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

Revision as of 23:40, 12 February 2012

Definition

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

Form of the differential equation

A (one-dimensional) second-order autonomous differential equation is a differential equation of the form:

$\frac{d^{2} x}{d t^{2}} = F (x, \frac{d x}{d t})$

Solution method and formula

We set a variable $v = d x / d t$ Then, we can rewrite $d t = (d x) / v$ . In particular, $d^{2} x / d t^{2} = (d / d t) (d x / d t) = d v / d t = d v / ((d x) / v) = v d v / d x$ . Plug this in:

$v \frac{d v}{d x} = F (x, v)$

Solve this to obtain the general solution for $v$ in terms of $x$ . Plug this expression in:

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

and solve this first-order differential equation. Note that if $v$ is not expressible as an explicit function of $x$ , but we instead have a relational solution $R (x, v) = 0$ , then solve the first-order differential equation:

$R (x, \frac{d x}{d t}) = 0$

Particular cases

Case where the function on the right depends only on $x$ and not on $d x / d t$

Consider a situation of the form:

$\frac{d^{2} x}{d t^{2}} = g (x)$

We do the same substitution $v = d x / d t$ and obtain:

$\frac{v d v}{d x} = g (x)$

This is now a separable differential equation relating $x$ and $v$ . Integrate and obtain:

$\int v d v = \int g (x) d x$

We thus get:

$\frac{v^{2}}{2} = \int g (x) d x$

In particular, if $G$ is an antiderivative for $g$ , then we get:

$v = \pm \sqrt{2 G (x) + C_{1}}$

where $C_{1} \in R$ is a parameter. Each choice of $C_{1}$ gives a different solution.

Plug this back in and get:

$\frac{d x}{d t} = \pm \sqrt{2 G (x) + C_{1}}$

(The $\pm$ indicates that there are in fact two differential equations and we need to take the union of their solution sets).

This is a first-order autonomous differential equation, and in particular a separable differential equation. Rearrange and get:

$\pm \int \frac{d x}{\sqrt{2 G (x) + C_{1}}} = \int d t$

An additional constant, $C_{2}$ , arises from this indefinite integration. The upshot is that the general solution relates $x$ to $t$ and has two parameters $C_{1}, C_{2}$ , as we might expect from the degree of the equation.

Case where the function on the right is multiplicatively separable

Consider a situation of the form:

$\frac{d^{2} x}{d t^{2}} = g (x) h (\frac{d x}{d t})$

We do the same substitution $v = d x / d t$ and obtain:

$\frac{v d v}{d x} = g (x) h (v)$

This is a separable differential equation and we can rearrange it to obtain:

$\int \frac{v d v}{h (v)} = \int g (x) d x$

We now perform the integration both sides. Suppose $H (v) = \int \frac{v d v}{h (v)}$ and $G (x) = \int g (x) d x$ . We get:

$H (v) = G (x) + C_{1}$

If $H$ can locally be inverted, we can write $v$ as an explicit function of $x$ We now plug this into the original differential equation and get:

$H (d x / d t) = G (x) + C_{1}$

This is now a first-order differential equation. If $H$ can locally be inverted, we can write $d x / d t$ as an explicit function of $x$ (locally) and then solve the resultant separable differential equation. Otherwise, there may be some other method available.

