--- title: Higher-Order Linear Equations module: Second-Order Linear Equations moduleNumber: 3 lessonNumber: 6 order: 306 summary: > The second-order theory extends directly to order $n$: the solution space is $n$-dimensional, spanned by any $n$ solutions with nonzero Wronskian. For constant coefficients the characteristic polynomial has degree $n$, and its roots (counted with multiplicity, real and complex) build the basis by the same rules as before. Coupled oscillators are the natural application that raises the order. topics: [Second-Order Linear Equations] draft: false sources: - book: Boyce ref: "Ch. 4 — §4.1 General Theory of nth-Order Linear Equations; §4.2 Homogeneous Equations with Constant Coefficients; §4.3 The Method of Undetermined Coefficients" - book: Simmons ref: "Ch. 3 §22 Higher-Order Linear Equations; Coupled Harmonic Oscillators" --- The second-order theory extends to order $n$ with no essential change. An $n$-th order linear equation has the same structure: an $n$-dimensional solution space, a Wronskian test for a basis, a characteristic polynomial whose roots build the general solution, and undetermined-coefficients and variation-of-parameters methods for forcing. The one new phenomenon is a root of multiplicity greater than two. Coupled oscillators are the physical setting that produces equations of order four and higher. ## The general theory at order n An $n$-th order linear equation, after dividing by the leading coefficient, is $$ L[y] = y^{(n)} + p_1(t)\, y^{(n-1)} + \cdots + p_{n-1}(t)\, y' + p_n(t)\, y = g(t), $$ with the $p_i$ and $g$ continuous on an interval $I$. Solving it should require $n$ integrations, each adding one constant, so an initial value problem now fixes the function _and its first $n - 1$ derivatives_ at a point. > **Theorem (Existence and Uniqueness, order $n$).** If $p_1, \ldots, p_n, g$ are > continuous on an open interval $I$ containing $t_0$, then the initial value > problem $L[y] = g$ with > $y(t_0) = y_0,\ y'(t_0) = y_0',\ \ldots,\ y^{(n-1)}(t_0) = y_0^{(n-1)}$ > has exactly one solution, defined throughout $I$. The homogeneous equation $L[y] = 0$ has an $n$-dimensional solution space. Any $n$ solutions $y_1, \ldots, y_n$ span it exactly when their **Wronskian**, the determinant of the functions and their first $n - 1$ derivatives, $$ W[y_1, \ldots, y_n] = \begin{vmatrix} y_1 & y_2 & \cdots & y_n \\ y_1' & y_2' & \cdots & y_n' \\ \vdots & \vdots & & \vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \cdots & y_n^{(n-1)} \end{vmatrix}, $$ is nonzero. As in the second-order case, this Wronskian is either identically zero or never zero on $I$, so a single-point test decides it. > **Theorem (Fundamental set and independence).** If $y_1, \ldots, y_n$ solve > $L[y] = 0$ on $I$ and $W[y_1, \ldots, y_n](t_0) \neq 0$ at some point, then > every solution is a linear combination $c_1 y_1 + \cdots + c_n y_n$. For > solutions of the equation, this is equivalent to $y_1, \ldots, y_n$ being > **linearly independent** on $I$. Such a set is a **fundamental set of solutions** and its combination is the general solution. Linear independence means no nontrivial constants satisfy $k_1 y_1 + \cdots + k_n y_n \equiv 0$; for arbitrary functions this does