Linear First-Order Equations

Suppose a tank holds brine whose concentration is stirred by an inflow. That inflow itself changes with time. How does the amount of salt evolve? This question, and hundreds like it in circuits, mixing, cooling, and population models, leads to a differential equation. In this equation, the unknown function and its derivative appear only to the first power. But the coefficients are allowed to depend on the independent variable. Such equations are not generally separable. The forcing term $g(t)$ and the coefficient $p(t)$ tangle $y$ and $t$ together. This resists factoring into $t$ -stuff times $y$ -stuff. The remarkable fact is that every such equation can be solved in closed form. This works by a single unifying trick. Multiply through by a cleverly chosen function called an integrating factor. It is chosen precisely so that the left side collapses into the derivative of a product. This converts the differential equation into something you can integrate directly. The trick is old — it goes back to Leibniz in the 1690s. It is one of the first places a student sees a general strategy. That strategy is "transform the equation into a shape you already know how to solve." The big picture is that linear first-order equations sit at the base of a tower. The integrating-factor idea reappears, generalized, when treating Exact Equations and general Integrating Factors for nonlinear equations. The explicit solution formula derived here is also a seed. From it, the variation-of-parameters method for higher-order linear equations grows. The entire linear theory is later reorganized using the The Laplace Transform: Definition and Existence. This transform converts differentiation into algebra. This article gives the precise definition of a linear first-order equation. It builds geometric intuition via slope fields. It proves the general solution formula rigorously (existence, uniqueness, and an explicit closed form). It works two asymmetric examples in full. It shows exactly which hypothesis cannot be dropped, and why.

Key takeaways

A linear first-order ODE has the form $y' + p(t)y = g(t)$ , with no powers or products of $y$ and $y'$ .
The integrating factor $\mu(t) = e^{\int p(t)\,dt}$ turns the left side into $(\mu y)'$ .
Multiplying by $\mu(t)$ and integrating gives an explicit formula for every solution.
Continuity of $p$ and $g$ on an interval guarantees a unique solution through any initial point in that interval.
The general solution is (particular solution) + (constant) $\times$ (homogeneous solution) — a preview of linear superposition.
Key idea: multiply by $\mu(t)=e^{\int p\,dt}$ so the left-hand side becomes a single derivative, $(\mu y)'$ , which integrates directly.

Where this lives today

The integrating-factor solution to $y'+p(t)y=g(t)$ is not a historical curiosity; it is a working tool across quantitative science.

Circuits. An RL circuit with resistance $R$ , inductance $L$ , and a time-varying source voltage $E(t)$ obeys $L\,i' + R\,i = E(t)$ , exactly a linear first-order equation with $p(t)=R/L$ , $g(t)=E(t)/L$ ; the integrating-factor formula gives the current $i(t)$ directly, including the transient decay $e^{-Rt/L}$ from the homogeneous part.

Population and pharmacokinetics. Drug concentration in the bloodstream with a time-dependent infusion rate $I(t)$ and constant elimination rate $k$ satisfies $C' + kC = I(t)$ ; clinicians use exactly this formula to predict blood concentration profiles for variable-rate IV infusions.

Newton's law of cooling with variable ambient temperature. If the surrounding temperature $T_a(t)$ oscillates (e.g. daily temperature cycles), the object's temperature obeys $T' + kT = kT_a(t)$ , again solved by the integrating factor, producing phase-shifted, damped oscillatory responses relevant to building thermal modeling.

Numerical analysis and control theory. The variation-of-constants formula proven here is the prototype for the matrix exponential solution $\dot{\mathbf x} = A(t)\mathbf x + \mathbf b(t)$ in linear systems and control theory, where $\mu(t)$ generalizes to a fundamental matrix.

Transform methods. The The Laplace Transform: Definition and Existence later re-derives and vastly generalizes this solution formula algebraically, converting the ODE into $s\hat y(s) - y_0 + p\ast \hat y = \hat g(s)$ -type algebraic relations for constant-coefficient cases, recovering the integrating-factor result as a special case and extending naturally to higher-order and coupled systems.

