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Second-Order Linear Non-Homogeneous Equations

Mathematics Algebra Difficulty 2/5

Beginning of the lesson

1. The Structure Theorem: General Solution = Homogeneous Family + One Particular Solution

The Big Picture

We study the second-order linear non-homogeneous ODE with constant coefficients:

$$a\,y'' + b\,y' + c\,y = f(x) \qquad (a \neq 0)$$

where $a, b, c$ are real constants and $f(x) \neq 0$ is the forcing function (also called the right-hand side or the non-homogeneous term).

Why does this work? — Linearity's Gift

Suppose $y_A$ and $y_B$ are both solutions of the non-homogeneous equation. Define $w = y_A - y_B$. Then:

$$a\,w'' + b\,w' + c\,w = a\,y_A'' + b\,y_A' + c\,y_A - \bigl(a\,y_B'' + b\,y_B' + c\,y_B\bigr) = f(x) - f(x) = 0$$

So $w = y_A - y_B$ satisfies the homogeneous equation. In other words, the difference of any two solutions of the non-homogeneous equation is a solution of the homogeneous equation. This means every solution of the non-homogeneous equation has the form $y_h + y_p$ for some choice of the constants $c_1, c_2$.

Callback: First-Order Rehearsal

Recall from first-order linear equations $y' + p(x)y = q(x)$: the general solution was $y = y_h + y_p$, where $y_h = Ce^{-\int p\,dx}$ and $y_p$ was found by an integrating factor. The same structural pattern now carries over to second order — it is the arc's load-bearing principle.

The Task Split

Task A — Already Owned

Find $y_h$: solve $a\,y'' + b\,y' + c\,y = 0$ via the characteristic equation $ar^2 + br + c = 0$. This was the previous station's business (distinct real roots, repeated roots, complex roots).

Task B — This Topic's Business

Find one particular solution $y_p$ of $a\,y'' + b\,y' + c\,y = f(x)$. Two methods: Undetermined Coefficients and Variation of Parameters.

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In the full lesson

  • The complete explanation
  • Worked examples
  • Common mistakes and how to avoid them
  • Test questions: 15
  • AI tutor that answers questions about the topic
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