not force a nonzero Wronskian, but for solutions of a common linear equation it does, which is what makes the Wronskian a reliable test here. $$ % caption: The homogeneous solution space at order $n$ is $n$-dimensional; any % $n$ solutions with nonzero Wronskian form a basis for it. \begin{tikzpicture}[>=stealth, font=\footnotesize, u/.style={draw, minimum width=15mm, minimum height=8mm, align=center, font=\scriptsize}] \definecolor{acc}{HTML}{2A6F97} \node[u] (y1) at (0,0) {$y_1$}; \node[u] (y2) at (1.9,0) {$y_2$}; \node[font=\small] (dots) at (3.5,0) {$\cdots$}; \node[u] (yn) at (5.1,0) {$y_n$}; \node[u, draw=acc, text=acc, thick, minimum width=52mm] (span) at (2.55,-1.8) {general solution $\ c_1 y_1 + \cdots + c_n y_n$}; \draw[->, thick] (y1) -- (span); \draw[->, thick] (y2) -- (span); \draw[->, thick] (yn) -- (span); \node[anchor=west, font=\scriptsize, text=black] at (6.0,0) {$W \neq 0$}; \end{tikzpicture} $$ ## Constant coefficients: the characteristic polynomial For constant coefficients $a_0 y^{(n)} + a_1 y^{(n-1)} + \cdots + a_n y = 0$, the substitution $y = e^{rt}$ again works, giving $L[e^{rt}] = e^{rt} Z(r)$ with the **characteristic polynomial** $$ Z(r) = a_0 r^n + a_1 r^{n-1} + \cdots + a_{n-1} r + a_n. $$ By the fundamental theorem of algebra $Z$ has $n$ roots (with multiplicity), possibly repeated or complex, and factors as $$ Z(r) = a_0 (r - r_1)(r - r_2)\cdots(r - r_n). $$ Each root contributes solutions by the same rules established for second order, now allowing higher multiplicity. $$ % caption: The degree-$n$ characteristic polynomial factors into root blocks; % each block, by its multiplicity and type, contributes basis solutions. \begin{tikzpicture}[>=stealth, font=\footnotesize, box/.style={draw, minimum width=22mm, minimum height=10mm, align=center, font=\scriptsize}] \definecolor{acc}{HTML}{2A6F97} \node[box, draw=acc, text=acc, thick] (Z) at (0,1.6) {$Z(r) = 0$}; \node[box] (b1) at (-4.4,-0.3) {simple real $r$\\ gives $e^{rt}$}; \node[box] (b2) at (-1.2,-0.3) {complex pair\\ gives $\cos, \sin$}; \node[box] (b3) at (2.0,-0.3) {double real $r$\\ gives $e^{rt}, t e^{rt}$}; \node[box] (b4) at (5.2,-0.3) {multiplicity $s$\\ up to $t^{s-1}e^{rt}$}; \draw[->, thick] (Z) -- (b1); \draw[->, thick] (Z) -- (b2); \draw[->, thick] (Z) -- (b3); \draw[->, thick] (Z) -- (b4); \end{tikzpicture} $$ > **Worked example (Distinct real roots).** Solve > $y^{(4)} + y''' - 7y'' - y' + 6y = 0$. The characteristic polynomial > $r^4 + r^3 - 7r^2 - r + 6 = 0$ factors with roots $1, -1, 2, -3$, so > $$ > y = c_1 e^{t} + c_2 e^{-t} + c_3 e^{2t} + c_4 e^{-3t}. > $$ > The dominant term as $t \to \infty$ is $e^{2t}$; the sign of $c_3$ decides > whether the solution runs to $+\infty$ or $-\infty$. Complex roots come in conjugate pairs (the coefficients are real), each pair $\lambda \pm i\mu$ contributing $e^{\lambda t}\cos\mu t$ and $e^{\lambda t}\sin\mu t$, just as at second order. A fourth-order equation can mix real and complex roots. > **Worked example (Mixed real and imaginary roots).