---
title: Complex Roots, Repeated Roots, and Reduction of Order
module: Second-Order Linear Equations
moduleNumber: 3
lessonNumber: 2
order: 302
summary: >
  When the characteristic equation has complex conjugate roots, Euler's formula
  converts the complex exponentials into a real fundamental set of decaying or
  growing oscillations. When it has a repeated root, one exponential is lost and
  reduction of order recovers the missing second solution as $t\,e^{rt}$. The
  same substitution $y = v(t)y_1(t)$ finds a second solution from any known one.
topics: [Second-Order Linear Equations]
draft: false
sources:
  - book: Boyce
    ref: "Ch. 3 — §3.3 Complex Roots of the Characteristic Equation; §3.4 Repeated Roots; Reduction of Order"
  - book: Simmons
    ref: "Ch. 3 §16 The Use of a Known Solution to Find Another"
---

The characteristic equation $ar^2 + br + c = 0$ has three outcomes, fixed by the
sign of the discriminant $b^2 - 4ac$. Two distinct real roots ($b^2 - 4ac > 0$)
give $y = c_1 e^{r_1 t} + c_2 e^{r_2 t}$. The remaining two cases break that
form. With complex conjugate roots ($b^2 - 4ac < 0$), Euler's formula converts
the complex exponentials into real-valued solutions. With a repeated real root
($b^2 - 4ac = 0$), only one exponential solution exists, and reduction of order
supplies the second by a method that applies far beyond constant coefficients.

$$
% caption: The discriminant partitions the characteristic roots into three
% cases, each producing a distinct form for the fundamental set.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={draw, minimum width=32mm, minimum height=15mm, align=center, font=\scriptsize}]
  \definecolor{acc}{HTML}{2A6F97}
  \node[box, draw=acc, text=acc, thick] (root) at (0,2.2) {$ar^2+br+c=0$};
  \node[box] (real) at (-4.6,-0.4) {$b^2-4ac>0$\\ real, distinct\\ $e^{r_1 t},\ e^{r_2 t}$};
  \node[box] (comp) at (0,-0.4) {$b^2-4ac<0$\\ complex conjugates\\ exp $\times$ $\cos,\ \sin$};
  \node[box] (rep) at (4.6,-0.4) {$b^2-4ac=0$\\ repeated real\\ $e^{rt},\ t\,e^{rt}$};
  \draw[->, acc, thick] (root) -- (real);
  \draw[->, acc, thick] (root) -- (comp);
  \draw[->, acc, thick] (root) -- (rep);
\end{tikzpicture}
$$

## Complex roots and Euler's formula

Suppose $b^2 - 4ac < 0$. The quadratic formula gives conjugate roots

$$
r_1 = \lambda + i\mu, \qquad r_2 = \lambda - i\mu, \qquad
\lambda = -\frac{b}{2a}, \quad \mu = \frac{\sqrt{4ac - b^2}}{2a},
$$

with $\lambda, \mu$ real and $\mu > 0$. Formally $y_1 = e^{(\lambda + i\mu)t}$
and $y_2 = e^{(\lambda - i\mu)t}$ are solutions, but they are complex-valued,
whereas the equation has real coefficients and we want real solutions.
**Euler's formula** defines the complex exponential through real functions.

> **Theorem (Euler's formula).** For real $\theta$,
> $$
> e^{i\theta} = \cos\theta + i\sin\theta.
> $$
> More generally $e^{(\lambda + i\mu)t} = e^{\lambda t}\big(\cos(\mu t) +
> i\sin(\mu t)\big)$.

The formula is obtained by substituting $i\theta$ into the Taylor series for the
exponential and separating the powers of $i$: the even powers reproduce the
cosine series and the odd powers, carrying one factor of $i$, reproduce the sine
series. Adopting it as the _definition_ of the complex exponential makes the
usual law $e^{(\lambda+i\mu)t} = e^{\lambda t}e^{i\mu t}$ hold and gives a
concrete complex number with real and imaginary parts built from elementary real
functions.

To extract real solutions, use a structural fact about the operator: because
$p$ and $q$ are real, the real and imaginary parts of a complex-valued solution
are each real-valued solutions on their own.

