5.2 First-order linear ODEs

We begin with the simplest non-trivial differential equation in physics. It has one derivative, one unknown function, and one parameter — and yet it is the model for an astonishing range of phenomena. By the end of this lesson you’ll know how to solve it, why the solution is an exponential, and how the single trick that handles it actually generalises to every linear ODE we’ll meet in this chapter.

The general form

A first-order linear ODE with constant coefficients looks like

\dot x \;+\; \alpha\, x \;=\; f(t),

where $\alpha$ is a constant and $f(t)$ is a given forcing function — some external signal or driving that pumps into the system. If $f(t) = 0$ we call the equation homogeneous, and that is where we begin. Setting $f = 0$ and $\alpha > 0$ gives

\dot x \;=\; -\alpha\, x.

In words: the rate of change of $x$ is proportional to $x$ itself, with a negative sign. If $x$ is positive, it shrinks; if negative, it grows back toward zero. Either way, the larger $|x|$ is, the faster it heads home.

decay rate α = 0.50 initial value x₀ = 1.00

t = 0.00 · x = x₀·e^−αt = 1.000

Slide $\alpha$ and watch the steepness change. The solution — which we’ll derive in a moment, but might as well show now — is

x(t) \;=\; x_0\, e^{-\alpha t},

with $x_0 = x(0)$ the initial value. The constant

\tau \;\equiv\; 1/\alpha

is the time constant: the time it takes the curve to fall by a factor of $1/e \approx 0.37$ . The dashed lines in the plot mark $\tau$ and $x_0/e$ .

Where this equation shows up

Exponential decay is everywhere because it describes the most generic possible loss process — a rate proportional to what is left. A short, incomplete tour:

RC circuit discharge. A capacitor of capacitance $C$ discharging through a resistor of resistance $R$ has voltage $V(t)$ satisfying $\dot V + V/(RC) = 0$ . The time constant is $\tau = RC$ .
Newton’s law of cooling. A hot body in a cool room loses heat at a rate proportional to the temperature difference: $\dot T = -k\, (T - T_\infty)$ . The substitution $x = T - T_\infty$ converts this to $\dot x = -k x$ — exactly our equation.
Radioactive decay. A sample of $N$ unstable atoms decays at rate $\lambda N$ : $\dot N = -\lambda N$ . The half-life $T_{1/2} = (\ln 2)/\lambda$ is a rescaled time constant.
First-order chemical reactions. The concentration $c$ of a reactant in a unimolecular reaction obeys $\dot c = -k c$ .
Drug clearance. Plasma concentration of a drug eliminated by first-order kinetics follows $\dot c = -k c$ ; the half-life is the pharmacologist’s analogue of the time constant.
Membrane voltage at rest. A passive neuronal membrane disturbed from rest relaxes back according to $\dot V = -(V - V_\text{rest})/\tau_m$ , with $\tau_m$ the membrane time constant.

Each of those systems has a different physical meaning for $\alpha$ — a leak rate, a cooling rate, a half-life — but the mathematics is identical. Recognising “this is exponential decay” instantly tells you the time constant, the asymptotic value, and the qualitative behaviour. That recognition pays off again and again across the rest of the bookshelf.

The trick that solves every linear ODE

How did we know to try $e^{-\alpha t}$ as the solution? We didn’t, exactly — we guessed. Substitute the ansatz $x(t) = e^{\lambda t}$ for some unknown number $\lambda$ into the equation:

\underbrace{\lambda\, e^{\lambda t}}_{\dot x} \;=\; \underbrace{-\alpha\, e^{\lambda t}}_{-\alpha x}.

For this to hold at all $t$ , the exponential factors on both sides cancel, leaving

\lambda \;=\; -\alpha.

So $x = e^{-\alpha t}$ works. By linearity, any constant multiple of it works too — and $x_0 e^{-\alpha t}$ is the unique solution that matches the initial condition $x(0) = x_0$ .

That guess-an-exponential move is the characteristic-equation method, and it works for every linear ODE with constant coefficients. The recipe is the same four steps every time:

Write the ODE.
Substitute $x = e^{\lambda t}$ .
The exponentials cancel and you are left with an algebraic equation in $\lambda$ — the characteristic equation.
Solve for $\lambda$ . Each root $\lambda_k$ gives one solution $e^{\lambda_k t}$ , and the general solution is the linear combination.

That is it. We’ve just demoted “solving an ODE” to “solving a polynomial,” which is to say to high-school algebra.

If you’ve never seen the recipe spelled out at full granularity, the collapsible block below walks it through on a concrete numerical example, one elementary step at a time, with the why attached to every move.

▶ Worked example: every step, no shortcuts

The problem. Solve

\dot x \;+\; 3\, x \;=\; 0, \qquad x(0) \;=\; 5.

Find the function $x(t)$ that satisfies the equation for all $t \geq 0$ and also matches the initial value $x(0) = 5$ .

Step 1 — Write down the ansatz. “Ansatz” is German for educated guess. Based on the recipe, we propose that the answer has the form

x(t) \;=\; e^{\lambda t},

where $\lambda$ is an unknown constant we are about to solve for. The whole point of the trick is to defer committing to a specific value of $\lambda$ and see what the equation forces it to be.

Step 2 — Differentiate the ansatz. The chain rule gives

\dot x(t) \;=\; \frac{d}{dt}\, e^{\lambda t} \;=\; \lambda\, e^{\lambda t}.

The factor of $\lambda$ comes down out of the exponent because $\lambda t$ has derivative $\lambda$ with respect to $t$ . This is the reason exponentials are the magic ansatz for linear constant-coefficient ODEs: differentiating an exponential gives you back the same exponential, multiplied by a constant. The structure of the equation is preserved under differentiation.

