# First-order first-degree autonomous differential equation

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## Definition

Following the convention for autonomous differential equation, we denote the dependent variable by $x$ and independent variable by $t$.

### Form of the differential equation

The differential equation is of the form:

$\frac{dx}{dt} = f(x)$

### Solution method and formula

We convert the differential equation to an integration problem:

$\int \frac{dx}{f(x)} = \int dt$

and carry out the integrations on both sides. If $p$ is an antiderivative of $1/f$, the solution will be:

$p(x) = t + C$

with $C$ the freely varying parameter over $\R$: every particular value of $C$ gives a solution. To express $x$ as a function of $t$, we need to invert $p$. If we can do so, we'd have:

$x = p^{-1}(t + C)$

as the general solution.

In addition, there may be stationary solutions. These are solutions that correspond to constant functions $x = k$ that satisfy $f(k) = 0$.

## Analysis

### Starting at time zero with value one

Suppose we want to solve the initial value problem for the differential equation:

$\frac{dx}{dt} = f(x)$

subject to the initial condition that at $t = 0$, $x = 1$.

We consider various possibilities for $f$ a function that sends 1 and higher numbers to positive numbers, and make cases based on the growth rate of $f$:

Nature of $f$ Nature of $x$ in terms of $t$?
constant function $f(x) := c$ linear function $x := 1 + ct$
$f(x) := x^\gamma, 0 < \gamma < 1$ $x = O(t^{1/(1 - \gamma)})$ (i.e., it grows roughly like a power function of $t$)
linear function $f(x) := mx, m > 0$ exponential function $x := \exp(mt)$
linear times logarithmic, something like $x (\ln x + 1)$ $x$ grows something like a doubly exponential function of $t$ (note: if we wanted growth between exponential and double exponential, we would need something like $x(\ln x + 1)^\gamma, 0 < \gamma < 1$, and if we wanted triple exponential growth, we would multiply by a double logarithmic term)
$f(x) := x^\gamma, \gamma > 1$ $x$ grows so fast in terms of $t$ that it reaches $\infty$ in finite time.