** Solve $y^{(4)} - y = 0$ > with $y(0) = \tfrac{7}{2}$, $y'(0) = -4$, $y''(0) = \tfrac{5}{2}$, > $y'''(0) = -2$. The characteristic polynomial factors as > $r^4 - 1 = (r^2 - 1)(r^2 + 1) = 0$, giving roots $1, -1, i, -i$ and > $$ > y = c_1 e^{t} + c_2 e^{-t} + c_3\cos t + c_4\sin t. > $$ > Evaluating $y, y', y'', y'''$ at $0$ produces four equations. Adding the > $y(0)$ and $y''(0)$ equations gives $c_1 + c_2 = 3$ and their difference gives > $c_3 = \tfrac{1}{2}$; the $y'(0)$ and $y'''(0)$ equations give $c_1 - c_2 = -3$ > and $c_4 = -1$. Hence $c_1 = 0$, $c_2 = 3$, and > $$ > y = 3e^{-t} + \tfrac{1}{2}\cos t - \sin t. > $$ > The growing term $e^{t}$ is absent because $c_1 = 0$: this data selects an > exponential decay onto a steady oscillation. $$ % caption: Roots of $r^4-1=0$ in the complex plane: a real pair on the axis % (one growing, one decaying) and an imaginary pair giving $\cos t,\ \sin t$. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{2A6F97} \draw[->, black] (-2.4,0) -- (2.4,0) node[right] {Re}; \draw[->, black] (0,-2.4) -- (0,2.4) node[above] {Im}; \fill[acc] (1.8,0) circle (2.6pt); \fill[acc] (-1.8,0) circle (2.6pt); \fill[acc] (0,1.8) circle (2.6pt); \fill[acc] (0,-1.8) circle (2.6pt); \node[anchor=south west, font=\scriptsize] at (1.8,0.05) {$r=1$}; \node[anchor=south east, font=\scriptsize] at (-1.8,0.05) {real pair}; \node[anchor=south west, font=\scriptsize] at (0.08,1.8) {$r=i$}; \node[anchor=north west, font=\scriptsize] at (0.08,-1.8) {conjugate}; \end{tikzpicture} $$ ## Repeated roots and multiplicity The one new feature is that a root can now repeat more than twice. If $r$ is a root of $Z$ of multiplicity $s$ (so $(r - r_1)^s$ divides $Z$), it contributes $s$ independent solutions $$ e^{rt}, \quad t e^{rt}, \quad t^2 e^{rt}, \quad \ldots, \quad t^{s-1} e^{rt}. $$ The rule generalizes the second-order repeated root $\{e^{rt}, t e^{rt}\}$: multiplicity $s$ multiplies the exponential by $1, t, \ldots, t^{s-1}$. A repeated _complex_ pair $\lambda \pm i\mu$ of multiplicity $s$ contributes $2s$ solutions, the sines and cosines each multiplied by $1, t, \ldots, t^{s-1}$. > **Worked example (Repeated complex pair).** Solve $y^{(4)} + 2y'' + y = 0$. The > characteristic polynomial is $r^4 + 2r^2 + 1 = (r^2 + 1)^2$, so > $r = \pm i$ each with multiplicity two. The general solution is > $$ > y = c_1\cos t + c_2\sin t + c_3\, t\cos t + c_4\, t\sin t. > $$ The table collects the contribution of each root type to the basis. | Root of $Z(r)$ | Multiplicity | Contribution to the basis | | --- | --- | --- | | real $r$ | $1$ | $e^{rt}$ | | real $r$ | $s$ | $e^{rt}, t e^{rt}, \ldots, t^{s-1}e^{rt}$ | | complex $\lambda \pm i\mu$ | $1$ | $e^{\lambda t}\cos\mu t,\ e^{\lambda t}\sin\mu t$ | | complex $\lambda \pm i\mu$ | $s$ | $t^{j}e^{\lambda t}\cos\mu t,\ t^{j}e^{\lambda t}\sin\mu t$, $j = 0,\ldots,s-1$ | Summing the contributions of all roots always yields exactly $n$ independent solutions, matching the dimension of the solution space. ## Nonhomogeneous higher-order equations Both nonhomogeneous methods carry over. **Undetermined coefficients** uses the same trial forms, with one adjustment: because a characteristic root can now have multiplicity up to $n$, a colliding trial may need to be multiplied by $t^s$ (rather than at most $t^2$) to clear all overlap with the homogeneous solution, where $s$ is the multiplicity of the offending root. **Variation of parameters** generalizes through the $n \times n$ Wronskian: the coefficient functions $u_i'$ solve an $n \times n$ linear system built from the fundamental set and the forcing, then integrate. Reduction of order also applies but is rarely useful beyond second order, since it only lowers the order by one and the reduced equation is still of order $n - 1$. > **Worked example (Triple root forces $t^3$).