Linear First-Order Equations

A linear first-order differential equation is the workhorse of applied mathematics. Whenever a quantity changes at a rate that depends linearly on its current value plus some outside forcing — salt dissolving in a stirred tank, current building up in a circuit, a bank balance earning interest while you make deposits — the model that falls out is $y' + p(x)y = q(x)$. What makes this class so beloved is that we have a completely reliable recipe, the integrating-factor method, that solves every equation of this form. No guessing, no luck: multiply by the right function and the left side collapses into a single derivative you can integrate directly.

This page teaches you not just to turn the crank, but to understand why the trick works, where it came from, and how to bend real problems into the shape it needs.

Learning Objectives

• Recognize when a first-order equation is linear and put it in the standard form $y' + p(x)y = q(x)$.
• Derive and apply the integrating factor $\mu(x) = e^{\int p(x)\,dx}$.
• Solve initial-value problems and interpret the transient and steady-state parts of a solution.
• Model and solve mixing (tank) problems and RL/RC electrical circuits.
• Distinguish linear equations from separable, Bernoulli, and exact equations.

Quick Answer

A linear first-order ODE has the standard form $y' + p(x)y = q(x)$, where $p$ and $q$ depend only on $x$ (never on $y$ or powers of $y$). To solve it, compute the integrating factor $\mu(x) = e^{\int p(x)\,dx}$ and multiply the whole equation by it. The left side then becomes exactly $\frac{d}{dx}\big[\mu(x)\,y\big]$, so integrating both sides gives $\mu(x)\,y = \int \mu(x)\,q(x)\,dx + C$. Divide by $\mu(x)$ to get $y$. The arbitrary constant $C$ is pinned down by an initial condition. This single method handles mixing problems, electrical circuits, cooling, and many growth models.

Where It Came From

By the late 1600s, Newton and Leibniz had given the world calculus, and mathematicians immediately began writing differential equations to describe motion, heat, and mechanics. But writing an equation is not solving it, and early solutions were a patchwork of clever, one-off substitutions. The pressing need was a general, reusable method for the equations that kept reappearing in physics.

The Bernoulli family drove much of this. In 1695 Jakob Bernoulli posed what we now call the Bernoulli equation, $y' + p(x)y = q(x)y^n$, and his brother Johann Bernoulli showed that a substitution reduces it to a linear equation — implicitly identifying the linear case $n = 0$ as the fundamental, solvable core. Johann's insight was essentially that multiplying by a well-chosen factor made the equation integrable.

Leonhard Euler, Johann's student, then systematized the idea in the 1730s–1760s into the integrating factor as we teach it today: a function $\mu(x)$ that turns a messy sum into a perfect derivative. Euler's genius was recognizing that the awkward combination $\mu y' + \mu p y$ is just $(\mu y)'$ provided $\mu' = p\mu$ — a condition that itself is a simple separable equation. This transformed a whole family of physically important equations from "solve by inspiration" into "solve by procedure," and it remains one of the first truly general techniques a student of differential equations learns. The motivation was never abstract: these were the equations of pendulums, resisted motion, and later electrical circuits, and industry and physics needed answers, not artistry.

The Standard Form and the Integrating Factor

An equation is linear in $y$ if $y$ and $y'$ appear only to the first power, are not multiplied together, and are not inside any nonlinear function. The standard form is

$y' + p(x)\,y = q(x)$

Always divide through by the coefficient of $y'$ first so that $y'$ stands alone — a common source of errors is forgetting this and using the wrong $p(x)$.

Why an integrating factor exists. We want a multiplier $\mu(x)$ so that

$\mu y' + \mu p\, y = \frac{d}{dx}(\mu y)$

By the product rule, $\frac{d}{dx}(\mu y) = \mu y' + \mu' y$. Matching the two expressions forces $\mu' = \mu p$, whose solution is

$\mu(x) = e^{\int p(x)\,dx}$

(We drop the constant of integration here because any single working $\mu$ suffices.) Once we multiply by this $\mu$, the equation becomes $\frac{d}{dx}(\mu y) = \mu q$, and integrating gives the master formula:

$y = \frac{1}{\mu(x)}\left[\int \mu(x)\,q(x)\,dx + C\right]$

Worked Example 1

Solve $y' + 2y = e^{-x}$ with $y(0) = 3$.

