What is differential equation second-order linear?

Example 5.11 (A Cauchy problem, start to finish)

Solve $y'' - 3y' + 2y = 0$ with $y(0) = 0$, $y'(0) = 1$. The characteristic polynomial $r^2 - 3r + 2 = (r - 1)(r - 2)$ has the real roots $1$ and $2$: general solution $y = \lambda\eu^{x} + \mu\eu^{2x}$. The two conditions give the linear system

so $\mu = 1$, $\lambda = -1$:

Check: $y(0) = 0$; $y' = 2\eu^{2x} - \eu^x$ has $y'(0) = 1$; and $y'' - 3y' + 2y = (4 - 6 + 2)\eu^{2x} + (-1 + 3 - 2)\eu^x = 0$. Note the shape of the answer: near $-\infty$ the slow mode $-\eu^x$ dominates; near $+\infty$ the fast mode $\eu^{2x}$ does. Reading solutions as superpositions of modes with different decay or growth rates is the profitable habit — it is how the transient/steady-state split of the weekend problem is organized.

Example 5.14 (Superposition in action)

Solve $y'' - y = \eu^{x} + 4$ on $\R$. Homogeneous: $\chi(r) = r^2 - 1$, roots $\pm1$, so $y_h = \lambda\eu^x + \mu\eu^{-x}$. Split the right-hand side and treat each piece by the method box. Piece $\eu^x$: here $\gamma = 1$ is a simple root of $\chi$, so try $y_1 = c\,x\,\eu^x$: then $y_1'' - y_1 = c(x + 2)\eu^x - cx\eu^x = 2c\,\eu^x$, giving $c = \frac12$. Piece $4$: $\gamma = 0$ is not a root; the constant $y_2 = -4$ works. By superposition (Theorem 5.12 (2)):

Note how the two pieces demanded different shapes ($m = 1$ versus $m = 0$): the multiplicity test is applied to each exponent separately, which is the whole point of splitting the right-hand side before guessing.

Example 5.15 (The multiplicity rule at work)

Solve $y'' + y' = x$ on $\R$. The right-hand side is $P(x)\eu^{0 \cdot x}$ with $P(x) = x$, and $\gamma = 0$ is a simple root of $\chi(r) = r^2 + r = r(r + 1)$: so $m = 1$, and the correct guess is $y_p = x\,(\alpha x + \beta) = \alpha x^2 + \beta x$, one degree higher than $P$. Substituting:

and identification with $x$ gives $\alpha = \frac12$, $\beta = -1$: $y_p = \frac{x^2}2 - x$. General solution: $y = \frac{x^2}2 - x + \lambda + \mu\,\eu^{-x}$. Had we guessed $y_p = \alpha x + \beta$ (ignoring the multiplicity), substitution would give $y_p'' + y_p' = \alpha$, a constant — no choice of $\alpha, \beta$ can match $x$, and the failure is structural: constants already solve the homogeneous equation, so they are invisible to the left-hand side. The factor $x^m$ exists precisely to climb out of the homogeneous solution space.