** Solve > $y''' - 3y'' + 3y' - y = 4e^{t}$. The characteristic polynomial is > $r^3 - 3r^2 + 3r - 1 = (r - 1)^3$, a triple root $r = 1$, so > $y_c = (c_1 + c_2 t + c_3 t^2)e^{t}$. The forcing $4e^{t}$ duplicates the > homogeneous solution, and so would $te^{t}$ and $t^2 e^{t}$; the trial must be > multiplied by $t^3$: $Y = At^3 e^{t}$. Applying $(D - 1)^3$ term by term gives > $(D - 1)^3\bigl[At^3 e^{t}\bigr] = 6A\,e^{t}$, so $6A = 4$, $A = \tfrac{2}{3}$ > and > $$ > y = c_1 e^{t} + c_2 t e^{t} + c_3 t^2 e^{t} + \tfrac{2}{3}t^3 e^{t}. > $$ > A second-order equation never needs more than $t^2$; the multiplicity-3 root > forces $t^3$. ## Coupled oscillators Higher-order equations appear naturally when several oscillators are linked. Two masses connected by springs (each mass to a wall, and to each other) have two displacements $u_1, u_2$ obeying coupled second-order equations. Eliminating one variable produces a single fourth-order equation in the other, whose characteristic polynomial factors into two quadratic blocks. The roots are two pairs of imaginary numbers $\pm i\omega_a$ and $\pm i\omega_b$, giving two **normal-mode frequencies**: the system oscillates as a superposition of two pure sinusoids, one where the masses move together and one where they move opposite. $$ % caption: Two masses coupled by springs; eliminating one displacement yields a % fourth-order equation with two normal-mode frequencies. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{2A6F97} % walls \draw[thick] (-0.2,-0.6) -- (-0.2,0.9); \foreach \y in {-0.5,-0.2,0.1,0.4,0.7} \draw[black] (-0.2,\y) -- (-0.45,\y+0.12); \draw[thick] (8.2,-0.6) -- (8.2,0.9); \foreach \y in {-0.5,-0.2,0.1,0.4,0.7} \draw[black] (8.2,\y) -- (8.45,\y+0.12); % baseline \draw[black] (-0.2,-0.6) -- (8.2,-0.6); % spring 1 \draw[black!70, thick] (-0.2,0.15) -- (0.3,0.15) -- (0.5,0.4) -- (0.8,-0.1) -- (1.1,0.4) -- (1.4,-0.1) -- (1.6,0.15) -- (2.0,0.15); % mass 1 \draw[thick, fill=black!8] (2.0,-0.3) rectangle (3.2,0.6); \node at (2.6,0.15) {$m_1$}; % spring 2 (coupling) \draw[black!70, thick] (3.2,0.15) -- (3.7,0.15) -- (3.9,0.4) -- (4.2,-0.1) -- (4.5,0.4) -- (4.8,-0.1) -- (5.0,0.15) -- (5.4,0.15); \node[black, anchor=south, font=\scriptsize] at (4.3,0.45) {coupling}; % mass 2 \draw[thick, fill=black!8] (5.4,-0.3) rectangle (6.6,0.6); \node at (6.0,0.15) {$m_2$}; % spring 3 \draw[black!70, thick] (6.6,0.15) -- (7.0,0.15) -- (7.2,0.4) -- (7.5,-0.1) -- (7.8,0.4) -- (8.0,-0.1) -- (8.1,0.15) -- (8.2,0.15); % displacement labels \draw[->] (2.6,-0.5) -- (3.1,-0.5); \node[anchor=north, font=\scriptsize] at (2.85,-0.55) {$u_1$}; \draw[->] (6.0,-0.5) -- (6.5,-0.5); \node[anchor=north, font=\scriptsize] at (6.25,-0.55) {$u_2$}; \end{tikzpicture} $$ > **Worked example (Two equal masses, three equal springs).