Numerical ODE solvers. Implicit and exponential integrators used in stiff ODE solvers (e.g. exponential Euler methods) are direct numerical descendants of the exact integrating-factor formula, using $e^{\int p}$ -type factors to stabilize step sizes for stiff linear parts of a system.

Ask the tutor: How does the integrating-factor method change if $p(t)$ and $g(t)$ are replaced by matrices and vectors in a linear system $\mathbf{x}' = A(t)\mathbf{x} + \mathbf{b}(t)$ ?

\frac{dy}{dt} + p(t)\,y = g(t),

Intuition

Picture the equation $y' + p(t) y = g(t)$ as a rule assigning a slope to every point $(t,y)$ in the plane: the slope at $(t,y)$ is $g(t) - p(t) y$ . A solution curve is a curve that, at every one of its points, has exactly the slope prescribed there — it "follows the arrows" of this slope field. Because the slope depends on $y$ linearly, the slope field has a rigid, orderly structure: along any vertical line $t = t_0$ , the slope is an affine (linear) function of $y$ , so the arrows tilt in a single consistent way as $y$ increases, never folding back on themselves. This rigidity is exactly why a closed-form solution exists for every choice of continuous $p$ and $g$ , unlike general nonlinear equations. The integrating factor $\mu(t) = e^{\int p(t)\,dt}$ can be understood as a reweighting of the unknown. Instead of tracking $y(t)$ directly, track the auxiliary quantity $\mu(t) y(t)$ . The differential equation, written in terms of this new quantity, becomes trivial: its derivative is just $\mu(t) g(t)$ , a known function of $t$ alone. Antidifferentiating recovers $\mu y$ , and dividing by $\mu$ recovers $y$ . The whole solution process is: multiply, recognize a product rule in reverse, integrate, divide.

Figure

Slope field for

y' + y = t

(so

p(t)=1

g(t)=t

), with the solution through

(0,1)

The theorem

p(t)

p

g(t) \equiv 0

Worked examples

Example 1 (variable coefficient, nonzero forcing, nonzero initial time). Solve

t y' - 2y = t^5 \sin(3t), \qquad y(\pi) = 0, \qquad t > 0.

Step 1 — standardize. Divide by $t$ (valid since $t>0$ ):

y' - \frac{2}{t} y = t^4 \sin(3t).

So $p(t) = -2/t$ , $g(t) = t^4\sin(3t)$ , both continuous on $I=(0,\infty)$ , which contains $t_0=\pi$ . Step 2 — integrating factor.

\int p(t)\,dt = \int -\frac{2}{t}\,dt = -2\ln t \quad (t>0),

so take

\mu(t) = e^{-2\ln t} = t^{-2}.

Check: $\mu' = -2t^{-3} = p(t)\mu(t) = (-2/t)(t^{-2})$ . ✓. Step 3 — multiply through.

t^{-2} y' - 2t^{-3} y = t^{-2}\cdot t^4 \sin(3t) = t^2 \sin(3t),

and the left side is $\dfrac{d}{dt}\left[t^{-2} y\right]$ by construction. Step 4 — integrate. We need $\int t^2 \sin(3t)\,dt$ . Integrate by parts twice. First: let $u=t^2$ , $dv = \sin(3t)\,dt$ , so $du = 2t\,dt$ , $v = -\tfrac13\cos(3t)$ :

\int t^2\sin(3t)\,dt = -\frac{t^2}{3}\cos(3t) + \frac{2}{3}\int t\cos(3t)\,dt.

Second, on $\int t\cos(3t)\,dt$ : let $u=t$ , $dv=\cos(3t)dt$ , $du=dt$ , $v=\tfrac13\sin(3t)$ :

\int t\cos(3t)\,dt = \frac{t}{3}\sin(3t) - \frac13\int \sin(3t)\,dt = \frac{t}{3}\sin(3t) + \frac{1}{9}\cos(3t).

Combine:

\int t^2\sin(3t)\,dt = -\frac{t^2}{3}\cos(3t) + \frac{2}{3}\left[\frac{t}{3}\sin(3t) + \frac19\cos(3t)\right] + C = -\frac{t^2}{3}\cos(3t) + \frac{2t}{9}\sin(3t) + \frac{2}{27}\cos(3t) + C.

t^{-2} y = -\frac{t^2}{3}\cos(3t) + \frac{2t}{9}\sin(3t) + \frac{2}{27}\cos(3t) + C.