Here $p(x) = 2$ and $q(x) = e^{-x}$. The integrating factor is

$\mu(x) = e^{\int 2\,dx} = e^{2x}$

Multiply through: $e^{2x}y' + 2e^{2x}y = e^{2x}e^{-x} = e^{x}$. The left side is $\frac{d}{dx}(e^{2x}y)$, so

$e^{2x}y = \int e^{x}\,dx = e^{x} + C$

Therefore $y = e^{-x} + Ce^{-2x}$. Apply $y(0) = 3$: $3 = 1 + C$, so $C = 2$. The solution is

$y = e^{-x} + 2e^{-2x}$

Check: $y' = -e^{-x} - 4e^{-2x}$, and $y' + 2y = -e^{-x} - 4e^{-2x} + 2e^{-x} + 4e^{-2x} = e^{-x}$. Correct.

Worked Example 2 (variable coefficient)

Solve $x y' + 3y = \dfrac{\sin x}{x}$ for $x > 0$.

First divide by $x$ to reach standard form: $y' + \frac{3}{x}y = \frac{\sin x}{x^2}$. So $p(x) = 3/x$ and

$\mu(x) = e^{\int 3/x\,dx} = e^{3\ln x} = x^{3}$

Multiply the standard-form equation by $x^3$: $x^3 y' + 3x^2 y = x\sin x$. The left side is $\frac{d}{dx}(x^3 y)$, so

$x^3 y = \int x\sin x\,dx = \sin x - x\cos x + C$

using integration by parts. Hence

$y = \frac{\sin x - x\cos x + C}{x^{3}}$

Transient and Steady State

Many solutions split into two meaningful pieces: a transient part that decays to zero and a steady-state part that persists. In Example 1, both terms decayed, but in forced problems with constant input, one term survives. This structure is central to engineering: it tells you the long-term behavior of a circuit or tank regardless of how it started, and the transient tells you how fast it settles. The constant $C$ (fixed by the initial condition) rides only on the transient part, which is why different starting states converge to the same steady state.

Real-World Applications

Mixing (tank) problems. A tank holds a solution; brine flows in, the well-stirred mixture flows out. If $Q(t)$ is the amount of dissolved substance, then

$\frac{dQ}{dt} = (\text{rate in}) - (\text{rate out})$

The rate out is (concentration in tank) × (outflow rate) $= \frac{Q}{V}\cdot r_{out}$, which is linear in $Q$ — exactly $Q' + \frac{r_{out}}{V}Q = (\text{rate in})$.

Worked Example 3 (mixing)

A 100 L tank starts with pure water. Brine containing 2 g/L of salt flows in at 5 L/min, and the well-mixed solution drains at 5 L/min (so volume stays 100 L). Find the salt amount $Q(t)$.

Rate in $= 2 \text{ g/L} \times 5 \text{ L/min} = 10$ g/min. Rate out $= \frac{Q}{100}\times 5 = \frac{Q}{20}$ g/min. So

$Q' + \frac{1}{20}Q = 10, \qquad Q(0) = 0$

Integrating factor $\mu = e^{t/20}$. Then $\frac{d}{dt}(e^{t/20}Q) = 10e^{t/20}$, so $e^{t/20}Q = 200e^{t/20} + C$, giving $Q = 200 + Ce^{-t/20}$. With $Q(0)=0$, $C = -200$:

$Q(t) = 200\big(1 - e^{-t/20}\big)$

The steady state is 200 g — which makes sense: 100 L at the inflow concentration of 2 g/L is exactly 200 g.

Electrical circuits. For an RL circuit with resistance $R$, inductance $L$, and voltage source $E(t)$, Kirchhoff's law gives $L I' + RI = E(t)$, i.e. $I' + \frac{R}{L}I = \frac{E}{L}$ — linear first-order.

Worked Example 4 (RL circuit)

An RL circuit has $L = 2$ H, $R = 4\ \Omega$, and a constant source $E = 12$ V, with $I(0) = 0$. Find $I(t)$.

Standard form: $I' + 2I = 6$. Integrating factor $\mu = e^{2t}$. Then $\frac{d}{dt}(e^{2t}I) = 6e^{2t}$, so $e^{2t}I = 3e^{2t} + C$, giving $I = 3 + Ce^{-2t}$. With $I(0) = 0$, $C = -3$:

$I(t) = 3\big(1 - e^{-2t}\big)$

The steady-state current is $3$ A $= E/R$ (Ohm's law), and the $e^{-2t}$ transient dies with time constant $L/R = 0.5$ s.

Common Mistakes

1. Forgetting to normalize to standard form. If the equation is $x y' + 3y = \dots$, students often read off $p = 3$ instead of $p = 3/x$. Why wrong: the integrating-factor derivation assumes the coefficient of $y'$ is exactly 1. Correction: always divide by that coefficient first, then read $p(x)$.

2. Adding a constant when computing $\mu$. Writing $\mu = e^{\int p\,dx + C} = C' e^{\int p\,dx}$ and carrying $C'$ around. Why wrong: it clutters the algebra and can hide sign errors; any nonzero constant multiple works, so we choose the simplest. Correction: take the constant of integration to be zero when finding $\mu$. Keep the $+C$ only for the final integration.