> **Theorem (Real and imaginary parts).** If $y = u(t) + iv(t)$ solves
> $y'' + p(t)y' + q(t)y = 0$ with $p, q$ real, then $u$ and $v$ each solve the
> equation.

Substituting $u + iv$ into $L$ and using that $p, q$ are real splits
$L[u + iv] = L[u] + iL[v]$. A complex number is zero exactly when its real and
imaginary parts both vanish, so $L[y] = 0$ forces $L[u] = 0$ and $L[v] = 0$.
Applying this to $y_1 = e^{\lambda t}\cos(\mu t) + i e^{\lambda t}\sin(\mu t)$
yields the real pair

$$
u(t) = e^{\lambda t}\cos(\mu t), \qquad v(t) = e^{\lambda t}\sin(\mu t).
$$

Their Wronskian is $W[u, v](t) = \mu\, e^{2\lambda t}$, which is nonzero since
$\mu \neq 0$, so $u, v$ form a fundamental set and the general solution is

$$
y = e^{\lambda t}\big(c_1 \cos(\mu t) + c_2 \sin(\mu t)\big). \tag{C}
$$

> **Worked example (Decaying oscillation).** Solve $y'' + y' + 9.25y = 0$ with
> $y(0) = 2$, $y'(0) = 8$. The characteristic equation $r^2 + r + 9.25 = 0$ has
> roots $r = -\tfrac{1}{2} \pm 3i$, so $\lambda = -\tfrac{1}{2}$, $\mu = 3$ and
> $y = e^{-t/2}(c_1\cos 3t + c_2\sin 3t)$. From $y(0) = 2$ we get $c_1 = 2$;
> differentiating and setting $y'(0) = 8$ gives $-\tfrac{1}{2}c_1 + 3c_2 = 8$,
> so $c_2 = 3$ and $y = e^{-t/2}(2\cos 3t + 3\sin 3t)$. The solution oscillates
> with period $2\pi/3$ inside a decaying envelope $\pm R e^{-t/2}$.

The sign of $\lambda$ controls the envelope and $\mu$ controls the oscillation
rate. When $\lambda < 0$ the amplitude shrinks (a **decaying oscillation**);
when $\lambda > 0$ it grows; when $\lambda = 0$ the roots are purely imaginary
$\pm i\mu$ and the solution is a pure sinusoid $c_1 \cos(\mu t) + c_2\sin(\mu t)$
of constant amplitude. The three regimes are the underdamped, unstable, and
undamped cases of a mechanical or electrical oscillator.

> **Worked example (Purely imaginary roots).** Solve $y'' + 9y = 0$ with
> $y(0) = 1$, $y'(0) = 6$. The characteristic equation $r^2 + 9 = 0$ has roots
> $r = \pm 3i$, so $\lambda = 0$, $\mu = 3$ and $y = c_1\cos 3t + c_2\sin 3t$.
> From $y(0) = 1$, $c_1 = 1$; from $y'(0) = 3c_2 = 6$, $c_2 = 2$. Thus
> $y = \cos 3t + 2\sin 3t$, a constant-amplitude sinusoid of amplitude
> $R = \sqrt{c_1^2 + c_2^2} = \sqrt 5$ and period $2\pi/3$.

$$
% caption: A complex-root solution $e^{\lambda t}\cos(\mu t)$ with $\lambda<0$:
% an oscillation trapped between the mirror-image envelopes $\pm e^{\lambda t}$.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{2A6F97}
  \draw[->, thick] (-0.3,0) -- (6.4,0) node[right] {$t$};
  \draw[->, thick] (0,-2.2) -- (0,2.3) node[above] {$y$};
  % envelopes +- e^{-0.35 t} scaled to start at 2
  \draw[black, dashed, thick] (0,2.0) .. controls (1.5,1.35) and (3.0,0.9) .. (6.2,0.42);
  \draw[black, dashed, thick] (0,-2.0) .. controls (1.5,-1.35) and (3.0,-0.9) .. (6.2,-0.42);
  \node[black, anchor=south, font=\scriptsize] at (5.4,0.52) {upper envelope};
  \node[black, anchor=north, font=\scriptsize] at (5.4,-0.52) {lower envelope};
  % oscillation touching envelopes
  \draw[acc, very thick]
    (0,2.0) .. controls (0.5,1.6) and (0.7,0.0) .. (1.0,-1.55)
    .. controls (1.25,-1.35) and (1.5,-0.1) .. (1.85,1.25)
    .. controls (2.1,1.15) and (2.35,0.1) .. (2.7,-1.02)
    .. controls (2.95,-0.95) and (3.2,-0.05) .. (3.55,0.82)
    .. controls (3.85,0.78) and (4.1,0.05) .. (4.45,-0.66)
    .. controls (4.75,-0.62) and (5.0,-0.05) .. (5.35,0.52)
    .. controls (5.65,0.5) and (5.9,0.05) .. (6.2,-0.4);
\end{tikzpicture}
$$