Step 5 — solve for $y$ .

y(t) = -\frac{t^4}{3}\cos(3t) + \frac{2t^3}{9}\sin(3t) + \frac{2t^2}{27}\cos(3t) + C t^2.

Step 6 — impose $y(\pi)=0$ . Since $3\pi$ has $\cos(3\pi) = \cos(\pi) = -1$ (as $3\pi = \pi + 2\pi$ , cosine period $2\pi$ ) and $\sin(3\pi)=0$ :

0 = -\frac{\pi^4}{3}(-1) + \frac{2\pi^3}{9}(0) + \frac{2\pi^2}{27}(-1) + C\pi^2 = \frac{\pi^4}{3} - \frac{2\pi^2}{27} + C\pi^2.

Solve for $C$ :

C\pi^2 = \frac{2\pi^2}{27} - \frac{\pi^4}{3} \quad\Longrightarrow\quad C = \frac{2}{27} - \frac{\pi^2}{3}.

Final answer.

y(t) = -\frac{t^4}{3}\cos(3t) + \frac{2t^3}{9}\sin(3t) + \frac{2t^2}{27}\cos(3t) + \left(\frac{2}{27} - \frac{\pi^2}{3}\right) t^2, \qquad t>0.

This solution is valid and unique on the entire interval $(0,\infty)$ by the theorem, since $p,g$ are continuous there.

Example 2 (mixing problem — a genuine modeling application). A 200-liter tank initially contains 40 kg of dissolved salt in 100 liters of brine. Starting at $t=0$ , brine containing $0.5\,\text{kg/L}$ of salt flows in at $5\,\text{L/min}$ , while the well-mixed solution drains out at only $3\,\text{L/min}$ (so the tank fills up over time). Find the amount of salt $Q(t)$ (in kg) for $0 \le t < 50$ (the tank overflows at $t=50$ , since it gains 2 L/min net and starts at 100 L, filling the remaining 100 L in 50 min). Setting up the equation. Volume in the tank at time $t$ : inflow rate $5$ L/min, outflow rate $3$ L/min, so net rate $+2$ L/min, giving

V(t) = 100 + 2t \quad \text{liters}.

Salt balance: rate in minus rate out.

\text{rate in} = 5 \cdot 0.5 = 2.5 \ \text{kg/min (constant)},

\text{rate out} = 3 \cdot \frac{Q(t)}{V(t)} = \frac{3Q}{100+2t} \ \text{kg/min},

since the outflow concentration equals the tank's current concentration $Q/V$ (well-mixed assumption). So

Q'(t) = 2.5 - \frac{3Q}{100+2t}, \qquad Q(0) = 40.

Rearrange to standard linear form:

Q' + \frac{3}{100+2t} Q = 2.5.

Here $p(t) = \dfrac{3}{100+2t}$ and $g(t) = 2.5$ , both continuous on $[0,50)$ (the denominator $100+2t \ge 100 > 0$ throughout), so the theorem applies with $I = [0,50)$ extended to an open interval containing it, and $t_0 = 0$ . Integrating factor.

\int p(t)\,dt = \int \frac{3}{100+2t}\,dt = \frac{3}{2}\ln(100+2t) \quad (\text{since } 100+2t>0),

\mu(t) = e^{\frac32 \ln(100+2t)} = (100+2t)^{3/2}.

Multiply through and integrate.

\frac{d}{dt}\left[(100+2t)^{3/2} Q\right] = 2.5\,(100+2t)^{3/2}.

Integrate the right side using the substitution $u = 100+2t$ , $du = 2\,dt$ :

\int 2.5(100+2t)^{3/2}\,dt = 2.5 \cdot \frac{1}{2}\int u^{3/2}\,du = 1.25 \cdot \frac{2}{5} u^{5/2} + C = \frac12 u^{5/2} + C = \frac12 (100+2t)^{5/2} + C.

(100+2t)^{3/2} Q(t) = \frac12 (100+2t)^{5/2} + C,

Q(t) = \frac12(100+2t) + C(100+2t)^{-3/2}.