** Let both masses equal > $m$ and all three springs have constant $k$. Newton's law for each mass gives > the coupled system > $$ > m u_1'' = -k u_1 + k(u_2 - u_1), \qquad m u_2'' = -k u_2 - k(u_2 - u_1). > $$ > Adding and subtracting the equations decouples them into the **normal > coordinates** $s = u_1 + u_2$ and $d = u_1 - u_2$: > $$ > m s'' = -k s, \qquad m d'' = -3k d. > $$ > Each is a simple harmonic oscillator, so the two normal-mode frequencies are > $$ > \omega_a = \sqrt{\frac{k}{m}} \quad(\text{masses move together}), \qquad > \omega_b = \sqrt{\frac{3k}{m}} \quad(\text{masses move opposite}). > $$ > The general motion is a superposition of the two modes, and the fourth-order > equation for $u_1$ alone has characteristic roots $\pm i\omega_a$, > $\pm i\omega_b$. $$ % caption: The two normal modes: in the slow mode the masses move together, in % the fast mode they move in opposition and compress the middle spring. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{2A6F97} % in-phase mode \node[black, anchor=east, font=\scriptsize] at (-0.2,1.2) {in phase}; \draw[thick, fill=black!8] (1.0,0.9) rectangle (1.8,1.5); \node[font=\scriptsize] at (1.4,1.2) {$m_1$}; \draw[thick, fill=black!8] (3.4,0.9) rectangle (4.2,1.5); \node[font=\scriptsize] at (3.8,1.2) {$m_2$}; \draw[->, acc, thick] (1.9,1.2) -- (2.5,1.2); \draw[->, acc, thick] (4.3,1.2) -- (4.9,1.2); % out-of-phase mode \node[black, anchor=east, font=\scriptsize] at (-0.2,-0.3) {out of phase}; \draw[thick, fill=black!8] (1.0,-0.6) rectangle (1.8,0.0); \node[font=\scriptsize] at (1.4,-0.3) {$m_1$}; \draw[thick, fill=black!8] (3.4,-0.6) rectangle (4.2,0.0); \node[font=\scriptsize] at (3.8,-0.3) {$m_2$}; \draw[->, acc, thick] (1.9,-0.3) -- (2.5,-0.3); \draw[->, acc, thick] (3.3,-0.3) -- (2.7,-0.3); \end{tikzpicture} $$ In general, a symmetric coupled system decouples into independent oscillators after the right change of variables, and each decoupled equation is a second-order harmonic oscillator. The change of variables that achieves this is exactly the diagonalization studied in the systems module. Rather than eliminate variables to raise the order, the coupled equations can be kept as a first-order [linear system](/differential-equations/systems/matrices-eigenvalues-review) in the vector $(u_1, u_2, u_1', u_2')$, where the normal modes reappear as eigenvectors of a matrix. The same reformulation turns any $n$-th order equation into a first-order system.[^boyce41][^boyce42][^simmons22] [^boyce41]: **Boyce**, _Elementary Differential Equations_, §4.1 — General Theory of nth-Order Linear Equations: the existence-uniqueness theorem (4.1.1), the $n \times n$ Wronskian, and the equivalence of linear independence and fundamental sets (4.1.3). [^boyce42]: **Boyce**, _Elementary Differential Equations_, §4.2–§4.3 — Homogeneous Equations with Constant Coefficients and Undetermined Coefficients: the degree-$n$ characteristic polynomial, the multiplicity-$s$ solutions $e^{rt}, \ldots, t^{s-1}e^{rt}$, repeated complex roots, and the higher-power resonance correction. [^simmons22]: **Simmons**, _Differential Equations with Applications and Historical Notes_, §22 — Higher-Order Linear Equations; Coupled Harmonic Oscillators: two masses linked by springs reduced to a fourth-order equation and its two normal-mode frequencies.