## The repeated root

Now $b^2 - 4ac = 0$, so the quadratic has a double root

$$
r_1 = r_2 = -\frac{b}{2a}.
$$

Only one exponential $y_1 = e^{-bt/(2a)}$ emerges, and a fundamental set needs a
second solution not proportional to it. **Reduction of order**, due to
d'Alembert, produces it: since any constant multiple $c\,y_1$ is a solution,
replace the constant by a function and look for a solution
$y = v(t)\, y_1(t)$.

> **Worked example (Repeated root by reduction of order).** Solve
> $y'' + 4y' + 4y = 0$. The characteristic equation $(r+2)^2 = 0$ gives the
> double root $r = -2$ and $y_1 = e^{-2t}$. Set $y = v(t)e^{-2t}$. Then
> $$
> y' = (v' - 2v)e^{-2t}, \qquad y'' = (v'' - 4v' + 4v)e^{-2t}.
> $$
> Substituting and collecting terms, every undifferentiated-$v$ and every
> $v'$ term cancels, leaving $v''(t)e^{-2t} = 0$, i.e. $v'' = 0$. Hence
> $v = c_1 t + c_2$, and
> $$
> y = (c_1 t + c_2)e^{-2t} = c_1\, t e^{-2t} + c_2\, e^{-2t}.
> $$

The new solution is $y_2 = t e^{-2t}$: the same exponential multiplied by $t$.
Its Wronskian with $y_1$ is $W[e^{-2t}, te^{-2t}] = e^{-4t} \neq 0$, so
$\{e^{rt}, te^{rt}\}$ is a fundamental set and the general solution for a
repeated root $r$ is

$$
y = (c_1 + c_2 t)\, e^{rt}. \tag{R}
$$

The cancellation is general. When $r$ is a double
root, both $ar^2 + br + c$ and its derivative $2ar + b$ vanish at $r$; since these
are the very coefficients of $v$ and $v'$, both terms drop out and leave
$v'' = 0$.

$$
% caption: The repeated-root solution $t\,e^{rt}$ for $r<0$: the extra factor
% $t$ lets it rise before the exponential pulls it back to zero.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{2A6F97}
  \draw[->, thick] (-0.3,0) -- (6.2,0) node[right] {$t$};
  \draw[->, thick] (0,-0.3) -- (0,2.6) node[above] {$y$};
  % e^{rt}
  \draw[black, thick] (0,2.2) .. controls (0.9,1.25) and (2.0,0.6) .. (3.2,0.33)
    .. controls (4.2,0.18) and (5.2,0.1) .. (6.0,0.07);
  \node[black, anchor=west, font=\scriptsize] at (2.5,0.55) {$e^{rt}$};
  % t e^{rt}: 0 at t=0, peak at t = -1/r, decay
  \draw[acc, very thick] (0,0) .. controls (0.4,0.9) and (0.9,1.55) .. (1.5,1.6)
    .. controls (2.4,1.6) and (3.4,0.9) .. (4.3,0.5)
    .. controls (5.0,0.3) and (5.6,0.2) .. (6.0,0.15);
  \node[acc, anchor=south, font=\scriptsize] at (1.5,1.62) {$t\,e^{rt}$};
\end{tikzpicture}
$$

