---
title: "Linear Differential Equations"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 5
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/5-linear-differential-equations
---

# Chapter 5 — Linear Differential Equations

First met in the High School volume, differential equations are treated here with full proofs and in greater generality: first-order linear equations with variable coefficients (solved completely by the variation-of-constants method) and second-order linear equations with constant coefficients, the model for oscillations. Both cases display the same structure: *general solution $=$ one particular solution $+$ general solution of the homogeneous equation*.

## 5.1 First-order linear equations

**Definition 5.1.**

Let $I$ be an interval and $a, b \colon I \to \R$ (or $\C$) be continuous. The equation

$$
(E)\colon\quad y' + a(x)\,y = b(x),
$$

in the unknown differentiable function $y \colon I \to \R$ (or $\C$), is a *first-order linear differential equation*. The equation $(H)\colon y' + a(x) y = 0$ is its *homogeneous* equation.

**Theorem 5.2 (Solving the homogeneous equation).**

Let $A$ be a primitive of $a$ on $I$ (it exists: [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration)). The solutions of $(H)$ on $I$ are exactly the functions

$$
y(x) = \lambda\, \eu^{-A(x)}, \qquad \lambda \in \R \text{ (or } \C).
$$

**Proof.** These functions are solutions: $y' = -\lambda A' \eu^{-A} = -a y$. Conversely, let $y$ solve $(H)$ and set $z(x) = y(x)\, \eu^{A(x)}$. Then

$$
z' = y' \eu^{A} + y\, a\, \eu^{A} = (y' + ay)\, \eu^{A} = 0 ,
$$

so $z$ is constant on the interval $I$, say $z = \lambda$: $y =
\lambda \eu^{-A}$. (Note the logic: no solution is lost, because *every* solution has been written in the announced form.) ∎

**Example 5.3 (A homogeneous equation with variable coefficient).**

Solve $y' + (\cos x)\,y = 0$ on $\R$. A primitive of $a(x) =
\cos x$ is $A(x) = \sin x$, so the solutions are

$$
y(x) = \lambda\,\eu^{-\sin x}, \qquad \lambda \in \R .
$$

Two readings. Every solution is periodic (period $2\pi$) and never vanishes unless $\lambda = 0$ — the sign of $\lambda$ is the sign of $y$ forever, since an exponential cannot cross zero. And the solution through $y(0) = y_0$ is $y_0\eu^{-\sin x}$: exactly one curve of the family through each initial point, the one-dimensional picture of [Theorem 5.4](#thm-b1-diffeq-voc) (2).

**Theorem 5.4 (Variation of constants; Cauchy problem).**

With the notation above:

1. The solutions of $(E)$ on $I$ are exactly $$y(x) = \Bigl(\lambda + \int_{x_0}^{x} b(t)\, \eu^{A(t)} \dd t  \Bigr)\, \eu^{-A(x)},  \qquad \lambda \in \R,$$ where $x_0 \in I$ is fixed. Equivalently: general solution of $(H)$ plus one particular solution of $(E)$.
2. For every $x_0 \in I$ and $y_0$ , the *Cauchy problem* “ $(E)$ and $y(x_0) = y_0$ ” has exactly one solution on $I$ .

**Proof.** (1) Following the method named *variation of constants*, look for solutions of the form $y = \mu(x)\, \eu^{-A(x)}$ with $\mu$ differentiable — no generality is lost, since every function on $I$ can be so written ($\mu = y\,\eu^{A}$). Substituting,

$$
y' + ay = \mu' \eu^{-A} - \mu a \eu^{-A} + a\mu\eu^{-A}
= \mu'\, \eu^{-A},
$$

so $y$ solves $(E)$ if and only if $\mu'(x) = b(x)\,\eu^{A(x)}$, if and only if $\mu(x) = \lambda + \int_{x_0}^x b(t)\eu^{A(t)}\dd t$ for some constant $\lambda$ (two primitives of the same continuous function on an interval differ by a constant).

(2) In the formula, $y(x_0) = \lambda\,\eu^{-A(x_0)}$: the condition $y(x_0) = y_0$ determines $\lambda = y_0 \eu^{A(x_0)}$ uniquely. ∎

**Example 5.5.**

Solve $y' + \dfrac{y}{x} = x^2$ on $I = \intoo{0}{+\infty}$. Here $a(x) = \frac 1x$, $A(x) = \ln x$, $\eu^{-A(x)} = \frac 1x$. Homogeneous solutions: $\frac{\lambda}{x}$. [Variation of constants](#thm-b1-diffeq-voc): $\mu'(x) = x^2 \cdot x = x^3$, so $\mu = \frac{x^4}{4} + \lambda$, and

$$
y(x) = \frac{x^3}{4} + \frac{\lambda}{x}, \qquad \lambda \in \R .
$$

With the initial condition $y(1) = 0$: $\lambda = -\frac14$. *Check:* $y' + \frac yx = \frac{3x^2}{4} - \frac{\lambda}{x^2} +
\frac{x^2}{4} + \frac{\lambda}{x^2} = x^2$.

**Example 5.6 (Guessing beats integrating).**

Solve $y' + 2x\,y = x$ on $\R$. [Variation of constants](#thm-b1-diffeq-voc) works ($A =
x^2$, $\mu' = x\,\eu^{x^2}$, $\mu = \frac12\eu^{x^2} + \lambda$), but observing that the *constant* $y_p = \frac12$ solves the equation ($0 + 2x\cdot\frac12 = x$) is faster. With the homogeneous solutions $\lambda\,\eu^{-x^2}$:

$$
y(x) = \frac12 + \lambda\,\eu^{-x^2}, \qquad \lambda \in \R .
$$

Every solution converges to $\frac12$ extremely fast as $x \to
\pm\infty$: the constant particular solution is an *equilibrium* that all solutions join. The insight: before launching the general method, spend ten seconds looking for an obvious particular solution (constant, monomial, multiple of the right-hand side); the structure theorem then finishes the job.

**Remark 5.7 (Intervals matter).**

The theorem lives on an *interval* where $a$ and $b$ are continuous. For $y' + \frac yx = 0$ on $\R^*$, the solutions are $\frac{\lambda}{x}$ on $\intoo{0}{+\infty}$ and $\frac{\mu}{x}$ on $\intoo{-\infty}{0}$ with *independent* constants: there is no reason for a single formula to glue across the singularity at $0$.

**Example 5.8 (A complex right-hand side, two real answers).**

Solve $y' - y = \cos x$ and $y' - y = \sin x$ in one stroke. Work in $\C$ with the right-hand side $\eu^{\iu x}$: trying $y_p =
c\,\eu^{\iu x}$ gives $c(\iu - 1)\eu^{\iu x} = \eu^{\iu x}$, so

$$
c = \frac1{\iu - 1} = \frac{-1 - \iu}2,
\qquad
y_p = -\frac{(1 + \iu)(\cos x + \iu\sin x)}2
= \frac{\sin x - \cos x}2 + \iu\,\frac{-\sin x - \cos x}2 .
$$

Since the equation has real coefficients, real and imaginary parts split: $\frac{\sin x - \cos x}2$ solves $y' - y = \cos x$, and $-\frac{\sin x + \cos x}2$ solves $y' - y = \sin x$ (check the first: derivative $\frac{\cos x + \sin x}2$, minus the function, gives $\cos x$). One complex line replaced two runs of [variation of constants](#thm-b1-diffeq-voc) — the same economy that [Method 5.13](#met-b1-diffeq-particular) systematizes for second order, and a recurring dividend of [Chapter 3](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#ch-b1-complex).

## 5.2 Second-order linear equations with constant coefficients

**Definition 5.9.**

Let $a, b \in \R$ and $f \colon I \to \R$ continuous. The equation

$$
(E)\colon\quad y'' + a\,y' + b\,y = f(x)
$$

is a *second-order linear equation with constant coefficients*; $(H)\colon y'' + ay' + by = 0$ is its homogeneous equation, and $\chi(r) = r^2 + ar + b$ its *characteristic polynomial*.

**Theorem 5.10 (Homogeneous solutions).**

Let $\Delta = a^2 - 4b$ be the discriminant of $\chi$. The real solutions of $(H)$ on $\R$ are:

1. if $\Delta > 0$ , with $r_1 \neq r_2$ the two real roots: $\;y = \lambda\, \eu^{r_1 x} + \mu\, \eu^{r_2 x}$ ;
2. if $\Delta = 0$ , with $r_0$ the double root: $\;y = (\lambda + \mu x)\, \eu^{r_0 x}$ ;
3. if $\Delta < 0$ , with roots $\alpha \pm \iu\omega$ ( $\omega > 0$ ): $\;y = \eu^{\alpha x} (\lambda \cos\omega x + \mu  \sin\omega x)$ ;

in each case with $(\lambda, \mu)$ running over $\R^2$.

