Structure of Nonhomogeneous Solutions
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 7.2: “Nonhomogeneous Linear Equations” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/7-2-nonhomogeneous-linear-equations |
| Textbook used in class | Stewart, Calculus, Section 17.2: “Nonhomogeneous Linear Equations” (introduction) |
Quick Reference
For $ay'' + by' + cy = g(x)$, the general solution is: $$y = y_h + y_p,$$ where:
- $y_h = c_1 y_1 + c_2 y_2$ is the complementary (homogeneous) solution, i.e., the general solution of $ay''+by'+cy=0$
- $y_p$ is any particular solution of the full nonhomogeneous equation
Motivation
When a forcing function $g(x)$ is present, the equation $ay''+by'+cy=g(x)$ describes a driven system: a spring being pushed, a circuit with a voltage source. The response of the system splits naturally into two parts. The homogeneous part $y_h$ captures how the system moves on its own (its natural oscillation or decay). The particular part $y_p$ captures the response forced by $g(x)$. Together they account for all possible behaviors.
Key Concept: Why $y_h + y_p$ Works
Let $L[y] = ay''+by'+cy$. If $L[y_p] = g$ and $L[y_h] = 0$, then: $$L[y_h + y_p] = L[y_h] + L[y_p] = 0 + g = g.$$
So any function of the form $y_h + y_p$ is a solution of the nonhomogeneous equation. Going the other direction: if $y$ is any solution, then $y - y_p$ satisfies $L[y-y_p] = L[y]-L[y_p] = g-g = 0$, so $y - y_p$ is a solution of the homogeneous equation and must equal $c_1 y_1 + c_2 y_2$ for some constants. Therefore every solution has the form $y_h + y_p$. The structure is exhaustive.
Worked Example
Verify that $y_p = \frac{1}{5}e^{3x}$ is a particular solution of $y'' - 4y = e^{3x}$, and write the general solution.
$y_p' = \frac{3}{5}e^{3x}$, $y_p'' = \frac{9}{5}e^{3x}$.
$y_p'' - 4y_p = \frac{9}{5}e^{3x} - \frac{4}{5}e^{3x} = \frac{5}{5}e^{3x} = e^{3x}$. Check.
Homogeneous equation $y''-4y = 0$: characteristic equation $r^2-4=0$, $r=\pm 2$. $y_h = c_1 e^{2x} + c_2 e^{-2x}$.
General solution: $$y = c_1 e^{2x} + c_2 e^{-2x} + \frac{1}{5}e^{3x}.$$
you need to find all particular solutions. Any single particular solution $y_p$ will do. The two-parameter family $c_1 y_1 + c_2 y_2$ in $y_h$ absorbs all the freedom. If you find $y_p$ by one method and someone else finds a different $y_p$ by another method, the two general solutions $y_h + y_{p,1}$ and $y_h + y_{p,2}$ describe the same family of functions, because $y_{p,1} - y_{p,2}$ is itself a solution of the homogeneous equation and is absorbed into $y_h$.
Leveled Practice
Problem 1. The general solution of $y'' + y = \cos x$ has the form $y_h + y_p$. Given that $y_h = c_1\cos x + c_2\sin x$, what must $y_p$ look like? (Hint: the usual guess for $\cos x$ on the right side will not work here -- why not?)
Show answer
The usual guess $y_p = A\cos x + B\sin x$ fails because these are already solutions of the homogeneous equation -- the particular solution must be modified by multiplying by $x$: $y_p = x(A\cos x + B\sin x)$. This is the modification rule, covered in the next lesson.