## Reduction of order in general

The substitution $y = v(t) y_1(t)$ is not tied to constant coefficients. Given
_any_ single solution $y_1$ of the variable-coefficient equation

$$
y'' + p(t) y' + q(t) y = 0,
$$

it produces a second, independent solution. Substituting $y = v y_1$ and using
that $y_1$ itself solves the equation, the $v$-only terms vanish and what remains
is

$$
y_1 v'' + (2 y_1' + p\, y_1)\, v' = 0.
$$

There is no undifferentiated $v$: the equation is first-order in $w = v'$. It
separates,

$$
\frac{w'}{w} = -\frac{2 y_1'}{y_1} - p, \qquad
w = \frac{1}{y_1^2}\exp\!\left(-\int p\, \d t\right),
$$

and one integration of $w$ recovers $v$, hence $y_2 = v\, y_1$. The order of the
equation has been reduced from two to one, which is where the method's name
comes from and why it works even when no characteristic equation is available.

$$
% caption: Reduction of order: substituting $y=v\,y_1$ into the second-order
% equation removes the $v$-term, giving a first-order equation for $w=v'$.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={draw, minimum width=38mm, minimum height=12mm, align=center, font=\scriptsize}]
  \definecolor{acc}{HTML}{2A6F97}
  \node[box] (a) at (0,0) {known solution $y_1$};
  \node[box] (b) at (0,-1.7) {substitute $y = v(t)\,y_1(t)$};
  \node[box, draw=acc, text=acc, thick] (c) at (5.6,-1.7) {$y_1 v'' + (2y_1' + p\,y_1)v' = 0$};
  \node[box] (d) at (5.6,-3.4) {first-order in $w = v'$};
  \node[box] (e) at (0,-3.4) {integrate $w$, then $y_2 = v\,y_1$};
  \draw[->, acc, thick] (a) -- (b);
  \draw[->, acc, thick] (b) -- (c);
  \draw[->, acc, thick] (c) -- (d);
  \draw[->, acc, thick] (d) -- (e);
\end{tikzpicture}
$$

> **Worked example (Reduction of order, variable coefficients).** The equation
> $t^2 y'' - 3t y' + 4y = 0$ ($t > 0$) has the solution $y_1 = t^2$. Writing
> $y = v\, t^2$ and reducing gives, after dividing through, $t v'' + v' = 0$,
> i.e. $(t w)' = 0$ with $w = v'$. Thus $w = 1/t$, $v = \ln t$, and the second
> solution is $y_2 = t^2 \ln t$. The pair $\{t^2, t^2\ln t\}$ is a fundamental
> set.

The repeated-root form $t e^{rt}$ is the constant-coefficient special case: with
$y_1 = e^{rt}$ and $p = -2r$, the integral gives $w = $ const and $v = t$.

## The three cases side by side

| Discriminant | Roots | Fundamental set | General solution |
| --- | --- | --- | --- |
| $b^2 - 4ac > 0$ | real, distinct $r_1, r_2$ | $e^{r_1 t},\ e^{r_2 t}$ | $c_1 e^{r_1 t} + c_2 e^{r_2 t}$ |
| $b^2 - 4ac < 0$ | complex $\lambda \pm i\mu$ | $e^{\lambda t}\cos\mu t,\ e^{\lambda t}\sin\mu t$ | $e^{\lambda t}(c_1\cos\mu t + c_2\sin\mu t)$ |
| $b^2 - 4ac = 0$ | repeated $r$ | $e^{rt},\ t e^{rt}$ | $(c_1 + c_2 t)e^{rt}$ |

Every row is a two-dimensional solution space with a nonzero Wronskian; the
cases differ only in how the second basis function is built.[^boyce33][^boyce34][^simmons16]

[^boyce33]: **Boyce**, _Elementary Differential Equations_, §3.3 — Complex Roots of the Characteristic Equation: Euler's formula via the exponential series, the real fundamental set $e^{\lambda t}\cos\mu t,\ e^{\lambda t}\sin\mu t$, and the decaying/growing oscillation examples.
[^boyce34]: **Boyce**, _Elementary Differential Equations_, §3.4 — Repeated Roots; Reduction of Order: d'Alembert's substitution $y = v(t)y_1(t)$, the $t e^{rt}$ second solution, and the general reduction-of-order formula for variable coefficients.
[^simmons16]: **Simmons**, _Differential Equations with Applications and Historical Notes_, §16 — The Use of a Known Solution to Find Another: obtaining a second independent solution from a first via the reduction substitution.