Second-order first-degree autonomous differential equation

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 and degree one) second-order autonomous differential equation is a differential equation of the form:

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

Solution method and formula

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

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

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

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

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:

$R(x,\frac{dx}{dt})=0$

Particular cases

Case where the function on the right depends only on $x$ and not on $dx/dt$

Consider a situation of the form:

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

We do the same substitution $v=dx/dt$ and obtain:

$\frac{vdv}{dx}=g(x)$

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

$\int vdv=\int g(x)dx$

We thus get:

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

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

$v=\pm \sqrt{2G(x)+C_1}$

where $C_1\in \mathbb{R}$ is a parameter. Each choice of $C_1$ gives a different solution.

Plug this back in and get:

$\frac{dx}{dt}=\pm \sqrt{2G(x)+C_1}$

(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:

$\pm \int \frac{dx}{\sqrt{2G(x)+C_1}}=\int dt$

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:

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

We do the same substitution $v=dx/dt$ and obtain:

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

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

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

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

$H(dx/dt)=G(x)+C_1$

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