Step 3 — Substitute into the ODE. Plug both $x$ and $\dot x$ into the equation $\dot x + 3 x = 0$ :

\underbrace{\lambda\, e^{\lambda t}}_{\dot x} \;+\; \underbrace{3 \cdot e^{\lambda t}}_{3 x} \;=\; 0.

Step 4 — Factor out the exponential. Both terms on the left share the factor $e^{\lambda t}$ . Pull it out:

e^{\lambda t}\, (\lambda + 3) \;=\; 0.

Step 5 — Cancel the exponential. For the product on the left to be zero, either $e^{\lambda t} = 0$ or $\lambda + 3 = 0$ . But $e^{\lambda t}$ is never zero — the exponential function is strictly positive for any real exponent, and even for complex exponents it has magnitude $e^{\mathrm{Re}\,\lambda \cdot t} \neq 0$ . So that factor cannot be what is making the product vanish. The only way the equation can hold is

\lambda + 3 \;=\; 0.

This is the characteristic equation for our ODE. Notice what just happened: we went from an equation about a function (the ODE) to an equation about a single number ( $\lambda$ ). The function-level structure has been absorbed into the algebra.

Step 6 — Solve for $\lambda$ . Trivially:

\lambda \;=\; -3.

Step 7 — Write the candidate solution. Plug $\lambda = -3$ back into the ansatz:

x(t) \;=\; e^{-3 t}.

This is one function that satisfies the ODE. Check by direct substitution: $\dot x = -3 e^{-3 t}$ , and $\dot x + 3 x = -3 e^{-3 t} + 3 e^{-3 t} = 0$ . ✓

Step 8 — Promote to the general solution using linearity. The ODE is linear, so for any constant $C$ the function $C\, e^{-3 t}$ also satisfies it. Verify: $\dot x = -3 C e^{-3 t}$ and $\dot x + 3 x = -3 C e^{-3 t} + 3 \cdot C e^{-3 t} = 0$ . ✓ This is the general solution — a one-parameter family of curves, with $C$ free:

x(t) \;=\; C\, e^{-3 t}.

It’s a one-parameter family because the ODE is first-order and “free constants $=$ order” (from 5.1). For a second-order ODE we’d get a two-parameter family.

Step 9 — Apply the initial condition. We were given $x(0) = 5$ . Evaluate the general solution at $t = 0$ :

x(0) \;=\; C\, e^{-3 \cdot 0} \;=\; C\, e^{0} \;=\; C \cdot 1 \;=\; C.

So $C = 5$ .

Step 10 — Write the final answer.

\boxed{\;x(t) \;=\; 5\, e^{-3 t}\;}

Step 11 — Sanity-check the answer. Three quick checks:

Initial condition. $x(0) = 5 e^0 = 5$ . ✓
ODE itself. $\dot x = -15 e^{-3t}$ , and $\dot x + 3 x = -15 e^{-3 t} + 3 \cdot 5 e^{-3 t} = -15 e^{-3 t} + 15 e^{-3 t} = 0$ . ✓
Qualitative shape. The coefficient in $\dot x + 3 x = 0$ is positive, so $\alpha = 3$ and the system should decay with time constant $\tau = 1/\alpha = 1/3$ . At $t = 1/3$ the answer should be $5 / e \approx 1.84$ . Check: $5 e^{-3 \cdot 1/3} = 5 e^{-1} \approx 1.84$ . ✓

That is the entire recipe. Eleven steps and we never had to know a single thing about ODEs beyond the shape of the ansatz. The recipe scales without modification to second- and higher-order linear ODEs — only steps 5–6 get more interesting, because the characteristic equation becomes a quadratic or higher-degree polynomial with more (and possibly complex) roots to combine. We’ll see that play out in 5.3.

The carousel below walks the same recipe across seven increasingly rich examples — including the second-order cases we’ll work through properly in 5.3.

1. the ODE ẋ + α x = 0 (α = 1)

2. substitute x = e^λt λ + α = 0

3. solve for λ λ = −α = −1

4. solution x(t) = e^(−α t)

One real negative root → pure exponential decay.

Click through. Watch how the qualitative shape of the answer falls directly out of where the root $\lambda$ lives in the complex plane. (For higher-degree characteristic polynomials, finding $\lambda$ becomes a numerical root-finding problem — see Foundations 10.6.)

One negative real $\lambda$ → exponential decay (this lesson).
One positive real $\lambda$ → exponential growth (unstable; e.g. unbounded chain reactions).
Pure-imaginary $\lambda = \pm i\omega$ → pure sinusoidal oscillation (SHM, the first stage of 5.3).
Complex $\lambda = -\gamma \pm i\omega$ → sinusoid inside a decaying envelope (damped oscillation).
Two distinct real negative roots → sum of two decay rates (overdamped).
Two distinct real roots of opposite signs → one grows, one decays (saddle; unstable except for a special initial line).
A repeated real root → the second solution gains a factor of $t$ : $(A + Bt)\, e^{\lambda t}$ . This is critical damping.

The last case is the one worth flagging here, because it looks like an exception to the recipe. When the characteristic polynomial has a double root, the recipe yields only one exponential — but a second-order ODE must have a two-parameter family of solutions. The missing piece turns out to be $t\, e^{\lambda t}$ , which you can verify by substitution. The “critically damped” regime of the damped oscillator is precisely this case, and we’ll meet it concretely in 5.3.

The central fact about linear constant-coefficient ODEs is in that bullet list: the character of the solution is encoded entirely in the position of the roots in the complex plane. The next lesson takes us from one root (first-order) to two roots (second-order), where the action gets interesting.