**Proof.** First observe that for $r \in \C$, $x \mapsto \eu^{rx}$ solves $(H)$ if and only if $\chi(r) = 0$ (substitute: $(r^2 + ar + b)\eu^{rx} =
0$). This is why exponentials are the natural first guess: differentiation acts on $\eu^{rx}$ as multiplication by the number $r$, so the differential equation becomes the numerical equation $\chi(r) = 0$ — the whole analytic problem is compressed into finding the roots of one quadratic.

The key step is a change of unknown that reduces the order. Let $r$ be a (possibly complex) root of $\chi$ and write $y = z\, \eu^{rx}$, which loses no generality. Then

$$
y'' + ay' + by
= \bigl(z'' + (2r + a) z' + \chi(r) z\bigr)\eu^{rx}
= \bigl(z'' + (2r + a) z'\bigr)\eu^{rx},
$$

so $(H)$ becomes the *first-order* equation $u' + (2r + a) u = 0$ for $u = z'$.

*Case $\Delta \neq 0$:* choose $r = r_1$; then $2r_1 + a = r_1 -
r_2$ (since $r_1 + r_2 = -a$). By [Theorem 5.2](#thm-b1-diffeq-homogeneous1), $z' = c\,\eu^{(r_2 - r_1)x}$ for some constant $c$; integrating on $\R$, $z = \mu\, \eu^{(r_2 - r_1)x}
+ \lambda$ with $\mu = \frac{c}{r_2 - r_1}$, and therefore $y = z\,
\eu^{r_1 x} = \lambda\,\eu^{r_1x} + \mu\,\eu^{r_2x}$. When $\Delta < 0$, the roots are $\alpha \pm \iu\omega$ and the complex solutions are $y = c_1\eu^{(\alpha+\iu\omega)x} +
c_2\eu^{(\alpha-\iu\omega)x}$ with $c_1, c_2 \in \C$. Which of these are real-valued? Since $\conj{\eu^{(\alpha+\iu\omega)x}} =
\eu^{(\alpha-\iu\omega)x}$, the conjugate of $y$ is $\conj{c_2}\,
\eu^{(\alpha+\iu\omega)x} + \conj{c_1}\,\eu^{(\alpha-\iu\omega)x}$, and $y = \conj y$ for all $x$ forces $c_2 = \conj{c_1}$ (the two exponentials are linearly independent: evaluate at two points, or compare at $x=0$ after dividing by $\eu^{\alpha x}$). Writing $c_1
= \frac{\lambda - \iu\mu}2$ with $\lambda, \mu$ real:

$$
y = 2\,\Re\Bigl(c_1\,\eu^{\alpha x}(\cos\omega x +
\iu\sin\omega x)\Bigr)
= \eu^{\alpha x}\bigl(\lambda\cos\omega x + \mu\sin\omega
x\bigr),
$$

and conversely every such function is a solution (real part of a complex solution of a real equation): the real solution space is as announced.

*Case $\Delta = 0$:* $r = r_0$, $2r_0 + a = 0$, so $z'' = 0$: $z = \lambda + \mu x$ and $y = (\lambda + \mu x)\eu^{r_0 x}$. ∎

**Example 5.11 (A Cauchy problem, start to finish).**

Solve $y'' - 3y' + 2y = 0$ with $y(0) = 0$, $y'(0) = 1$. The [characteristic polynomial](#def-b1-diffeq-linear2) $r^2 - 3r + 2 = (r - 1)(r - 2)$ has the real roots $1$ and $2$: general solution $y = \lambda\eu^{x} +
\mu\eu^{2x}$. The two conditions give the linear system

$$
\lambda + \mu = 0, \qquad \lambda + 2\mu = 1 ,
$$

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

$$
y(x) = \eu^{2x} - \eu^{x} .
$$

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.

![The three regimes of y'' + ay' + by = 0 with decaying solutions: damped oscillation (complex roots), critical return (double root), overdamped decay (two real roots). Which regime occurs is read off the sign of = a2 - 4b alone — before solving anything.](https://one-course.com/images/onecourse/chapters/math-3/b1-diffeq/fig-42a605097459.svg)

*The three regimes of $y'' + ay' + by = 0$ with decaying solutions: damped oscillation (complex roots), critical return (double root), overdamped decay (two real roots). Which regime occurs is read off the sign of $\Delta = a^2 - 4b$ alone — before solving anything.*

**Theorem 5.12 (Structure and Cauchy problem).**

1. If $y_p$ is one particular solution of $(E)$ , the solutions of $(E)$ are exactly $y_p + y_h$ , $y_h$ running over the solutions of $(H)$ .
2. (Superposition) If $y_1$ solves $y'' + ay' + by = f_1$ and $y_2$ solves $y'' + ay' + by = f_2$ , then $y_1 + y_2$ solves the equation with right-hand side $f_1 + f_2$ .
3. For all $x_0 \in I$ and $(y_0, y_0')$ , the [Cauchy problem](#thm-b1-diffeq-voc) “ $(E)$ , $y(x_0) = y_0$ , $y'(x_0) = y_0'$ ” has exactly one solution on $I$ . *(Existence granted a particular solution; uniqueness in full.)*

**Proof.** (1) $y$ solves $(E)$ iff $y - y_p$ solves $(H)$, by linearity of $y
\mapsto y'' + ay' + by$. (2) is the same linearity.

(3) By (1) it suffices to prove that the constants $(\lambda,
\mu)$ can always be adjusted, uniquely, to any data $(y_0, y_0')$. Translating the variable, assume $x_0 = 0$. In case (1) of [Theorem 5.10](#thm-b1-diffeq-homogeneous2), $y = \lambda\eu^{r_1x} +
\mu\eu^{r_2x}$ gives

$$
y(0) = \lambda + \mu, \qquad y'(0) = r_1\lambda + r_2\mu :
$$

a linear system in $(\lambda, \mu)$ whose determinant is $r_2 -
r_1 \neq 0$; solving it explicitly, $\mu = \frac{y_0' -
r_1y_0}{r_2 - r_1}$ and $\lambda = y_0 - \mu$: exactly one solution. In case (2), $y(0) = \lambda$ and $y'(0) = r_0\lambda +
\mu$: the system is triangular with determinant $1$, solved by $\lambda = y_0$, $\mu = y_0' - r_0y_0$. In case (3), $y(0) =
\lambda$ and $y'(0) = \alpha\lambda + \omega\mu$: determinant $\omega \neq 0$, solved by $\lambda = y_0$, $\mu = \frac{y_0' -
\alpha y_0}\omega$. In each case the [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $(\lambda, \mu) \mapsto
(y(x_0), y'(x_0))$ is a linear bijection — the language of [Chapter 20](https://one-course.com/books/math/3/en/chapter/20-linear-maps#ch-b1-linmaps) will compress this case check into one sentence. ∎

**Method 5.13 (Particular solution for f(x)=P(x) eγxf(x) = P(x)\,\eu^{\gamma x}f(x)=P(x)eγx).**

When the right-hand side is $P(x)\,\eu^{\gamma x}$ with $P$ a polynomial and $\gamma \in \R$ (this covers polynomials, exponentials, and via complex $\gamma$ or superposition, $\cos$ and $\sin$): look for a particular solution of the form

$$
y_p(x) = x^{m}\, Q(x)\, \eu^{\gamma x},
\qquad
m = \text{multiplicity of } \gamma \text{ as a root of } \chi
\ (m = 0, 1 \text{ or } 2),
$$

with $Q$ a polynomial of the same degree as $P$, whose coefficients are found by substitution and identification. For $f = K\cos\omega x$ (or $\sin$), solve with right-hand side $K\eu^{\iu\omega x}$ and take the real (resp. imaginary) part.

**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](#thm-b1-diffeq-structure2) (2)):

$$
y = \frac{x\,\eu^x}2 - 4 + \lambda\,\eu^x + \mu\,\eu^{-x},
\qquad (\lambda, \mu) \in \R^2 .
$$

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:

$$
y_p'' + y_p' = 2\alpha + (2\alpha x + \beta)
= 2\alpha x + (2\alpha + \beta) ,
$$

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.

**Remark 5.16 (The thirty-second insurance policy).**

Every solved equation in this chapter ends with a substitution check, and this is not decorative. A differential-equation computation chains many small steps (a primitive, a product rule, two constants), and a single sign error propagates invisibly; substituting the final formula back into the equation catches essentially all of them at the cost of one differentiation. Cultivate the reflex in three layers: check the *particular solution* alone (the homogeneous part cancels anyway), check the *initial conditions* on the full solution, and when a parameter is present, check a *degenerate value* (does the formula for general $\Omega$ reproduce the known answer at $\Omega = 0$?). The habit costs half a minute; it converts “probably right” into “verified”.

