A first-order linear equation is any equation in which the unknown and its rate each appear once, to the first power, multiplied by something that may depend on t however it likes. Dividing through by the coefficient of y′ puts it in the standard form y′ + p(t)·y = q(t), and everything the rest of this course does starts from reading p and q off that line. The word linear is about y, never about t.
Core ideas
Three ideas, and none of them is a technique for solving anything yet. They decide which equations the techniques apply to, and what the two functions the techniques are built from are called.
An equation a(t)·y′ + b(t)·y = c(t) is divided, term by term, by a(t). That gives y′ + p(t)·y = q(t) with p = b/a and q = c/a. The division is the only step, and it has to reach the right-hand side as well as the left. It is legitimate wherever a(t) is not zero, which is why t·y′ − 2y = 0 is divided on the stretch where t > 0 or the stretch where t < 0, and never across t = 0.
The equation is linear when y and y′ each appear on their own, to the first power, and are never multiplied together or put inside a function. The coefficients may be any functions of t at all. So y′ = 2y + t²·y is linear, because it is y′ + (−2 − t²)·y = 0, and y′ + y² = t is not, because of the y². A term with no y in it is forcing, and it goes on the right.
When q = 0 the equation is homogeneous: nothing pushes on it from outside, and the only thing acting on y is p. The word does not mean easy or constant. It names the equation whose solutions can be scaled, and the next lessons show that every forced equation carries its homogeneous partner inside it.
The material
The definition, the division, what the word linear rules out and what it allows, and the one property of the homogeneous case the rest of the course uses.
An equation that can be written a(t)·y′ + b(t)·y = c(t) with a(t) not identically zero. Its standard form is y′ + p(t)·y = q(t), where p = b/a and q = c/a.
It is homogeneous when q is zero, and forced (or non-homogeneous) when it is not. The function q is the forcing.
Standard form is a convention with a reason. The formulas on this course — for a steady state, for an integrating factor, for a particular solution — are all written for y′ with coefficient 1, and every one of them goes wrong by exactly the leftover factor if the coefficient is something else. Dividing first is cheap. Dividing afterwards, in the middle of a formula, is where the errors are.
The unknown y and its rate y′ each appear once, alone, to the first power. That excludes y², y·y′, sin(y) and 1/y, and nothing else. The factors multiplying y and y′ are unrestricted: t², 1/t, sin(t), e^t. A term with no y in it is forcing and is free to be anything.
Moving a term across the equals sign changes its sign, and that is the commonest slip in reading p. The equation y′ = 4 − 2y has the unknown on both sides; collecting it on the left gives y′ + 2y = 4, so p = 2 and q = 4, not p = −2. The sign matters later: a positive constant p makes the solutions of the homogeneous equation die away, and a negative one makes them grow.
If y solves y′ + p(t)·y = 0, then so does C·y for every constant C.
The derivative of C·y is C·y′, because a constant factor comes straight out of a derivative.
So (C·y)′ + p·(C·y) = C·(y′ + p·y) = C·0 = 0. Nothing about p was used, only that y and y′ each appear once and to the first power. For y′ = y² the same step fails: y = 1/(1 − t) solves it and 2·y does not.
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Take t·y′ − 2y = 0 and divide by t: y′ − (2/t)·y = 0, so p(t) = −2/t and q = 0. It is homogeneous.
A solution is y = t². Substitute: t·(2t) − 2·t² = 0. By the scaling property C·t² is a solution for every C, and the lab prints exactly that as the homogeneous solution.
The lab at the foot of the page does the division and prints several things this lesson has not derived. Two of them matter now. μ, the integrating factor, is the subject of the next lesson but one; it is e^(3t) for p = 3 and 1/t² for p = −2/t. y_h is the homogeneous solution, and across the three presets it is always C divided by μ. The particular solution, the steady state and the full solution are later lessons' business. Read all of them as outputs for now; “The Integrating Factor” explains the first two.
In y′ = 2y + t²·y the t² sits beside the y. It is a coefficient of y, not a power of y, so the equation is y′ + (−2 − t²)·y = 0 with p(t) = −2 − t².
The lab refuses it, and the reason it gives is the lab's limit and not a verdict on the equation: the lab solves a constant p or one of the form a/t, and says that p is outside them. A refusal from the lab never means the equation is not linear unless it says so in those words.
Interactive
μ, y_h, y_p and the constant, exactly
Method
Collect, divide, read, then name it. The check at the end is the one that catches the equations that are not linear at all.
Move the terms with no y in them to the right. Moving a term changes its sign, so y′ = 4 − 2y becomes y′ + 2y = 4, and the +2 is the p.
Every term, including the right-hand side. If the coefficient is 2, the 6y becomes 3y and the 4t becomes 2t. If it is t, you divide by t and say on which side of zero the equation lives.
p is whatever now multiplies y; q is whatever is left on the right. Write both down as functions of t, with their signs.
If q is zero it is homogeneous. If not it is forced, and the homogeneous equation with the same p is its partner — the one “Homogeneous Plus Particular” builds the solution from.
If y or y′ is squared, multiplied by the other, in a denominator or inside a sine, an exponential or a logarithm, it is not linear and none of this course applies. A t anywhere else does not matter.
The first preset is 2y′ + 6y = 4t. The coefficient of y′ is 2, so everything is divided by 2.
The division changed 6y to 3y and 4t to 2t together. Leaving the right-hand side at 4t would have described a different equation, and the lab's μ and y_h would not have noticed, because neither depends on q at all. That is the reason the status line reports q as well as p.
For a rehearsal, read the other two presets before the lab does. The third, y′ = 4 − 2y, collects to y′ + 2y = 4; its p is the positive 2, and the lab agrees with μ = e^(2t). The second, t·y′ − 2y = 0, divides by t to give p = −2/t; predict the sign of the exponent in μ before you pick it, and then compare.
You should be able to collect the terms with the unknown on the left, divide every term by the coefficient of y′, name p and q with their signs, say whether the equation is homogeneous, and say why an equation with a power or product of y is not linear and an equation with a power of t is.
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