Second-Order Linear DEs
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 7.1: “Second-Order Linear Equations” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/7-1-second-order-linear-equations |
| Textbook used in class | Stewart, Calculus, Section 17.1: “Second-Order Linear Equations” |
Quick Reference
A second-order linear DE has the form $$a y'' + b y' + c y = g(x),$$ where $a, b, c$ may be constants or functions of $x$, and $a \neq 0$.
It is homogeneous when $g(x) = 0$ and nonhomogeneous otherwise.
Superposition: if $y_1$ and $y_2$ both satisfy the homogeneous equation, so does $c_1 y_1 + c_2 y_2$ for any constants $c_1, c_2$.
The general solution of the homogeneous equation requires two linearly independent solutions.
Motivation
A mass on a spring obeys Newton’s second law: force equals mass times acceleration. The spring pulls back proportional to displacement; friction pulls back proportional to velocity. Writing this out gives $m x'' + c x' + k x = F(t)$ -- a second-order linear DE with constant coefficients. Every quantity you can observe about the spring’s motion (how fast it oscillates, whether it dies out, whether it resonates) comes from solving this equation. The same equation governs current in an RLC circuit, the bending of a beam, and the motion of a pendulum for small angles.
Key Concept: Superposition
The defining feature of a linear equation is that it satisfies superposition: if you add two solutions, you get another solution. More precisely, for $Ly = ay'' + by' + cy$:
- $L(y_1) = 0$ and $L(y_2) = 0$ implies $L(c_1 y_1 + c_2 y_2) = c_1 L(y_1) + c_2 L(y_2) = 0$.
This is why you can write a general solution as a linear combination. It is also why two initial conditions (one for $y(x_0)$ and one for $y'(x_0)$) determine a unique solution: you have two free constants $c_1, c_2$ and two equations to pin them down.
Key Concept: Linear Independence
Two functions $y_1$ and $y_2$ are linearly dependent if one is a constant multiple of the other. In that case, $c_1 y_1 + c_2 y_2 = (c_1 + c_2 \cdot k) y_1$ is really only one function, and you cannot satisfy arbitrary initial conditions with it. The Wronskian $W = y_1 y_2' - y_2 y_1'$ detects this: if $W \neq 0$ at some point, the solutions are linearly independent.
Worked Example
Verify that $y_1 = e^{2x}$ and $y_2 = e^{-x}$ are both solutions of $y'' - y' - 2y = 0$, and that $y = c_1 e^{2x} + c_2 e^{-x}$ is the general solution.
For $y_1 = e^{2x}$: $y_1' = 2e^{2x}$, $y_1'' = 4e^{2x}$. $y_1'' - y_1' - 2y_1 = 4e^{2x} - 2e^{2x} - 2e^{2x} = 0$. Check.
For $y_2 = e^{-x}$: $y_2' = -e^{-x}$, $y_2'' = e^{-x}$. $y_2'' - y_2' - 2y_2 = e^{-x} + e^{-x} - 2e^{-x} = 0$. Check.
Wronskian: $W = e^{2x}(-e^{-x}) - e^{-x}(2e^{2x}) = -e^x - 2e^x = -3e^x \neq 0$. Linearly independent.
By superposition, $y = c_1 e^{2x} + c_2 e^{-x}$ is the general solution.
any two solutions give a general solution. Two solutions of a second-order linear DE give a general solution only if they are linearly independent. For example, $y_1 = e^{2x}$ and $y_2 = 3e^{2x}$ are both solutions of $y'' - 4y' + 4y = 0$ (check: they are not -- let me pick a better example). The key check is $W \neq 0$. If $W = 0$, the pair is dependent and $c_1 y_1 + c_2 y_2$ reduces to a one-parameter family, not the full two-parameter general solution.
Leveled Practice
Problem 1. Show that $y_1 = \cos x$ and $y_2 = \sin x$ are solutions of $y'' + y = 0$ and compute the Wronskian.
Show answer
$y_1'' = -\cos x$, so $y_1'' + y_1 = -\cos x + \cos x = 0$. Same for $\sin x$.
$W = \cos x \cdot \cos x - \sin x \cdot (-\sin x) = \cos^2 x + \sin^2 x = 1 \neq 0$.
They are linearly independent; $y = c_1\cos x + c_2\sin x$ is the general solution.