**Remark 5.17 (Common pitfalls).**

1. *Normalize first.* The formulas assume the equation reads $y' + a(x)y = b(x)$ — coefficient $1$ on $y'$ . For $xy' - 2y = x^3$ , divide by $x$ (on an interval avoiding $0$ ) before identifying $a$ and $b$ , as in [Exercise 5.2](#exo-b1-diffeq-2) .
2. *One constant per dimension, fixed at the end.* The general first-order solution carries one constant, the second-order one two; initial conditions are imposed on the *complete* solution $y_p + y_h$ , never on $y_h$ alone — imposing them before adding $y_p$ is the most frequent structural error.
3. *Mind the multiplicity.* A particular-solution guess that solves the homogeneous equation is invisible to the left-hand side; the factor $x^m$ of [Method 5.13](#met-b1-diffeq-particular) is not optional ( [Example 5.15](#ex-b1-diffeq-multiplicity) ).
4. *Intervals are part of the answer.* Solutions live on intervals where the coefficients are continuous; gluing across a singularity can create spurious constants ( [Exercise 5.12](#exo-b1-diffeq-12) ) or destroy uniqueness. “Solve on $\R^*$ ” means two independent problems.

**Example 5.18 (Off-resonance forcing).**

Solve $y'' + 4y = \sin x$ on $\R$. Natural frequency $2$, forcing frequency $1$: since $\iu$ is *not* a root of $\chi(r) = r^2
+ 4$, the multiplicity is $m = 0$ and a plain sinusoid suffices. Trying $y_p = \alpha\sin x$ (no cosine needed: the equation has no $y'$ term, and $\sin$ regenerates $\sin$):

$$
y_p'' + 4y_p = -\alpha\sin x + 4\alpha\sin x = 3\alpha\sin x ,
$$

so $\alpha = \frac13$ and the general solution is

$$
y = \frac{\sin x}3 + \lambda\cos 2x + \mu\sin 2x .
$$

Every solution stays bounded: a superposition of two oscillations at the frequencies $1$ (forced) and $2$ (natural). Compare with the next example, where forcing *at* the natural frequency changes the answer’s very shape.

**Example 5.19 (A forced oscillation).**

Solve $y'' + y = \cos x$, $y(0) = 0$, $y'(0) = 0$.

*Homogeneous:* $\chi(r) = r^2 + 1$, roots $\pm\iu$: $y_h =
\lambda\cos x + \mu \sin x$.

*Particular:* right-hand side $\Re(\eu^{\iu x})$ with $\gamma =
\iu$ a simple root of $\chi$: try $z_p = c\, x\, \eu^{\iu x}$ ($c \in \C$). Then $z_p'' + z_p = c\,(2\iu)\eu^{\iu x}$, which equals $\eu^{\iu x}$ for $c = \frac{1}{2\iu} = -\frac\iu2$. So $z_p =
-\frac{\iu}{2} x (\cos x + \iu \sin x)$ and $y_p = \Re(z_p) =
\frac{x \sin x}{2}$.

*General solution:* $y = \frac{x\sin x}{2} + \lambda\cos x +
\mu\sin x$. Conditions: $y(0) = \lambda = 0$; $y' = \frac{\sin x +
x\cos x}{2} + \mu\cos x$, so $y'(0) = \mu = 0$. Answer: $y =
\frac{x\sin x}{2}$ — an oscillation whose amplitude grows linearly: the *resonance* phenomenon, caused by forcing the system at its natural frequency.

![Resonance: the solution y = x x/2 of y'' + y = x oscillates between the lines y = ± x2 (dashed), with ever-growing amplitude.](https://one-course.com/images/onecourse/chapters/math-3/b1-diffeq/fig-7a7af5de9757.svg)

*[Resonance](#ex-b1-diffeq-oscillation): the solution $y = \frac{x \sin x}{2}$ of $y'' + y =
\cos x$ oscillates between the lines $y = \pm\frac x2$ (dashed), with ever-growing amplitude.*

**Remark 5.20 (Interlude: linearity is a geometry).**

Look back at the shape of every solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) in this chapter: a special solution plus a space of homogeneous solutions with one free constant (first order) or two (second order). The chapters on linear algebra (Chapters [18](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#ch-b1-vspaces), [19](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#ch-b1-findim) and [20](https://one-course.com/books/math/3/en/chapter/20-linear-maps#ch-b1-linmaps)) will supply the exact vocabulary: the [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $L(y) = y'' + ay' + by$ is *linear*, its homogeneous solutions form the *kernel* of $L$, a vector space whose *dimension* equals the order of the equation — that is the honest content of “one constant per order” — and the solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of $L(y) = f$ is an *affine subspace*, a translate of the kernel. Even the Cauchy [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $(\lambda, \mu)
\mapsto (y(x_0), y'(x_0))$ of [Theorem 5.12](#thm-b1-diffeq-structure2) is a linear bijection between two planes, i.e. an invertible $2
\times 2$ system ([Chapter 21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices)). Nothing in this chapter will need to be redone — it will only need to be renamed, and the renaming is the best possible warm-up for linear algebra: every abstract definition there has already earned its living here.

**Remark 5.21 (Where this chapter is used).**

The structure theorem — solutions of $(E)$ form “a particular solution plus the solutions of $(H)$” — is the first appearance of a pattern that Chapters [18](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#ch-b1-vspaces) and [20](https://one-course.com/books/math/3/en/chapter/20-linear-maps#ch-b1-linmaps) will name: the solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of $(H)$ is the *kernel* of the linear [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $y
\mapsto y'' + ay' + by$, and the solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of $(E)$ is an affine translate of it. The [characteristic polynomial](#def-b1-diffeq-linear2) reappears as the [characteristic polynomial](#def-b1-diffeq-linear2) of a matrix in [Chapter 21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices): a second-order equation is a $2 \times 2$ first-order system in disguise, a viewpoint the Year 2 volume systematizes. The integrals demanded by [variation of constants](#thm-b1-diffeq-voc) are supplied by [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration), and the weekend problem below — the driven damped oscillator — is the model case for every oscillation question in the sciences, from circuits to suspension bridges.

## 5.3 Exercises

**Exercise 5.1 ★.**

Solve on $\R$: $\;y' + 2y = \eu^{3x}$; then the [Cauchy problem](#thm-b1-diffeq-voc) $y(0) =
1$.

**Solution of Exercise 5.1.**

Homogeneous: $y_h = \lambda\,\eu^{-2x}$. Particular: try $y_p =
c\,\eu^{3x}$ ($3$ is not a root of $r + 2$): $3c + 2c = 1$, $c =
\frac15$. General solution: $y = \frac{\eu^{3x}}{5} +
\lambda\,\eu^{-2x}$. With $y(0) = 1$: $\frac15 + \lambda = 1$, $\lambda = \frac45$, so $y = \frac{\eu^{3x} + 4\,\eu^{-2x}}{5}$.

**Exercise 5.2 ★.**

Solve on $\intoo{0}{+\infty}$: $\;x y' - 2y = x^3$ *(put the equation in the normalized form first)*.

**Solution of Exercise 5.2.**

On $\intoo{0}{+\infty}$, divide by $x$: $y' - \frac{2}{x}\,y = x^2$. Here $A(x) = -2\ln x$, $\eu^{-A(x)} = x^2$: homogeneous solutions $\lambda x^2$. [Variation of constants](#thm-b1-diffeq-voc): $\mu'(x) = x^2 \cdot x^{-2} =
1$, so $\mu = x + \lambda$ and

$$
y(x) = x^3 + \lambda x^2, \qquad \lambda \in \R .
$$

*Check:* $x(3x^2 + 2\lambda x) - 2(x^3 + \lambda x^2) = x^3$.

**Exercise 5.3 ★.**

Solve on $\R$, giving the real general solution: $\;y'' - 3y' + 2y = 0$; $\;y'' + 4y' + 4y = 0$; $\;y'' - 2y' + 5y =
0$.

**Solution of Exercise 5.3.**

$y'' - 3y' + 2y = 0$: roots $1$ and $2$; $\;y = \lambda\,\eu^{x} +
\mu\,\eu^{2x}$.

$y'' + 4y' + 4y = 0$: double root $-2$; $\;y = (\lambda + \mu
x)\,\eu^{-2x}$.

$y'' - 2y' + 5y = 0$: roots $1 \pm 2\iu$; $\;y = \eu^{x}(\lambda\cos
2x + \mu\sin 2x)$.

**Exercise 5.4 ★.**

Solve $y'' - y = x^2$ on $\R$, then the [Cauchy problem](#thm-b1-diffeq-voc) $y(0) = 0$, $y'(0) = 1$.