Apply the initial condition $Q(0)=40$ .

40 = \frac12(100) + C(100)^{-3/2} = 50 + \frac{C}{1000},

C = 1000(40-50) = -10000.

Final answer.

\boxed{Q(t) = \frac12(100+2t) - \frac{10000}{(100+2t)^{3/2}} = (50+t) - \frac{10000}{(100+2t)^{3/2}}}, \qquad 0 \le t < 50.

Sanity checks: at $t=0$ , $Q(0) = 50 - 10000/1000 = 50-10 = 40$ ✓, matching the given initial condition. As $t\to\infty$ (formally, ignoring the overflow constraint), $Q(t) \approx 50+t$ while the equilibrium concentration $2.5/3 \cdot V(t)$ -type behavior is approached; the correction term $10000(100+2t)^{-3/2} \to 0$ , so the solution is asymptotic to $50+t$ , i.e. concentration $Q/V \to (50+t)/(100+2t) \to 1/2 = 0.5$ kg/L, matching the inflow concentration — exactly the physically expected long-run behavior since outflow becomes a vanishing fraction of the ever-larger tank relative to a constant inflow rate of pure input concentration. This qualitative check confirms the algebra.

p

Scope, edge cases & common traps

Situation	What happens
$p, g$ continuous on all of $\mathbb{R}$	Solution exists and is unique for all $t \in \mathbb{R}$ ; no finite escape time (linear equations with everywhere-continuous coefficients never blow up in finite time).
$p$ has a discontinuity (e.g. jump, pole) at $t=a$	Solve separately on each maximal interval of continuity not containing $a$ ; matching solutions across $a$ requires extra physical/continuity conditions, and uniqueness across $a$ can fail (see counterexample).
$g$ piecewise continuous (e.g. a switched forcing)	Solve on each piece, then match values at the switch point to fix constants — standard in circuits with a switch closing at $t=t_1$ .
Forgetting $C$	The homogeneous solution $Ce^{-\int p\,dt}$ must be retained; omitting it silently discards the entire one-parameter family of solutions and violates the initial condition in general.
Sign error in $\int p\,dt$	$\mu(t) = e^{\int p\,dt}$ , not $e^{-\int p\,dt}$ ; the sign is easy to flip and instantly breaks $(\ast)$ in the proof.
Dividing by $t$ (or another vanishing factor) to reach standard form	Valid only away from the zero; changes the natural domain of the problem — always re-state the interval $I$ after this step.

The mathematician behind it

Origins · 1 minGottfried Wilhelm Leibniz and how this was discovered

Gottfried Wilhelm Leibniz

1646–1716 · Co-inventor of calculus, Hanover

Leibniz developed the calculus notation and many of its early techniques, including the method for solving linear first-order differential equations by an integrating factor, in correspondence and papers during the 1690s. Working alongside Jacob and Johann Bernoulli on a flood of new differential equations arising from mechanics and geometry, Leibniz recognized that multiplying an equation by a well-chosen factor could reduce it to something directly integrable — the germinal idea behind essentially all of the elementary solution techniques that followed, including separation of variables and, much later, the systematic linear theory built on matrix exponentials and transforms.

History

- 1675–1676: Leibniz develops the fundamental notation and rules of calculus, including

\int

and

d

- 1691: Leibniz publishes a general method for solving what we now call linear first-order ODEs, using an integrating factor.

- 1694–1697: The Bernoulli brothers extend these methods, including to the nonlinear Bernoulli equation, reducible to the linear case by substitution.

- 1743: Leonhard Euler formalizes and generalizes integrating-factor and exact-equation techniques in his systematic treatment of differential equations.

- 1800s: The theory is subsumed into the broader existence-uniqueness framework (Picard–Lindelöf) and, later, into linear systems theory and transform methods.

Why is the equation called "linear" if

g(t)

can be a wild nonlinear function of

t

"Linear" refers strictly to how the unknown $y$ and its derivative $y'$ enter the equation, not to how the independent variable $t$ enters. The left side $y' + p(t)y$ is a linear (degree-one, no products) combination of $y$ and $y'$ ; the coefficients $p(t)$ and the forcing term $g(t)$ can be arbitrarily nonlinear or even non-elementary functions of $t$ alone, and this does not affect linearity in $y$ . This distinction matters because linearity in $y$ is exactly what makes superposition, the integrating-factor trick, and closed-form solvability possible — none of that requires $g$ or $p$ to be simple functions of $t$ .

