First-order first-degree autonomous differential equation: Difference between revisions

Latest revision as of 21:51, 5 July 2012

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:

$\frac{d x}{d t} = f (x)$

Solution method and formula

We convert the differential equation to an integration problem:

$\int \frac{d x}{f (x)} = \int d t$

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 $f (k) = 0$ .

Related notions

Separable differential equation: A slightly more general type of first-order differential equation.
Second-order autonomous differential equation of degree one: Although such equations cannot always be solved, they can always be reduced to first-order differential equations.

Analysis

Starting at time zero with value one

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

$\frac{d x}{d t} = f (x)$

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

@@ Line 51: / Line 51: @@
 | <math>f(x) := x^\gamma, 0 < \gamma < 1</math> || <math>x = O(t^{1/(1 - \gamma)})</math> (i.e., it grows roughly like a power function of <math>t</math>)
 |-
 | linear function <math>f(x) := mx, m > 0</math> || exponential function <math>x := \exp(mt)</math>
 |-
-| linear times logarithmic, something like <math>x (\ln x + 1)</math> || <math>x</math> grows something like a doubly exponential function of <math>t</math>
+| linear times logarithmic, something like <math>x (\ln x + 1)</math> || <math>x</math> grows something like a doubly exponential function of <math>t</math> (note: if we wanted growth between exponential and double exponential, we would need something like <math>x(\ln x + 1)^\gamma, 0 < \gamma < 1</math>, and if we wanted triple exponential growth, we would multiply by a double logarithmic term)
 |-
 | <math>f(x) := x^\gamma, \gamma > 1</math> || <math>x</math> grows so fast in terms of <math>t</math> that it reaches <math>\infty</math> in finite time.
 |}