@@ Line 7: / Line 7: @@
 A (one-dimensional) '''second-order autonomous differential equation''' is a differential equation of the form:
-<math>\frac{d^2x}{dt^2} = f(x)</math>
+<math>\frac{d^2x}{dt^2} = F(x, \frac{dx}{dt})</math>
 ===Solution method and formula===
@@ Line 13: / Line 13: @@
 We set a variable <math>v = dx/dt</math> Then, we can rewrite <math>dt = (dx)/v</math>. In particular, <math>d^2x/dt^2 = (d/dt)(dx/dt) = dv/dt = dv/((dx)/v) = vdv/dx</math>. Plug this in:
-<math>v \frac{dv}{dx} = f(x)</math>
+<math>v \frac{dv}{dx} = F(x, v)</math>
+Solve this to obtain the general solution for <math>v</math> in terms of <math>x</math>. Plug this expression in:
+<math>\frac{dx}{dt} = v(x)</math>
+and solve this first-order differential equation. Note that if <math>v</math> is not expressible as an explicit function of <math>x</math>, but we instead have a relational solution <math>R(x,v) = 0</math>, then solve the first-order differential equation:
+<math>R(x, \frac{dx}{dt}) = 0</math>
+==Particular cases==
+===Case where the function on the right depends only on <math>x</math> and not on <math>dx/dt</math>===
+Consider a situation of the form:
+<math>\frac{d^2x}{dt^2} = g(x)</math>
+We do the same substitution <math>v = dx/dt</math> and obtain:
+<math>\frac{v \, dv}{ \, dx} = g(x)</math>
 This is now a [[separable differential equation]] relating <math>x</math> and <math>v</math>. Integrate and obtain:
-<math>\int v \, dv = \int f(x) \, dx</math>
+<math>\int v \, dv = \int g(x) \, dx</math>
 We thus get:
-<math>\frac{v^2}{2} = \int f(x) \, dx</math>
+<math>\frac{v^2}{2} = \int g(x) \, dx</math>
-In particular, if <math>F</math> is an antiderivative for <math>f</math>, then we get:
+In particular, if <math>G</math> is an antiderivative for <math>g</math>, then we get:
-<math>v = \sqrt{2F(x) + C_1}</math>
+<math>v = \pm \sqrt{2G(x) + C_1}</math>
 where <math>C_1 \in \R</math> is a parameter. Each choice of <math>C_1</math> gives a different solution.
@@ Line 31: / Line 51: @@
 Plug this back in and get:
-<math>\frac{dx}{dt} = \sqrt{2F(x) + C_1}</math>
+<math>\frac{dx}{dt} = \pm \sqrt{2G(x) + C_1}</math>
+(The <math>\pm</math> indicates that there are in fact two differential equations and we need to take the union of their solution sets).
 This is a first-order autonomous differential equation, and in particular a [[separable differential equation]]. Rearrange and get:
-<math>\int \frac{dx}{\sqrt{2F(x) + C_1}} = \int dt</math>
+<math>\pm \int \frac{dx}{\sqrt{2G(x) + C_1}} = \int dt</math>
 An additional constant, <math>C_2</math>, arises from this indefinite integration. The upshot is that the general solution relates <math>x</math> to <math>t</math> and has two parameters <math>C_1,C_2</math>, as we might expect from the degree of the equation.
+===Case where the function on the right is multiplicatively separable===
+Consider a situation of the form:
+<math>\frac{d^2x}{dt^2} = g(x)h\left(\frac{dx}{dt}\right)</math>
+We do the same substitution <math>v = dx/dt</math> and obtain:
+<math>\frac{v \, dv}{ \, dx} = g(x)h(v)</math>
+This is a [[separable differential equation]] and we can rearrange it to obtain:
+<math>\int \frac{v \, dv}{h(v)} = \int g(x) \, dx</math>
+We now perform the integration both sides. Suppose <math>H(v) = \int \frac{v \, dv}{h(v)}</math> and <math>G(x) =\int g(x) \, dx</math>. We get:
+<math>H(v) = G(x) + C_1</math>
+If <math>H</math> can locally be inverted, we can write <math>v</math> as an explicit function of <math>x</math>
+We now plug this into the original differential equation and get:
+<math>H(dx/dt) = G(x) + C_1</math>
+This is now a [[first-order differential equation]]. If <math>H</math> can locally be inverted, we can write <math>dx/dt</math> as an explicit function of <math>x</math> (locally) and then solve the resultant [[separable differential equation]]. Otherwise, there may be some other method available.