Skip to content

Home Learn Practice About

Sign in Create account

Math Infinitum

Mapping the lesson.

Loading the workspace…

Mathematics, made tangible.

Explore

Home
Learn
Practice
About

Elsewhere

YouTube
Instagram
TikTok
Ko-fi
Telegram
Twitter

© 2026 Math InfinitumPrivacy Policy Development / Legacy prototype

First-order linear and exact equations · Differential Equations | Math Infinitum

Learn
Differential Equations
First-order linear and exact equations

Differential Equations · Chapter 03

First-order linear and exact equations

Two shapes of first-order equation that a method always solves. A linear equation becomes the derivative of a product after one well-chosen multiplication; an exact one is the derivative of a function along a curve, and its solutions are that function's level curves.

Two shapes of first-order equation that a method always solves. A linear equation becomes the derivative of a product after one well-chosen multiplication; an exact one is the derivative of a function along a curve, and its solutions are that function's level curves.

y^{'} + p (t) y = q (t)

μ = e^{\int p (t) d t}

M d x + N d y = 0

M_{y} = N_{x}

d Φ = 0

Φ (x, y) = C

v = y^{1 - n}

(μ y)^{'} = μ q

Lessons

Build the foundation

01First-order linear equationsA first-order equation is linear when the unknown and its derivative appear only to the first power, each multiplied by a function of the input. That restriction sounds technical, and it is the most useful thing to know about an equation: every first-order linear equation can be solved, by one method, on any interval where its coefficients are continuous. The first step is always the same — divide by the coefficient of the derivative and read off the two functions that are left.Read lesson →
02The integrating factorMultiply a linear equation by the right function of the input and its left-hand side becomes the derivative of a single product. After that one integration finishes the job. The right function is the exponential of the integral of the coefficient of the unknown, and it is called the integrating factor; this lesson is about using it, and the next is about why it works.Read lesson →
03Why the integrating factor worksThe integrating factor can look like a trick handed down from nowhere. It is not. Ask which multiplier would turn the left-hand side into the derivative of the multiplier times the unknown. The product rule answers: the multiplier has to solve a linear equation of its own, and its solutions are the multiples of the exponential of the integral of the coefficient. The method is the product rule run backwards, and running it backwards also proves that nothing has been lost.Read lesson →
04Initial value problemsA linear equation has a whole family of solutions, one for each value of the constant, and an initial condition picks one. The order of work matters: find the general solution first, then use the condition to fix the constant — never fix a constant in the middle of a derivation. Afterwards check two things: that the answer satisfies both the equation and the condition, and on which interval it is valid.

Read lesson →

05Exact differential equationsSome first-order equations are what you get by differentiating a relation. If F(x, y) = C describes a curve, differentiating along it gives an equation whose solutions are exactly the level curves of F: the rate of change of F in x, plus its rate of change in y times the slope, is zero. An exact equation is one that can be recognised as coming from such an F, and solving it means recovering F.Read lesson →

06Testing exactnessFinding a potential to prove an equation is exact would be going round in a circle. There is a test that needs two derivatives. If M and N are the rates of change of a potential F in x and in y, then the rate of M in y and the rate of N in x are the two mixed derivatives of F, taken in opposite orders, and for a smooth function those agree. So an exact equation must pass that test. On a rectangle, or any region without holes, the converse is true as well, and the test is decisive.Read lesson →

07Finding a potential functionOnce the test passes, the potential is recovered in two integrations and one comparison. Integrate M in x, holding y fixed: that gives F up to a function of y alone, because anything that depends only on y vanishes when differentiated in x. Then differentiate the result in y and match it with N, which pins down the missing function. The solutions are the level curves of F.Read lesson →

08Bernoulli and useful substitutionsSome nonlinear equations are linear equations in disguise, and the right change of variable takes the disguise off. The most useful case is Bernoulli's: a linear equation with a power of the unknown on its right-hand side, which a power of the unknown taken as the new variable turns into a linear equation that the integrating factor solves. The general lesson matters more than the formula — look for the substitution that makes the equation one you already know how to solve.Read lesson →

09Chapter practiceBring the chapter together: put a linear equation in standard form and solve it with an integrating factor, test an equation for exactness and recover its potential, and reduce an equation to one of these by a substitution. No worked example sits above the answers.Read lesson →