(x) = 3|y(x)|^{2/3} \qquad \text{for all} \quad x \in \mathbb{R}. $$

Since $y$ is continuous, the sets $U_+ = \{ x \in \mathbb{R} : f(x) > 0 \}$ and $U_- = \{ x \in \mathbb{R} : f(x) < 0 \}$ are open. Also, any open sets of of $\mathbb{R}$ can be written as a disjoint union of at most countably many open intervals, and so, it suffices to focus on each of the maximal open subintervals of $U_{\pm}$.

1. Let $I \subseteq U_+$ be a non-empty open subinterval which is maximal in the sense that no larger interval is a subset of $U_+$. Then $y(x) > 0$ on $I$, and so, $$ \frac{1}{3}y(x)^{-\frac{2}{3}}y'(x) = 1 \qquad \text{on} \quad I.$$ Since the left-hand side is the derivative of $y(x)^{\frac{1}{3}}$, it follows that $$ y(x)^{\frac{1}{3}} = x - c_I \qquad\text{and hence} \qquad y(x) = (x - c_I)^3 $$ for some constant $ c_I $ whose choice may depend on $I$. Since $x \mapsto (x-c_I)^3$ is positive only on $(c_I, \infty)$, we know that $I \subseteq (c_I, \infty)$. Moreover, by the maximality of $I$, we can deduce that $I = (c_I, \infty)$. Since any two such intervals must overlap, again by the maximality, we conclude that $U_+$ can have at most one such maximal subinterval, i.e., we must have either

   • $U_+$ is empty, or
   • $U_+ = (c_+, \infty)$ for some constant $c_+$ and $y(x) = (x - c_+)^3$ on $U_+$.

   The case $U_+ = \varnothing $ can be conveniently absorbed into the second case by allowing $c_+ = +\infty$.

2. By a similar reasoning, we can find that either

   • $U_-$ is empty, or
   • $U_- = (-\infty, c_-)$ for some constant $c_-$ and $y(x) = (x - c_-)^3$ on $U_-$.

   Similarly as before, the case $U_- = \varnothing $ can be conveniently absorbed into the second case by allowing $c_- = -\infty$.

3. Moreover, if both $U_+$ and $U_-$ are non-empty, then we must have $c_- \leq c_+$ since $U_+ \cap U_- = \varnothing$.

Combining these observations, we conclude that any differentiable solution of the equation must be of the form

$$ y(x) = \begin{cases} (x - c_+)^3, & x \geq c_+ \\ 0, & c_- < x < c_+ \\ (x - c_-)^3, & x \leq c_- \end{cases} $$

for some $-\infty \leq c_- \leq c_+ \leq \infty$. Conversely, any function of this form solves the given differential equation.