What Differential Equation Describes Newtonian Motion?

Newtonian mechanics describes motion through differential equations: acceleration is related to the net force acting on a body. The law alone does not give a specific trajectory, however. You must also specify how the force depends on the body’s position, velocity, and time, and give its initial position and velocity.

Newton’s law as a differential equation

For a particle of constant mass $m$ moving in an inertial reference frame, Newton’s second law is

$$m\ddot{\mathbf r}(t)=\mathbf F(\mathbf r(t),\dot{\mathbf r}(t),t),$$

where $\mathbf r(t)$ is the position, $\dot{\mathbf r}(t)$ is the velocity, $\ddot{\mathbf r}(t)$ is the acceleration, and $\mathbf F$ is the net force. The dots mean derivatives with respect to time. Because position is differentiated twice, this is a second-order ordinary differential equation.

In one dimension, the same law is

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

The right-hand side is a force law. For example, a spring obeying Hooke’s law has $F=-kx$, while gravity near Earth’s surface is approximately $F=-mg$ if upward is chosen as the positive direction.

For a system of $N$ particles, write one vector equation for each particle:

$$m_i\ddot{\mathbf r}_i=\mathbf F_i.$$

The force on each particle may depend on the positions and velocities of several particles. Constraints, such as motion along a track, can also restrict the allowed positions and require additional equations or constraint forces.

Why initial conditions are needed

A second-order equation generally needs two initial conditions for each independent coordinate. For one-dimensional motion, these are usually the initial position and velocity:

$$x(0)=x_0,\qquad \dot{x}(0)=v_0.$$

Together, the force law and these initial values select a particular solution from the family of possible trajectories. Under the usual smoothness conditions on the force, that solution is locally unique. Without a force law, Newton’s law says how acceleration relates to force but does not predict what the motion will be.

You can also rewrite the second-order equation as two first-order equations by introducing velocity $v=\dot{x}$:

$$\dot{x}=v,\qquad \dot{v}=\frac{1}{m}F(x,v,t).$$

This form is especially useful in numerical simulations and in the phase-space description of mechanics, where the state is specified by position and velocity.

Worked example: a mass on a spring

Consider a mass $m$ attached to an ideal spring with spring constant $k$, moving without friction along a horizontal line. If $x=0$ is the spring’s equilibrium position, Hooke’s law gives $F=-kx$. Newton’s law therefore becomes

$$m\ddot{x}=-kx,$$

or

$$\ddot{x}+\omega^2x=0,\qquad \omega=\sqrt{\frac{k}{m}}.$$

The general solution is

$$x(t)=A\cos(\omega t)+B\sin(\omega t),$$

where $A$ and $B$ are constants set by the initial conditions. Suppose the mass starts at $x(0)=0.10\,\mathrm{m}$ with velocity $\dot{x}(0)=0$, and let $m=0.50\,\mathrm{kg}$ and $k=8.0\,\mathrm{N/m}$. Then

$$\omega=\sqrt{\frac{8.0}{0.50}}=4.0\,\mathrm{s}^{-1}.$$

From $x(0)=A$, we get $A=0.10\,\mathrm{m}$. Differentiating the solution gives $\dot{x}(t)=-A\omega\sin(\omega t)+B\omega\cos(\omega t)$. Thus $\dot{x}(0)=B\omega=0$, so $B=0$. The motion is

$$x(t)=0.10\cos(4.0t)\,\mathrm{m},$$

with time measured in seconds. The force law produced the differential equation; the initial position and velocity determined which solution describes this particular motion.

Common mistakes

• Leaving out the force law: $m\ddot{x}=F$ is not a complete solvable model until $F$ is specified or otherwise known.
• Providing only a starting position: A second-order equation normally needs both position and velocity at the initial time.
• Using a force with the wrong sign: Signs depend on the chosen coordinate direction. For a spring, the force points toward equilibrium; for gravity, it points downward.
• Applying the law outside its assumptions: The usual equation assumes an inertial frame and constant mass. Accelerating frames require additional inertial forces, and variable-mass systems need a careful momentum balance.
• Assuming every solution has a simple formula: Many force laws do not yield elementary closed-form solutions. Numerical integration can still approximate the motion if the model and initial data are specified.