**Solution of Exercise 5.4.**

Homogeneous: roots $\pm 1$, $y_h = \lambda\,\eu^x + \mu\,\eu^{-x}$. Particular with polynomial right-hand side ($\gamma = 0$ not a root): $y_p = ax^2 + bx + c$; substituting, $2a - (ax^2 + bx + c) = x^2$ gives $a = -1$, $b = 0$, $c = 2a = -2$: $y_p = -x^2 - 2$. General solution $y = -x^2 - 2 + \lambda\eu^x + \mu\eu^{-x}$.

Cauchy: $y(0) = -2 + \lambda + \mu = 0$ and $y'(0) = \lambda - \mu =
1$: $\lambda = \frac32$, $\mu = \frac12$. So $y = -x^2 - 2 + \frac{3\eu^x + \eu^{-x}}{2}$.

**Exercise 5.5 ★★.**

Solve on $\intoo{-\frac\pi2}{\frac\pi2}$: $\;y' + y\tan x = \sin 2x$.

**Solution of Exercise 5.5.**

$a(x) = \tan x$, $A(x) = -\ln(\cos x)$ (valid: $\cos > 0$ on the interval), $\eu^{-A} = \cos x$: homogeneous solutions $\lambda\cos x$. [Variation of constants](#thm-b1-diffeq-voc): $\mu'(x) = \sin 2x \cdot \frac{1}{\cos x} =
2\sin x$, so $\mu = -2\cos x + \lambda$ and

$$
y(x) = -2\cos^2 x + \lambda \cos x .
$$

*Check:* $y' = 4\cos x \sin x - \lambda\sin x$ and $y\tan x =
-2\cos x\sin x + \lambda \sin x$; their sum is $2\cos x\sin x =
\sin 2x$, as required.

**Exercise 5.6 ★★.**

Solve $y'' - 4y' + 3y = (2x + 1)\,\eu^{x}$ on $\R$. *(Mind the multiplicity: is $1$ a root of the [characteristic polynomial](#def-b1-diffeq-linear2)?)*

**Solution of Exercise 5.6.**

$\chi(r) = r^2 - 4r + 3 = (r-1)(r-3)$: $\gamma = 1$ is a simple root ($m = 1$). Try $y_p = x(ax + b)\,\eu^x$. With $u = ax^2 + bx$,

$$
y_p'' - 4y_p' + 3y_p = \bigl(u'' + (2 - 4)u' + \chi(1) u\bigr)\eu^x
= \bigl(2a - 2(2ax + b)\bigr)\eu^x .
$$

Identify with $(2x + 1)\eu^x$: $-4a = 2$ and $2a - 2b = 1$, so $a =
-\frac12$, $b = -1$. General solution:

$$
y = -\Bigl(\frac{x^2}{2} + x\Bigr)\eu^{x} + \lambda\,\eu^{x} +
\mu\,\eu^{3x}, \qquad (\lambda, \mu) \in \R^2 .
$$

**Exercise 5.7 ★★.**

Solve $y'' + 4y = \sin 2x + x$ on $\R$ *(superposition; treat each right-hand side separately)*.

**Solution of Exercise 5.7.**

Homogeneous: $y_h = \lambda\cos 2x + \mu\sin 2x$.

Right-hand side $x$ ($\gamma = 0$ not a root): $y_1 = ax + b$ with $4(ax + b) = x$: $y_1 = \frac x4$.

Right-hand side $\sin 2x = \Im(\eu^{2\iu x})$, $2\iu$ simple root of $r^2 + 4$: try $z = c\,x\,\eu^{2\iu x}$; then $z'' + 4z =
4\iu c\,\eu^{2\iu x}$, equal to $\eu^{2\iu x}$ for $c =
\frac{1}{4\iu} = -\frac{\iu}{4}$. So $z = -\frac{\iu x}{4}(\cos 2x +
\iu \sin 2x)$ and $y_2 = \Im(z) = -\frac{x\cos 2x}{4}$.

By superposition:

$$
y = \frac{x}{4} - \frac{x\cos 2x}{4} + \lambda\cos 2x + \mu\sin 2x .
$$

**Exercise 5.8 ★★.**

A cup of coffee at temperature $T_0 = 80\,^\circ$C sits in a room at $20\,^\circ$C. Newton’s law of cooling states $T' = -k\,(T - 20)$ with $k > 0$. Solve for $T(t)$, and given that the coffee is at $50\,^\circ$C after $10$ minutes, find when it reaches $25\,^\circ$C.

**Solution of Exercise 5.8.**

The equation $T' + kT = 20k$ has constant particular solution $20$ and homogeneous solutions $\lambda\eu^{-kt}$: $T(t) = 20 +
\lambda\,\eu^{-kt}$, and $T(0) = 80$ gives $\lambda = 60$:

$$
T(t) = 20 + 60\,\eu^{-kt} .
$$

$T(10) = 50$: $\eu^{-10k} = \frac12$, so $k = \frac{\ln 2}{10}$. Then $T(t) = 25$ requires $\eu^{-kt} = \frac{5}{60} = \frac{1}{12}$, i.e.

$$
t = \frac{\ln 12}{k} = 10\,\frac{\ln 12}{\ln 2} \approx 35.8
\text{ minutes.}
$$

**Exercise 5.9 ★★★.**

(Damped oscillator) For $\varepsilon \geq 0$, consider $y'' + 2\varepsilon y' + y = 0$.

1. Solve for $\varepsilon \in \intco{0}{1}$ , $\varepsilon = 1$ , and $\varepsilon > 1$ .
2. Show that for $\varepsilon > 0$ every solution tends to $0$ at $+\infty$ , and that for $\varepsilon = 0$ the nonzero solutions do not.
3. For $\varepsilon \in \intoo{0}{1}$ , show that the zeros of a nonzero solution are regularly spaced, with gap $\frac{\pi}{\sqrt{1 - \varepsilon^2}}$ .

**Solution of Exercise 5.9.**

1. $\chi(r) = r^2 + 2\varepsilon r + 1$ , $\Delta = 4(\varepsilon^2  - 1)$ . For $\varepsilon \in \intco{0}{1}$ : roots $-\varepsilon \pm  \iu\sqrt{1 - \varepsilon^2}$ , so $y = \eu^{-\varepsilon t}\bigl(\lambda\cos\omega t +  \mu\sin\omega t\bigr)$ with $\omega = \sqrt{1 -  \varepsilon^2}$ . For $\varepsilon = 1$ : double root $-1$ , $y = (\lambda + \mu  t)\,\eu^{-t}$ . For $\varepsilon > 1$ : real roots $r_\pm = -\varepsilon \pm  \sqrt{\varepsilon^2 - 1}$ , both $< 0$ , and $y =  \lambda\eu^{r_+t} + \mu\eu^{r_-t}$ .
2. For $\varepsilon \in \intoo{0}{1}$ : $\abs y \leq  \eu^{-\varepsilon t}(\abs\lambda + \abs\mu) \to 0$ . For $\varepsilon = 1$ : $(\lambda + \mu t)\eu^{-t} \to 0$ (exponential beats polynomial, [Proposition 4.6](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules) ). For $\varepsilon > 1$ : both exponentials decay since $r_\pm < 0$ (indeed $\sqrt{\varepsilon^2 - 1} < \varepsilon$ ). For $\varepsilon =  0$ : $y = \lambda\cos t + \mu\sin t$ has constant amplitude $\sqrt{\lambda^2 + \mu^2} \neq 0$ unless $y = 0$ .
3. Write $\lambda\cos\omega t + \mu\sin\omega t =  R\cos(\omega t - \varphi)$ with $R = \sqrt{\lambda^2 + \mu^2}  > 0$ . The zeros of $y$ are those of $\cos(\omega t -  \varphi)$ (the factor $\eu^{-\varepsilon t}$ never vanishes): $\omega t - \varphi \equiv \frac\pi2 \pmod \pi$ , an arithmetic progression with gap $\frac{\pi}{\omega} =  \frac{\pi}{\sqrt{1 - \varepsilon^2}}$ .

