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.

Example

Separable example

Consider the following differential equation:

$\frac{d^{2}x}{d{t}^2}=2x\frac{dx}{dt}$

We describe how to solve this using the technique discussed here. First, set $v=dx/dt$. We get:

$v\frac{dv}{dx}=2xv$

Thus, we get:

$v=0\text{ or }\frac{dv}{dx}=2x$

The case $v=0$ solves to gives $x=C_0,C_0\in \mathbb{R}$

The case:

$\frac{dv}{dx}=2x$

solves to give:

$v=x^2+C_1$

Now, we plug back the original substitution to get:

$\frac{dx}{dt}=x^2+C_1$

We now make cases based on the sign of $C_1$.

Case $C_1=0$

In this case, the differential equation becomes:

$\frac{dx}{dt}=x^2$

This has a stationary solution $x=0$ (which, incidentally, is already included in the prior solution family of all constant solutions) and the general solution can be obtained as:

$\int \frac{dx}{x^2}=\int dt$

This gives:

$\frac{-1}{x}=t+C_2$

This solves to give:

$x=\frac{-1}{t+C_2}$

To verify that this is a solution, we compute, using the chain rule for differentiation and differentiation rule for power functions:

$\frac{dx}{dt}=\frac{1}{(t+C_2{)}^2}$

and:

$\frac{d^{2}x}{d{t}^2}=\frac{-2}{(t+C_2{)}^3}$

We can now plug these into the original differential equation and verify.

Case $C_1>0$

Let $k=\sqrt{C_1}$. The differential equation becomes:

$\frac{dx}{dt}=x^2+k^2$

There are no stationary solutions. The general solution is given by:

$\int \frac{dx}{x^2+k^2}=\int dt$

This solves to give:

$\frac{1}{k}\arctan (\frac{x}{k})=t+C_2$

Rearranging, we get:

$x=k\tan (k(t+C_2)),k>0,C_2\in \mathbb{R}$

Note that a priori, applying tan to both sides causes us to lose the information that

$k(t+C_2)$

is in the range of

$\arctan$

. However, this constraint is "artificial" anyway and can be got rid of by readjusting the value of

$C_2$

. Note that arc tangent constraints are artificial, but arc sine constraints are genuine, because the sine function repeats itself within a period but the tangent function does not.

To verify that this is a solution, we can compute:

$\frac{dx}{dt}=k^2{\sec }^2(k(t+C_2))$

$\frac{d^{2}x}{d{t}^2}=2k^3{\sec }^2(k(t+C_2))\tan (k(t+C_2))$

We can now plug in and verify the original differential equation.

Case $C_1<0$

Let $l=\sqrt{-C_1}$. The differential equation becomes:

$\frac{dx}{dt}=x^2-l^2$

There are stationary solutions corresponding to $x=l$ and $x=-l$. However, all stationary solutions are already included in another solution family, so we can ignore these. We continue solving:

$\int \frac{dx}{x^2-l^2}=\int dt$

We get:

$\frac{1}{2l}\ln \left|\frac{x-l}{x+l}\right|=t+C_2$

This rearranges to gives:

$\ln \left|\frac{x-l}{x+l}\right|=2l(t+C_2)$

Exponentiate both sides to get:

$\frac{x-l}{x+l}=\exp (2l(t+C_2))\text{ or }\frac{x-l}{x+l}=-\exp (2l(t+C_2))$

The solutions are respectively:

$x=l(\frac{1+\exp (2l(t+C_2))}{1-\exp (2l(t+C_2))})\text{ or }x=l(\frac{1-\exp (2l(t+C_2))}{1+\exp (2l(t+C_2))})$

All cases

The following are the solution families:

• $x=C_0,C_0\in \mathbb{R}$ (this is a singular solution family)
• $x=\frac{-1}{t+C_2},C_2\in \mathbb{R}$
• $x=k\tan (k(t+C_2)),k>0,C_2\in \mathbb{R}$
• $x=l(\frac{1+\exp (2l(t+C_2))}{1-\exp (2l(t+C_2))}),l>0,C_2\in \mathbb{R}$
• $x=l(\frac{1-\exp (2l(t+C_2))}{1+\exp (2l(t+C_2))}),l>0,C_2\in \mathbb{R}$