How to derive a (first-order) differential equation from a slope field A first order differential equation has a slope field shown in the following.

Question:
a) Suggest, with reasons, the simplest first order differential equation consistent with the slope field shown.
b) Suggest a possible general solution for your differential equation.
So what are the things that I should look for or things that I need to keep in mind while trying to derive a differential equation from a slope field?
In the first glance, I think it resembles a cubic? But that is just from observation, how can I confirm this?
 A: Observe that each of the vectors $F(x, y) = \langle P(x, y), Q(x, y) \rangle$ lies tangent to a vertical translation of some cubic function $f(x) = ax^3$ with $a > 0,$ i.e., each of the vectors in the slope field lies on some line tangent to $ax^3 + C$ for some real number $C.$ Observe that the slope of each tangent line is $f'(x) = 3ax^2,$ and this information is encoded in the vector $\langle 1, 3ax^2 \rangle.$ Our intuition tells us that $F(x, y) = \langle 1, 3ax^2 \rangle.$ Comparing with the graph of the slope field only strengthens this: we have that $F(0, y) = \langle 1, 0 \rangle = \langle 1, 3ax^2 \rangle|_{x = 0}$ by investigating the behavior of the slope field along the $y$-axis. Further, as $|x|$ increases, the slopes of the vectors in the slope field increase, and the arrows always point in the $(x \geq 0, y \geq 0)$-direction, from which it follows that (up to a scalar) $P(x, y) = 1$ and $Q(x, y) = g(x),$ where $g(x) \geq 0$ is some even function. Compare the given graph with the graph of $F(x, y) = \langle 1, x^2 \rangle$ to convince yourself.
Generally, this is the approach I would take if your slope field resembles a family of vertical translations of some function $f(x).$ Each of the vectors in the slope field lies tangent to a vertical translation of $f(x),$ hence each of the vectors in the slope field can be written as $\langle 1, f'(x) \rangle,$ where $a$ is a constant that depends on the shape of the graph. Given that your slope field resembles some other family of curves (e.g., concentric ellipses, horizontal translations of hyperbola, etc.), each of the vectors in the slope field lies tangent to these curves, so it suffices to find $y'$ in the implicit equation $f(x, y) = C$ that describes the curves. For instance, if you graph the slope field $F(x, y) = \langle y, x \rangle$, you will see that it resembles horizontal translations of the hyperbola $x^2 - y^2.$ But this is because the vector $\langle y, x \rangle$ encodes the slope $y' = m = \frac x y,$ and one can prove that this is precisely the slope of any line tangent to the curve $x^2 - y^2 = C.$ Likewise, if you graph the slope field $F(x, y) = \langle -y, x \rangle$, you will see that it resembles concentric circles because we obtain $y' = m = \frac x {-y}$ from the equations $x^2 + y^2 = C.$
Edit: One can think about a two-dimensional vector $\langle a, b \rangle$ in the following manner. Every line $\ell$ in $\mathbb R^2$ is uniquely determined by its $y$-intercept $(0, y_0)$ and its slope $m = \frac b a.$ Consequently, the line $\ell$ is uniquely determined by the vector $\mathbf m = \langle a, b \rangle$ (called the direction vector) and the vector $\mathbf b = \langle 0, y_0 \rangle$ (the $y$-intercept) in the sense that the line $\ell$ defined by $y = \frac b a x + y_0$ and the ray $\mathbf r(t) = \mathbf m t + \mathbf b = \langle at, bt + y_0 \rangle$ coincide. Particularly, for any point $(x, y)$ on the line $\ell,$ we have that $\mathbf r \bigl(\frac x a \bigr) = \langle x, \frac b a x + y_0 \rangle$ gives the ray originating at the origin and passing through the point $(x, y).$ Of course, if $a = 0,$ then the line $\ell$ is simply the $y$-axis, and the ray $\mathbf r(t) = \langle 0, bt + y_0 \rangle$ originates at the origin and passes through the point $(0, y_0).$
Consequently, the vector $\langle a, b \rangle$ can be thought of as describing a family of parallel lines whose slopes are $m = \frac b a$ whenever $a \neq 0$ and which are vertical whenever $a = 0.$ Particularly, given a differentiable function $f(x),$ the vector $\langle 1, f'(x) \rangle$ gives rise to a family of parallel lines whose slopes are $f'(x),$ i.e., the family of lines tangent to the $f(x) + C$ for any real number $C.$
A: You do not mention if you are supposed to give an ode in the plane or in 1D? So assume that we are looking for an ode of the form $\frac{dy}{dx}=v(x,y)$:
Obs 1: Visibly the slope field is invariant under vertical translations. So if the coordinates are $(x,y)$ you may chose a vector field $v(x)$, independent of $y$.
Obs 2: Slopes (values of $v$)  are non-negative, zero only for $x=0$ and increases when going away from zero. So a reasonable funciton with these properties is $v(x)=x^2$. Integral lines for $\frac{dy}{dx} = v(x)$ are $y=x^3/3+C$, with $C$ a constant.
The integral lines are cubics but the vector field is quadratic.