**Exercise 5.10 ★★★.**

Find all functions $f \colon \R \to \R$, twice differentiable, such that

$$
\forall x, y \in \R, \qquad f(x + y) + f(x - y) = 2 f(x) f(y),
$$

with $f(0) \ne 0$ and $f$ not constant. *Hint: fix $y$, differentiate twice with respect to $x$ at $0$; show $f(0) = 1$ and $f'' = c f$ for some constant $c$; then solve according to the sign of $c$ and check which solutions satisfy the functional equation.*

**Solution of Exercise 5.10.**

Set $x = y = 0$: $2f(0) = 2f(0)^2$, and $f(0) \neq 0$ gives $f(0) =
1$. Fix $x$ and differentiate the equation twice with respect to $y$:

$$
f''(x+y) + f''(x-y) = 2 f(x) f''(y) .
$$

Setting $y = 0$: $\;2f''(x) = 2 f(x) f''(0)$, that is

$$
f''(x) = c\,f(x), \qquad c = f''(0).
$$

*Case $c = \omega^2 > 0$:* $f(x) = \lambda\cosh\omega x +
\mu\sinh\omega x$; $f(0) = 1$ gives $\lambda = 1$. Plugging into the functional equation and using the addition formulas ([Proposition 4.18](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-hyprules)), the equation forces $\mu = 0$ (compare the coefficients of $\sinh\omega x \sinh\omega y$ or evaluate at $x = y$): $f = \cosh\omega x$, which does satisfy $\cosh(x+y) + \cosh(x-y) = 2\cosh x\cosh y$.

*Case $c = -\omega^2 < 0$:* similarly $f(x) = \cos\omega x$ ($\omega \neq 0$), which satisfies the equation.

*Case $c = 0$:* $f$ affine with $f(0) = 1$: $f(x) = 1 + \mu x$; the equation forces $\mu = 0$, excluded ($f$ not constant).

Conclusion: the solutions are $f(x) = \cos\omega x$ and $f(x) =
\cosh\omega x$, $\omega > 0$.

**Exercise 5.11 ★★.**

(Euler equation) Solve $x^2 y'' - x y' + y = 0$ on $\intoo{0}{+\infty}$. *Hint: set $z(t) = y(\eu^t)$, i.e. substitute $x = \eu^t$, and show that $z$ satisfies a linear equation with constant coefficients.*

**Solution of Exercise 5.11.**

Set $z(t) = y(\eu^t)$, so that $y(x) = z(\ln x)$ for $x > 0$. Then

$$
y'(x) = \frac{z'(\ln x)}x,
\qquad
y''(x) = \frac{z''(\ln x) - z'(\ln x)}{x^2} ,
$$

and substituting into the equation:

$$
x^2y'' - xy' + y = \bigl(z'' - z'\bigr) - z' + z
= z'' - 2z' + z = 0 .
$$

[Characteristic polynomial](#def-b1-diffeq-linear2) $(r - 1)^2$: double root $1$, so $z(t) =
(\lambda + \mu t)\,\eu^t$ and, back in the variable $x = \eu^t$:

$$
y(x) = (\lambda + \mu\ln x)\,x,
\qquad \lambda, \mu \in \R .
$$

**Exercise 5.12 ★★★.**

Consider the equation $x\,y' = 2y$ on the whole real line, in the unknown differentiable function $y \colon \R \to \R$.

1. Solve on $\intoo{0}{+\infty}$ and on $\intoo{-\infty}{0}$ .
2. Show that for *any* constants $a, b \in \R$ , the function equal to $ax^2$ for $x \geq 0$ and to $bx^2$ for $x < 0$ is differentiable on $\R$ and solves the equation everywhere.
3. Conclude that the solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) on $\R$ is a two-parameter family, and explain why this does not contradict the uniqueness in [Theorem 5.4](#thm-b1-diffeq-voc) .

**Solution of Exercise 5.12.**

1. In normalized form $y' - \frac2x\,y = 0$ on each interval: $A(x) = -2\ln\abs x$ , so the solutions are $y = a x^2$ on $\intoo0{+\infty}$ and $y = b x^2$ on $\intoo{-\infty}0$ , with independent constants ( [Theorem 5.2](#thm-b1-diffeq-homogeneous1) ).
2. Let $y = ax^2$ for $x \geq 0$ and $bx^2$ for $x < 0$ . On each open half-line $y$ is differentiable with $xy' = 2y$ . At $0$ : the difference quotients $\frac{y(h) - y(0)}h =  ah$ or $bh$ tend to $0$ , so $y'(0) = 0$ exists, and the equation at $x = 0$ reads $0 \cdot y'(0) = 2y(0) = 0$ : satisfied. So $y$ solves the equation on all of $\R$ .
3. The solutions on $\R$ are exactly these glued functions: a *two* -parameter family for a first-order equation. There is no contradiction with [Theorem 5.4](#thm-b1-diffeq-voc) , whose hypotheses fail here: written as $y' + a(x)y = 0$ , the coefficient $a(x) = -\frac2x$ is not continuous at $0$ — indeed not defined — so $\R$ is not an interval on which the theorem applies. The singularity at $0$ disconnects the two half-lines, and the value $y(0) = 0$ is forced, carrying no information across. Every Cauchy datum at $x_0 \neq 0$ determines the solution only on the half-line containing $x_0$ .

## 5.4 Problem: The driven damped oscillator

**Problem 5.1.**

One equation governs a mass on a spring in a viscous medium, the charge in an RLC circuit, and a building swaying in the wind:

$$
(E_\Omega)\colon\quad
x'' + 2\lambda x' + \omega_0^2\,x = A\cos(\Omega t),
$$

with $\lambda \geq 0$ the damping, $\omega_0 > 0$ the natural frequency, and $A > 0$, $\Omega > 0$ the amplitude and frequency of the forcing. This problem extracts its complete behavior: the decay of transients, the unique periodic steady state, the [resonance](#ex-b1-diffeq-oscillation) curve and its sharpness (the *[quality factor](#pb-b1-diffeq-1)*), the beats of the undamped case, and the energy balance that sustains the oscillation. Unless stated otherwise, $0 < \lambda < \omega_0$ (underdamped regime) and we write $\omega_d = \sqrt{\omega_0^2 - \lambda^2}$.

**Part I — The free oscillator.** Here $A = 0$.

1. Solve the homogeneous equation $(H)$ for $0 < \lambda <  \omega_0$ , and for $\lambda = 0$ . (The regimes $\lambda  \geq \omega_0$ were treated in [Exercise 5.9](#exo-b1-diffeq-9) ; quote them.)
2. Show that for every $\lambda > 0$ , all solutions of $(H)$ tend to $0$ at $+\infty$ — in all three regimes.
3. Define the energy $\mathcal E(t) = \frac12 x'(t)^2 +  \frac12\omega_0^2\,x(t)^2$ along a solution of $(H)$ . Show $\mathcal E'(t) = -2\lambda\,x'(t)^2 \leq 0$ , and deduce (without solving anything) that the [Cauchy problem](#thm-b1-diffeq-voc) “ $(H)$ , $x(t_0) = x'(t_0) = 0$ ” has only the zero solution, for every $\lambda \geq 0$ .
4. For $0 < \lambda < \omega_0$ , write the nonzero solution as $x(t) = R\,\eu^{-\lambda t}\cos(\omega_d t - \varphi)$ and let $T_d = \frac{2\pi}{\omega_d}$ be the pseudo-period. Show that $x(t + T_d) = \eu^{-\lambda T_d}\,x(t)$ : each swing is the previous one shrunk by the constant factor $\eu^{-\delta}$ , $\delta = \frac{2\pi\lambda}{\omega_d}$ (the *logarithmic decrement* ). Compute $\delta$ for $\omega_0 = 1$ , $\lambda = 0.1$ .
5. Define the *[quality factor](#pb-b1-diffeq-1)* $Q =  \dfrac{\omega_0}{2\lambda}$ . Show that after the time $\frac1\lambda$ (one amplitude $\eu$ -folding), the oscillator has completed $\frac{\omega_d}{2\pi\lambda}$ pseudo-periods, which for weak damping ( $\lambda \ll \omega_0$ ) is approximately $\frac Q\pi$ : the [quality factor](#pb-b1-diffeq-1) counts, up to $\pi$ , the oscillations survived before the amplitude decays by $\eu$ .