What if I cannot compute

\int p(t)\,dt

\int \mu(t) g(t)\,dt

in closed form?

The theorem guarantees existence and uniqueness regardless of whether the integrals are elementary; the formula $\phi(t) = \mu(t)^{-1}[y_0\mu(t_0) + \int_{t_0}^t \mu(s)g(s)\,ds]$ is a rigorous, exact representation of the solution even when the integrals must be left as definite integrals or evaluated numerically. This is common with equations like $y' + ty = e^{t^2}\sin t$ : the solution is entirely legitimate and well-defined, just not expressible via elementary functions, and one reports it either as a definite integral or approximates it numerically.

How does this relate to separable equations?

Separable Equations are those where $y'=f(t)h(y)$ can be split as $dy/h(y) = f(t)\,dt$ . A linear equation $y'+p(t)y=g(t)$ is separable only in the special homogeneous case $g\equiv0$ , since then $y'/y = -p(t)$ separates cleanly. Once $g\not\equiv0$ , the equation is generically not separable — you cannot isolate all $y$ -dependence on one side — which is exactly why the integrating-factor method, a genuinely different technique, is needed for the nonhomogeneous case.

Does the integrating-factor method generalize beyond first order?

Yes, in spirit. For higher-order linear equations, the analogous tool is variation of parameters, which uses a fundamental set of homogeneous solutions in place of the single factor $\mu(t)$ . For systems $\mathbf x' = A(t)\mathbf x + \mathbf b(t)$ , the scalar $\mu(t)=e^{\int p\,dt}$ generalizes to a matrix-valued fundamental solution $\Phi(t)$ , and the analogous formula $\mathbf x(t) = \Phi(t)\left[\mathbf x_0 + \int \Phi(s)^{-1}\mathbf b(s)\,ds\right]$ holds under suitable hypotheses. The scalar case here is the base of that entire hierarchy.

Can

p(t)

g(t)

be complex-valued, or can

t

range over an unbounded interval like

[0,\infty)

Yes to both. The proof only used continuity, the Fundamental Theorem of Calculus, and basic algebra, none of which require real-valuedness or a bounded interval; it goes through verbatim for complex-valued $p,g$ (interpreting the exponential via its complex power series) and for $I=(a,\infty)$ , $(-\infty,b)$ , or all of $\mathbb R$ . This generality is exactly what is exploited later when linear equations with complex or oscillatory forcing (e.g. $g(t)=e^{i\omega t}$ ) are solved via the same formula.

Is the integrating factor unique?

No — any positive constant multiple of $\mu(t)=e^{\int p\,dt}$ is also a valid integrating factor, since constant multiples don't affect whether $(\ast)$ holds (both sides simply get scaled by the same constant). What is unique, given the equation, is the family $\{Ce^{\int p\,dt} : C\neq 0\}$ ; conventionally one fixes $C=1$ (or, as in the theorem, picks the antiderivative vanishing at $t_0$ ) for simplicity, but the final solution $y(t)$ that results is identical regardless of that choice.

What goes wrong if I try to apply this method to a genuinely nonlinear equation like

y' + p(t)y^2 = g(t)

The key identity $(\ast)$ in the proof crucially used that $y$ and $y'$ appear to the first power, so that $\mu(t)[y'+p(t)y]$ literally equals $\frac{d}{dt}[\mu(t)y]$ by the product rule. If $y^2$ replaces $y$ , no single multiplier $\mu(t)$ turns $\mu(t)[y' + p(t)y^2]$ into an exact derivative of a simple product, because the product rule applied to $\mu y$ produces a term linear in $y$ , not $y^2$ . Such equations (e.g. the Bernoulli equation $y'+p(t)y = g(t)y^n$ ) require a nonlinear substitution, like $v=y^{1-n}$ , to convert them into a linear equation first — a different, though related, technique building directly on this one.