3. Dropping the $\frac{1}{\mu}$ at the end. After getting $\mu y = \int \mu q\,dx + C$, forgetting to divide the whole right side (including $C$) by $\mu$. Why wrong: the constant is also divided, so $C/\mu(x)$ is generally a function of $x$, not a constant — this is the transient term. Correction: solve for $y$ by dividing everything by $\mu(x)$.

Comparison and Connections

Linear first-order equations sit among several sibling techniques. Knowing which tool matches which equation saves enormous effort.

Type	Form	Method
Linear	$y' + p(x)y = q(x)$	Integrating factor
Separable	$y' = f(x)g(y)$	Separate variables, integrate
Bernoulli	$y' + p(x)y = q(x)y^n$	Substitute $v = y^{1-n}$, then linear
Exact	$M\,dx + N\,dy = 0$, $M_y = N_x$	Find potential $F(x,y)$

Note that a linear equation with $q(x) = 0$ (called homogeneous) is also separable — the two methods agree there. The Bernoulli equation, historically the parent of this whole story, reduces to the linear case, so mastering the linear method unlocks it too.

Practice Questions

Recall

State the standard form of a linear first-order ODE and write the formula for the integrating factor.

Answer: $y' + p(x)y = q(x)$; integrating factor $\mu(x) = e^{\int p(x)\,dx}$.

Understanding

Explain why multiplying by $\mu(x) = e^{\int p\,dx}$ makes the left side a single derivative.

Guidance: Because $\mu' = p\mu$, the expression $\mu y' + \mu p y = \mu y' + \mu' y$, which is exactly $\frac{d}{dx}(\mu y)$ by the product rule.

Application

Solve $y' - y = e^{2x}$, $y(0) = 1$.

Answer: $\mu = e^{-x}$; $\frac{d}{dx}(e^{-x}y) = e^{x}$, so $e^{-x}y = e^{x} + C$, giving $y = e^{2x} + Ce^{x}$. With $y(0)=1$: $1 = 1 + C \Rightarrow C = 0$. So $y = e^{2x}$.

Analysis

In Worked Example 3, suppose the drain rate were only 4 L/min while inflow stayed 5 L/min. Why is the resulting equation no longer constant-coefficient, and what changes?

Guidance: The volume grows: $V(t) = 100 + t$. Rate out $= \frac{Q}{100+t}\times 4$, so $Q' + \frac{4}{100+t}Q = 10$ — still linear, but now $p(t) = 4/(100+t)$ varies, giving $\mu = (100+t)^4$. The method still works; the coefficient is just no longer constant.

FAQ

Is $y' + y^2 = x$ linear? No. The $y^2$ term makes it nonlinear. Linear means $y$ and $y'$ appear only to the first power and never multiplied together.

Can $p(x)$ and $q(x)$ be any functions? They can be any functions of $x$ alone. The method always formally works, but the integrals $\int p\,dx$ and $\int \mu q\,dx$ may not have closed forms — then you leave the answer as an integral or evaluate numerically.

What if I can't do the final integral? Leave it in integral form: $y = \frac{1}{\mu}\left[\int_{x_0}^{x}\mu(t)q(t)\,dt + C\right]$. This is still a complete, exact solution.

Why do circuits and tanks give the same kind of equation? Both describe a quantity whose rate of change is a constant input minus a term proportional to the current amount. That "proportional loss" is what produces the linear $p(x)y$ term and the exponential decay of the transient.

Does the integrating factor ever equal zero? No. Since $\mu = e^{\int p\,dx}$ is an exponential, it is always strictly positive, so dividing by it at the end is always safe.

Quick Revision

• Standard form: $y' + p(x)y = q(x)$ — divide out the coefficient of $y'$ first.
• Integrating factor: $\mu(x) = e^{\int p(x)\,dx}$ (drop the constant here).
• After multiplying: $\frac{d}{dx}(\mu y) = \mu q$, so $\mu y = \int \mu q\,dx + C$.
• Solution: $y = \frac{1}{\mu}\left[\int \mu q\,dx + C\right]$; divide the $C$ too.
• Mixing: $Q' + \frac{r_{out}}{V}Q = c_{in}r_{in}$. RL circuit: $I' + \frac{R}{L}I = \frac{E}{L}$.
• Steady state = long-term value; transient $\propto e^{-\int p\,dx}$ decays away.

Related Topics

Prerequisites

• Differential Equations overview

Related Topics

• Separable equations (a special case when $q = 0$)
• Bernoulli equations (reduce to linear via substitution)

Next Topics

• Second-order linear differential equations
• Systems of linear differential equations