**Part II — The steady state.** Now $A > 0$ and $\lambda > 0$.

6. Look for a particular solution as the real part of $z\,\eu^{\iu\Omega t}$ with $z \in \C$ ([Method 5.13](#met-b1-diffeq-particular)). Show that this works with $$z = \frac{A}{\omega_0^2 - \Omega^2 + 2\iu\lambda\Omega} .$$
7. Deduce the steady state in amplitude–phase form: $x_p(t) = R(\Omega)\cos\bigl(\Omega t -  \varphi(\Omega)\bigr)$ with $$R(\Omega) = \frac{A}{\sqrt{(\omega_0^2 - \Omega^2)^2 +  4\lambda^2\Omega^2}},  \qquad  \tan\varphi = \frac{2\lambda\Omega}{\omega_0^2 -  \Omega^2},  \quad \varphi \in \intoo0\pi .$$
8. Interpret the two extreme regimes: compute the limits of $R$ and $\varphi$ as $\Omega \to 0^+$ (quasi-static response $A/\omega_0^2$ , phase $0$ ) and as $\Omega \to  +\infty$ ( $R \sim A/\Omega^2 \to 0$ , phase $\to \pi$ : the mass moves opposite to a too-fast forcing).
9. Show that *every* solution of $(E_\Omega)$ is $x_p$ plus a solution of $(H)$ , hence converges to the steady state $x_p$ as $t \to +\infty$ , whatever the initial conditions: after the transient dies, the oscillator has no memory of how it started.
10. Show that $x_p$ is the *only* periodic solution of $(E_\Omega)$ .
11. Work one [Cauchy problem](#thm-b1-diffeq-voc) to the end: for $x'' + 2x' + 2x =  \cos t$ with $x(0) = x'(0) = 0$, show that the solution is $$x(t) = \frac{\cos t + 2\sin t}5  - \eu^{-t}\,\frac{\cos t + 3\sin t}5 ,$$ and identify transient and steady parts.

**Part III — The [resonance](#ex-b1-diffeq-oscillation) curve.** Study of $\Omega \mapsto R(\Omega)$ on $\intoo0{+\infty}$.

12. Setting $u = \Omega^2$ and $g(u) = (\omega_0^2 - u)^2 +  4\lambda^2 u$, show: if $2\lambda^2 < \omega_0^2$, then $R$ attains a strict maximum at the *[resonance](#ex-b1-diffeq-oscillation) frequency* $\Omega_r = \sqrt{\omega_0^2 - 2\lambda^2}$, with $$R_{\max} = R(\Omega_r)  = \frac{A}{2\lambda\sqrt{\omega_0^2 - \lambda^2}} .$$
13. Show that $\dfrac{R_{\max}}{R(0)} = Q\,\bigl(1 -  \tfrac{\lambda^2}{\omega_0^2}\bigr)^{-1/2} \geq Q$ : at [resonance](#ex-b1-diffeq-oscillation) , the forcing is amplified by (essentially) the [quality factor](#pb-b1-diffeq-1) .
14. Prove that the *velocity* amplitude $V(\Omega) =  \Omega\,R(\Omega)$ is maximal exactly at $\Omega =  \omega_0$ (not at $\Omega_r$ ), and that the phase there is $\varphi(\omega_0) = \frac\pi2$ : at $\Omega = \omega_0$ the velocity is exactly in phase with the force.
15. (Bandwidth) Solve $g(u) = 2\,g(u_r)$ exactly, where $u_r =  \omega_0^2 - 2\lambda^2$ , and deduce that the two frequencies $\Omega_\pm$ where $R = R_{\max}/\sqrt2$ satisfy $\Omega_+^2 - \Omega_-^2 =  4\lambda\sqrt{\omega_0^2 - \lambda^2}$ ; conclude that for weak damping the bandwidth is $\Omega_+ - \Omega_-  \approx 2\lambda$ , i.e. $Q \approx  \frac{\omega_0}{\Omega_+ - \Omega_-}$ : sharp [resonance](#ex-b1-diffeq-oscillation) peaks are high- $Q$ systems.
16. Numerical portrait for $\omega_0 = 1$ , $\lambda = 0.05$ ( $Q = 10$ ), $A = 1$ : compute $\Omega_r$ , $R_{\max}$ , the static response $R(0)$ , and the approximate bandwidth.
17. Show that if $2\lambda^2 \geq \omega_0^2$ , then $R$ is strictly decreasing on $\intoo0{+\infty}$ : heavily damped systems have no [resonance](#ex-b1-diffeq-oscillation) peak at all.

**Part IV — No damping: beats and [resonance](#ex-b1-diffeq-oscillation).** Here $\lambda = 0$.

18. For $\Omega \neq \omega_0$ , find the general solution of $x'' + \omega_0^2 x = A\cos(\Omega t)$ .
19. Solve the [Cauchy problem](#thm-b1-diffeq-voc) $x(0) = x'(0) = 0$ and transform the answer into the product form $$x(t) = \frac{2A}{\omega_0^2 - \Omega^2}\,  \sin\Bigl(\frac{(\omega_0 - \Omega)t}2\Bigr)  \sin\Bigl(\frac{(\omega_0 + \Omega)t}2\Bigr) .$$
20. For $\Omega$ close to $\omega_0$ , read the product as a fast oscillation at frequency $\frac{\omega_0 + \Omega}2$ modulated by a slow envelope at frequency $\frac{\abs{\omega_0 - \Omega}}2$ : the *beats* . Give the period of the envelope and the maximal amplitude, and note how both blow up as $\Omega  \to \omega_0$ .
21. Fix $t$ and let $\Omega \to \omega_0$ in question 19’s formula: show the limit is $$x_\infty(t) = \frac{A\,t\,\sin(\omega_0 t)}{2\omega_0} ,$$ and check directly that $x_\infty$ solves the resonant equation $x'' + \omega_0^2 x = A\cos(\omega_0 t)$ with $x(0) = x'(0) = 0$ (compare [Example 5.19](#ex-b1-diffeq-oscillation)): [resonance](#ex-b1-diffeq-oscillation) is the limit of ever-slower, ever-larger beats.
22. Contrast the two fates of [resonance](#ex-b1-diffeq-oscillation) : linear growth $\frac{At}{2\omega_0}$ without damping, versus saturation at $R_{\max} \approx Q\,\frac{A}{\omega_0^2}$ with weak damping. In one sentence: what physical mechanism converts the first into the second?

**Part V — Energy balance and synthesis.**

23. In the steady state of Part II, compute the average over one period $\frac{2\pi}\Omega$ of (a) the power injected by the forcing, $P_{\mathrm{in}}(t) = A\cos(\Omega t)  \cdot x_p'(t)$ , and (b) the power dissipated by the damping, $P_{\mathrm{diss}}(t) = 2\lambda\,x_p'(t)^2$ . Show both averages equal $\lambda\,R^2\Omega^2$ : the forcing feeds in exactly what the damping burns — this is why the steady state is steady.
24. Where exactly did the problem use: (i) the structure theorem [Theorem 5.12](#thm-b1-diffeq-structure2) ; (ii) the complex exponential method; (iii) a real-variable function study in the style of [Chapter 4](https://one-course.com/books/math/3/en/chapter/4-standard-functions#ch-b1-functions) ? One sentence each.
25. Synthesis: describe the full behavior [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) of $(E_\Omega)$ — free versus forced, damped versus undamped, the role of $Q$ as the single dimensionless dial tuning peak height, bandwidth and transient lifetime — and mention where the story continues: first-order $2 \times 2$ systems ( [Chapter 21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices) and the Year 2 volume) and the decomposition of a general periodic forcing into sinusoids (Fourier series, in the Year 3 volume), for which this problem’s sinusoidal case is the fundamental building block.

**Solution of Problem 5.1.**

**1.** $\chi(r) = r^2 + 2\lambda r + \omega_0^2$, $\Delta =
4(\lambda^2 - \omega_0^2) < 0$ for $0 < \lambda < \omega_0$: roots $-\lambda \pm \iu\omega_d$, so by [Theorem 5.10](#thm-b1-diffeq-homogeneous2)

$$
x(t) = \eu^{-\lambda t}\bigl(\lambda_1\cos\omega_d t +
\mu_1\sin\omega_d t\bigr), \qquad (\lambda_1, \mu_1) \in \R^2 .
$$

For $\lambda = 0$: $x = \lambda_1\cos\omega_0 t +
\mu_1\sin\omega_0 t$. The critical ($\lambda = \omega_0$) and overdamped ($\lambda > \omega_0$) regimes are those of [Exercise 5.9](#exo-b1-diffeq-9) (after rescaling time): $(\lambda_1 +
\mu_1 t)\eu^{-\lambda t}$, resp. combinations of $\eu^{r_\pm t}$ with $r_\pm = -\lambda \pm \sqrt{\lambda^2 - \omega_0^2}$.

**2.** Underdamped: $\abs x \leq \eu^{-\lambda
t}(\abs{\lambda_1} + \abs{\mu_1}) \to 0$. Critical: $(\lambda_1 +
\mu_1 t)\eu^{-\lambda t} \to 0$ since exponentials beat polynomials ([Proposition 4.6](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules)). Overdamped: $r_- < r_+ = -\lambda + \sqrt{\lambda^2 - \omega_0^2} < 0$ because $\sqrt{\lambda^2 - \omega_0^2} < \lambda$; both exponentials decay.

