First-Order Linear DEs
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 4.5: “First-Order Linear Equations” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/4-5-first-order-linear-equations |
| Textbook used in class | Stewart, Calculus, Section 9.5: “Linear Equations” (introduction) |
Quick Reference
A first-order linear DE has the form: $$\frac{dy}{dx} + P(x)\,y = Q(x),$$ where $P$ and $Q$ are functions of $x$ only (no $y$ on the right side, and $y$ appears at most to the first power).
This is the standard form. If the equation does not look like this, rewrite it: move all $y$ and $y'$ terms to the left, divide by the coefficient of $y'$ to get $1\cdot y' + \cdots$.
Motivation
Separable equations require the right side to factor as (function of $x$)(function of $y$). When the equation involves $y$ mixed with an $x$-function -- for example, $dy/dx = x - y$ -- the right side is a sum, not a product, and separation fails. First-order linear equations are exactly this class: they involve $y$ multiplied by an $x$-function and added to another $x$-function. A new technique, the integrating factor, handles them all.
First-order linear DEs model a wide range of forced systems: a tank being filled and drained simultaneously, a circuit with a varying voltage source, a drug entering and leaving the bloodstream at different rates. In each case the “input” (an $x$-function) drives the system, and the response (the solution $y$) is what we seek.
Key Concept: Recognizing Standard Form
The defining features of a first-order linear DE in standard form are:
- The equation is first order (no $y''$ or higher).
- $y$ and $y'$ appear only to the first power (linear in $y$ and $y'$).
- The coefficient of $y'$ is 1 (or can be made 1 by dividing).
Examples of first-order linear equations:
- $y' + 3y = e^x$ (already standard)
- $2y' - xy = \sin x$ (divide by 2: $y' - (x/2)y = (\sin x)/2$)
- $xy' + y = x^2$ (divide by $x$: $y' + y/x = x$)
Not first-order linear:
- $y' = y^2$ (nonlinear: $y$ appears squared)
- $y'' + y = 0$ (second order)
- $y' = x + y$ (first-order linear -- rewrite as $y' - y = x$, which IS in standard form)
Worked Example
Rewrite $x\,y' - 3y = x^4\cos x$ in standard form and identify $P(x)$ and $Q(x)$.
Divide by $x$ (assuming $x \neq 0$): $$y' - \frac{3}{x}\,y = x^3\cos x.$$
Standard form: $y' + P(x)y = Q(x)$ with $P(x) = -3/x$ and $Q(x) = x^3\cos x$.
This cannot be separated: the right side involves $x$ only and the left side mixes $y'$ with $(3/x)y$. The integrating factor method handles it.
a first-order linear equation in $y$ is always separable. An equation linear in $y$ has the form $y' + P(x)y = Q(x)$. If $Q(x) = 0$ (homogeneous), it is separable: $y'/y = -P(x)$. But if $Q(x) \neq 0$ (nonhomogeneous), the right side $Q(x) - P(x)y$ is a sum of an $x$-term and a $y$-term, which does not factor as a product. Separation fails, and the integrating factor is required. The equation $y' + y = x$ is linear but not separable.
Leveled Practice
Problem 1. Decide whether each equation is first-order linear. If so, rewrite it in standard form.
(a) $y' = x^2 + 2xy$ (b) $y' + y^2 = x$ (c) $x^2 y' + xy = e^x$
Show answer
(a) $y' - 2xy = x^2$. First-order linear: $P(x) = -2x$, $Q(x) = x^2$.
(b) Not linear: $y^2$ makes it nonlinear.
(c) Divide by $x^2$: $y' + y/x = e^x/x^2$. First-order linear: $P(x) = 1/x$, $Q(x) = e^x/x^2$.