I have this following ODE:
$$ \dot{x} = \frac{2x^4 + 2t}{3x^4 + 14} $$
where $x(0) = 0$. Let $\phi : \mathbb{R} \to \mathbb{R}$ be a solution to this ODE. How do I analyze the points where $\phi$ is decreasing or increasing? Its local maximum or minimum?
I've tried to obtain the solution to the ODE and then analyze its derivatives, but I don't think that this is the correct way to approach this problem. Is there a proper way to solve it?
On
Hint Suppose that a solution $x(t)$ has a critical point at time $t_0$, that is, that $\dot x(t_0) = 0$. To determine the character of that critical point, we can apply the Second Derivative Test. Computing gives $$\ddot x(t) = f(t) \dot x(t) + \frac{2}{3 x(t)^4 + 14}$$ for some function $f(t)$ defined on all of $\Bbb R$, so $$\ddot x(t_0) = \frac{2}{3 x(t_0)^4 + 14} > 0 ,$$ and in particular any critical point is a local minimum.
So far we have not used the initial condition, i.e., that the particular solution $\phi$ satisfies $\phi(0) = 0$ (so the above analysis incidentally applies to any initial condition). Does that condition give us information about the time of some minimum of $\phi$?
The naivest way is to consider $t \in [-10,10]$, set $\Delta t \equiv \color{green}{\pm} 10/20$ and apply Euler's method:
$$ \begin{cases} t_0 = 0 \\ x_0 = 0 \\ \end{cases} \quad \quad \quad \quad \begin{cases} t_i = t_{i-1} + \Delta t \\ x_i = x_{i-1} + \frac{2x_{i-1}^4+2t_{i-1}}{3x_{i-1}^4+14}\,\Delta t \\ \end{cases} \quad \quad \quad \quad i = 1,2,\dots,20 $$
$\quad\quad\quad\quad\quad$
where I first applied $\color{green}{\pm}$ with the $\color{blue}{\text{minus sign}}$, then with the $\color{red}{\text{plus sign}}$.