**3.** Along a solution of $(H)$, using $x'' = -2\lambda x' -
\omega_0^2 x$:

$$
\mathcal E'(t) = x'x'' + \omega_0^2 x x'
= x'\bigl(-2\lambda x' - \omega_0^2 x\bigr) + \omega_0^2 xx'
= -2\lambda\,x'^2 \leq 0 .
$$

If $x(t_0) = x'(t_0) = 0$ then $\mathcal E(t_0) = 0$; $\mathcal E$ is nonnegative and nonincreasing, so $\mathcal E \equiv 0$ on $\intco{t_0}{+\infty}$, forcing $x \equiv 0$ there; for $t \leq
t_0$, run the same argument on $\tilde x(t) = x(2t_0 - t)$, which solves the equation with damping $-\lambda$ but still has $\tilde{\mathcal E}(t_0) = 0$ and $\tilde{\mathcal E}' =
+2\lambda\tilde x'^2 \geq 0$ with $\tilde{\mathcal E} \geq 0$; nonnegative, nondecreasing and zero at the right end of $\intoc{-\infty}{t_0}$ means zero throughout. So $x \equiv 0$ on $\R$ — an energy proof of uniqueness, valid for all $\lambda
\geq 0$.

**4.** $x(t + T_d) = R\,\eu^{-\lambda t}\eu^{-\lambda
T_d}\cos(\omega_d t + 2\pi - \varphi) = \eu^{-\lambda T_d}x(t)$. The shrink factor per pseudo-period is $\eu^{-\delta}$ with $\delta = \lambda T_d = \frac{2\pi\lambda}{\omega_d}$. For $\omega_0 = 1$, $\lambda = 0.1$: $\omega_d = \sqrt{0.99} =
0.99499$, so $\delta = \frac{0.62832}{0.99499} = 0.6315$: each swing keeps $\eu^{-0.63} \approx 53\%$ of its amplitude.

**5.** The amplitude factor is $\eu^{-\lambda t}$, which decays by $\eu$ over $t = \frac1\lambda$. That interval contains $\frac{1/\lambda}{T_d} = \frac{\omega_d}{2\pi\lambda}$ pseudo-periods. For $\lambda \ll \omega_0$, $\omega_d \approx
\omega_0$ and this is $\approx \frac{\omega_0}{2\pi\lambda} =
\frac Q\pi$. A $Q = 300$ guitar string rings for about a hundred periods; a $Q = 1$ door damper does not complete one.

**6.** Substituting $z\,\eu^{\iu\Omega t}$ into the left-hand side gives $z\,(-\Omega^2 + 2\iu\lambda\Omega +
\omega_0^2)\,\eu^{\iu\Omega t}$, which equals $A\,\eu^{\iu\Omega
t}$ exactly for $z = \frac{A}{\omega_0^2 - \Omega^2 +
2\iu\lambda\Omega}$ (the denominator is nonzero: its imaginary part is $2\lambda\Omega > 0$). Since the coefficients are real, the real part $x_p = \Re\bigl(z\eu^{\iu\Omega t}\bigr)$ solves the equation with right-hand side $\Re\bigl(A\eu^{\iu\Omega t}\bigr) =
A\cos\Omega t$.

**7.** Write $\omega_0^2 - \Omega^2 + 2\iu\lambda\Omega =
\sqrt{D}\,\eu^{\iu\varphi}$ with $D = (\omega_0^2 - \Omega^2)^2 +
4\lambda^2\Omega^2$ and $\varphi \in \intoo0\pi$ (the imaginary part $2\lambda\Omega$ is positive), so $\tan\varphi =
\frac{2\lambda\Omega}{\omega_0^2 - \Omega^2}$. Then $z =
\frac{A}{\sqrt D}\eu^{-\iu\varphi}$ and

$$
x_p(t) = \Re\Bigl(\frac A{\sqrt D}\,\eu^{\iu(\Omega t -
\varphi)}\Bigr) = R\cos(\Omega t - \varphi),
\qquad R = \frac A{\sqrt D} .
$$

**8.** As $\Omega \to 0^+$: $D \to \omega_0^4$, so $R \to
A/\omega_0^2$ and $\tan\varphi \to 0^+$ with $\varphi \in
\intoo0{\frac\pi2}$: $\varphi \to 0$. The mass follows the force quasi-statically, displaced by force/stiffness. As $\Omega \to
+\infty$: $D \sim \Omega^4$, so $R \sim A/\Omega^2 \to 0$, and $\varphi \to \pi$ (the complex number $\omega_0^2 - \Omega^2 +
2\iu\lambda\Omega$ goes to the second quadrant with argument $\to
\pi$): the mass barely moves, and in opposition of phase — inertia dominates.

**9.** By [Theorem 5.12](#thm-b1-diffeq-structure2) (1), every solution is $x = x_p + x_h$ with $x_h$ solving $(H)$; by question 2, $x_h(t)
\to 0$, so $x(t) - x_p(t) \to 0$: all solutions converge to the same steady state. The initial conditions only shape the transient.

**10.** If $x$ is a periodic solution, $x - x_p = x_h$ is a periodic solution of $(H)$ which tends to $0$ at $+\infty$; a periodic function with limit $0$ is identically $0$ (its values on one period repeat forever, so every value is a limit of a subsequence tending to $0$). Hence $x = x_p$.

**11.** Here $\lambda = 1$, $\omega_0^2 = 2$, $\Omega = 1$, $A = 1$: $z = \frac1{2 - 1 + 2\iu} = \frac{1 - 2\iu}5$, so

$$
x_p = \Re\Bigl(\frac{(1 - 2\iu)(\cos t + \iu\sin t)}5\Bigr)
= \frac{\cos t + 2\sin t}5 .
$$

Homogeneous: roots of $r^2 + 2r + 2$ are $-1 \pm \iu$: $x_h =
\eu^{-t}(C\cos t + S\sin t)$. Conditions: $x(0) = \frac15 + C = 0$ gives $C = -\frac15$; differentiating, $x'(0) = \frac25 - C + S =
0$ gives $S = C - \frac25 = -\frac35$. Hence

$$
x(t) = \underbrace{\frac{\cos t + 2\sin t}5}_{\text{steady}}
- \underbrace{\eu^{-t}\,\frac{\cos t + 3\sin
t}5}_{\text{transient}} ,
$$

the transient dying like $\eu^{-t}$.

**12.** Expanding, $g(u) = u^2 - 2(\omega_0^2 - 2\lambda^2)u
+ \omega_0^4 = (u - u_r)^2 + g(u_r)$ with $u_r = \omega_0^2 -
2\lambda^2$ and

$$
g(u_r) = \omega_0^4 - u_r^2 = (\omega_0^2 - u_r)(\omega_0^2 +
u_r) = 2\lambda^2\bigl(2\omega_0^2 - 2\lambda^2\bigr) =
4\lambda^2(\omega_0^2 - \lambda^2) .
$$

If $2\lambda^2 < \omega_0^2$, then $u_r > 0$ is an admissible squared frequency: $g$ has a strict minimum there, so $R = A/\sqrt
g$ has a strict maximum at $\Omega_r = \sqrt{u_r} =
\sqrt{\omega_0^2 - 2\lambda^2}$, with $R_{\max} = A/\sqrt{g(u_r)}
= \frac{A}{2\lambda\sqrt{\omega_0^2 - \lambda^2}}$.

**13.** $R(0) = A/\omega_0^2$, so

$$
\frac{R_{\max}}{R(0)}
= \frac{\omega_0^2}{2\lambda\sqrt{\omega_0^2 - \lambda^2}}
= \frac{\omega_0}{2\lambda}\cdot
\frac{\omega_0}{\sqrt{\omega_0^2 - \lambda^2}}
= Q\,\Bigl(1 - \frac{\lambda^2}{\omega_0^2}\Bigr)^{-1/2}
\geq Q .
$$

For weak damping the correction factor is close to $1$: [resonance](#ex-b1-diffeq-oscillation) multiplies the static displacement by essentially $Q$.

**14.** $V(\Omega)^2 = \frac{A^2 u}{g(u)}$ with $u =
\Omega^2$. Its derivative has the sign of $g(u) - u\,g'(u) =
(\omega_0^2 - u)^2 + 4\lambda^2 u - u\bigl(2(u - \omega_0^2) +
4\lambda^2\bigr) = (\omega_0^2 - u)^2 + 2u(\omega_0^2 - u) =
(\omega_0^2 - u)(\omega_0^2 + u)$, positive for $u < \omega_0^2$ and negative beyond: strict maximum exactly at $\Omega =
\omega_0$, for every damping. There $\tan\varphi$ blows up with $\varphi \in \intoo0\pi$: $\varphi = \frac\pi2$, and $x_p'(t) =
-R\Omega\sin(\Omega t - \frac\pi2) = R\Omega\cos(\Omega t)$ is exactly in phase with the force: optimal power transfer.

**15.** $R = R_{\max}/\sqrt2 \iff g(u) = 2g(u_r) \iff (u -
u_r)^2 = g(u_r) = 4\lambda^2(\omega_0^2 - \lambda^2)$, giving

$$
u_\pm = u_r \pm 2\lambda\sqrt{\omega_0^2 - \lambda^2},
\qquad
\Omega_+^2 - \Omega_-^2 = 4\lambda\sqrt{\omega_0^2 - \lambda^2} .
$$

Then $\Omega_+ - \Omega_- = \frac{\Omega_+^2 -
\Omega_-^2}{\Omega_+ + \Omega_-}$, and for $\lambda \ll \omega_0$ both $\Omega_\pm \approx \omega_0$: $\Omega_+ - \Omega_- \approx
\frac{4\lambda\omega_0}{2\omega_0} = 2\lambda$, so $\frac{\omega_0}{\Omega_+ - \Omega_-} \approx
\frac{\omega_0}{2\lambda} = Q$. Measuring a [resonance](#ex-b1-diffeq-oscillation) peak’s width measures its [quality factor](#pb-b1-diffeq-1).

**16.** $Q = 10$; $\Omega_r = \sqrt{1 - 2(0.05)^2} =
\sqrt{0.995} = 0.9975$; $R_{\max} = \frac1{2 \times 0.05
\sqrt{1 - 0.0025}} = \frac1{0.1 \times 0.99875} = 10.01$; static response $R(0) = 1$; bandwidth $\approx 2\lambda = 0.1$. A tall thin spike of height $\approx Q$ over a plateau of height $1$.

**17.** If $2\lambda^2 \geq \omega_0^2$ then $u_r \leq 0$ and $g'(u) = 2(u - u_r) > 0$ for all $u > 0$: $g$ increases strictly on $\intoo0{+\infty}$, so $R = A/\sqrt g$ decreases strictly from $R(0) = A/\omega_0^2$: the response is largest at zero frequency and there is no peak.

**18.** $\gamma = \iu\Omega$ is not a root of $r^2 +
\omega_0^2$ (as $\Omega \neq \omega_0$), so [Method 5.13](#met-b1-diffeq-particular) with $m = 0$ gives $x_p =
\frac{A\cos\Omega t}{\omega_0^2 - \Omega^2}$ (substitute and check: $-\Omega^2 + \omega_0^2$ times the cosine). General solution:

$$
x(t) = \frac{A\cos\Omega t}{\omega_0^2 - \Omega^2}
+ \lambda_1\cos\omega_0 t + \mu_1\sin\omega_0 t .
$$

**19.** $x(0) = 0$ forces $\lambda_1 = -\frac A{\omega_0^2 -
\Omega^2}$, and $x'(0) = 0$ forces $\mu_1 = 0$:

$$
x(t) = \frac{A}{\omega_0^2 - \Omega^2}\,
\bigl(\cos\Omega t - \cos\omega_0 t\bigr)
= \frac{2A}{\omega_0^2 - \Omega^2}\,
\sin\Bigl(\frac{(\omega_0 - \Omega)t}2\Bigr)
\sin\Bigl(\frac{(\omega_0 + \Omega)t}2\Bigr) ,
$$

by the product formula $\cos a - \cos b =
2\sin\frac{b - a}2\sin\frac{b + a}2$ applied with $a = \Omega t$, $b = \omega_0 t$.

**20.** For $\Omega$ near $\omega_0$, the second sine oscillates at the fast frequency $\frac{\omega_0 + \Omega}2
\approx \omega_0$, while the first is a slow envelope of frequency $\frac{\abs{\omega_0 - \Omega}}2$: the amplitude of the fast oscillation waxes and wanes with envelope period $\frac{2\pi}{\abs{\omega_0 - \Omega}}$ (two beats per envelope period), reaching maxima $\frac{2A}{\abs{\omega_0^2 - \Omega^2}}$. As $\Omega \to \omega_0$, the beats become both slower (period $\to \infty$) and taller (amplitude $\to \infty$).

**21.** Fix $t$. As $\Omega \to \omega_0$:

$$
\frac{2A}{\omega_0^2 - \Omega^2}\sin\Bigl(\frac{(\omega_0 -
\Omega)t}2\Bigr)
= \frac{2A}{\omega_0 + \Omega}\cdot
\frac{\sin\bigl(\frac{(\omega_0 - \Omega)t}2\bigr)}
{\omega_0 - \Omega}
\longrightarrow \frac{2A}{2\omega_0}\cdot\frac t2
= \frac{At}{2\omega_0} ,
$$

while $\sin\bigl(\frac{(\omega_0 + \Omega)t}2\bigr) \to
\sin(\omega_0 t)$: the limit is $x_\infty(t) =
\frac{At\sin\omega_0 t}{2\omega_0}$. Direct check: with $C =
\frac A{2\omega_0}$, $x_\infty = Ct\sin\omega_0 t$ has $x_\infty'' = 2C\omega_0\cos\omega_0 t - C\omega_0^2 t
\sin\omega_0 t$, so $x_\infty'' + \omega_0^2 x_\infty =
2C\omega_0\cos\omega_0 t = A\cos\omega_0 t$, with $x_\infty(0) =
0$ and $x_\infty'(0) = C\sin 0 + C\omega_0 \cdot 0 \cdot \cos 0 =
0$ — for $\omega_0 = A = 1$ this is exactly [Example 5.19](#ex-b1-diffeq-oscillation). [Resonance](#ex-b1-diffeq-oscillation) is the degeneration of beats: the first swell of the envelope, stretched to infinite length.

**22.** Without damping the resonant amplitude grows linearly and without bound; with damping $\lambda > 0$ the growth saturates at $R_{\max} \approx Q\,\frac A{\omega_0^2}$. The mechanism: dissipation removes energy at a rate growing with the amplitude (question 23), so the build-up stops exactly when the damping burns energy as fast as the forcing supplies it.

**23.** With $x_p' = -R\Omega\sin(\Omega t - \varphi)$, over a period the averages $\langle\cos^2\rangle =
\langle\sin^2\rangle = \frac12$ and $\langle\sin\cos\rangle = 0$ give:

$$
\langle P_{\mathrm{diss}}\rangle
= 2\lambda\,R^2\Omega^2\,\langle\sin^2(\Omega t -
\varphi)\rangle = \lambda R^2\Omega^2 ;
$$

and expanding $\sin(\Omega t - \varphi) = \sin\Omega t\cos\varphi
- \cos\Omega t\sin\varphi$:

$$
\langle P_{\mathrm{in}}\rangle
= -AR\Omega\,\bigl\langle\cos\Omega t\,\sin(\Omega t -
\varphi)\bigr\rangle
= AR\Omega\,\frac{\sin\varphi}2 .
$$

Since $\sin\varphi = \frac{2\lambda\Omega}{\sqrt D} =
\frac{2\lambda\Omega R}A$, this is $\frac{AR\Omega}2 \cdot
\frac{2\lambda\Omega R}A = \lambda R^2\Omega^2$: injected and dissipated powers balance exactly — the defining property of a steady regime.

**24.** (i) The structure theorem split every solution into steady state plus transient (questions 9–11) and reduced uniqueness to the homogeneous problem. (ii) The complex method turned the search for a particular solution into one division of complex numbers (question 6), with amplitude and phase read off a [modulus](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-field) and an argument. (iii) The [resonance](#ex-b1-diffeq-oscillation) curve is a pure function study — a quadratic in $u = \Omega^2$, its minimum, its level [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) — in the style of [Chapter 4](https://one-course.com/books/math/3/en/chapter/4-standard-functions#ch-b1-functions) (questions 12–17).

**25.** Free and damped: decaying pseudo-oscillations, lifetime $\frac1\lambda$, about $\frac Q\pi$ swings. Forced and damped: transients die, and a unique sinusoidal steady state survives at the forcing frequency, with amplitude peaking near $\omega_0$ (height $\approx Q \times$ static, width $\approx
\frac{\omega_0}Q$) and phase sweeping from $0$ to $\pi$ through $\frac\pi2$ at $\omega_0$. Free and undamped: perpetual oscillation. Forced and undamped: beats, degenerating into linearly growing [resonance](#ex-b1-diffeq-oscillation) at exact tuning. One dimensionless number, $Q = \frac{\omega_0}{2\lambda}$, tunes everything — peak height, bandwidth, and transient lifetime are three readings of the same dial. The sequel: rewriting $x'' + 2\lambda x' +
\omega_0^2x$ as a first-order system opens the matrix methods of [Chapter 21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices) and the Year 2 volume, and decomposing an arbitrary periodic forcing into sinusoids (Fourier series, Year 3 volume) makes this problem’s single-frequency analysis the universal building block: solve for each frequency, superpose.