---
title: "Differential Equations"
book: "High School Mathematics"
subject: math
language: en
chapter: 26
exercises: 9
source: https://one-course.com/books/math/2/en/chapter/26-differential-equations
---

# Chapter 26 — Differential Equations

A [differential equation](#def-g12-diffeq-ode) relates a [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) to its [derivatives](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative). Physics, chemistry, biology and economics express their laws in this form: the rate of cooling of a body, the decay of a radioactive nucleus, the growth of a population are all statements about $y'$. This chapter solves completely the linear first-order [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) with constant coefficients.

## 26.1 The equation $y' = ay$

**Definition 26.1 (Differential equation).**

A *differential equation* is an [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) whose unknown is a [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $y$, involving $y$ and its [derivatives](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative). A *solution* on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $I$ is a [differentiable](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) satisfying the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) at every point of $I$.

**Theorem 26.2 (Solutions of y′=ayy' = ayy′=ay).**

Let $a \in \R$. The solutions on $\R$ of the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $y' = ay$ are exactly the [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function)

$$
y(x) = C\,\eu^{ax}, \qquad C \in \R .
$$

For every pair $(x_0, y_0)$ there is a unique solution with $y(x_0) = y_0$.

**Proof.** Each $y = C\eu^{ax}$ satisfies $y' = Ca\,\eu^{ax} = ay$. Conversely, let $y$ be any solution and set $z(x) = y(x)\,\eu^{-ax}$. Then

$$
z'(x) = y'(x)\,\eu^{-ax} - a\,y(x)\,\eu^{-ax}
= \bigl(y'(x) - a y(x)\bigr)\eu^{-ax} = 0,
$$

so $z$ is constant, say $C$, and $y(x) = C\eu^{ax}$. The initial condition $y(x_0) = y_0$ forces $C = y_0 \eu^{-a x_0}$, uniquely. ∎

**Remark 26.3.**

The exponential is thus *characterized* by the simplest of all [differential equations](#def-g12-diffeq-ode): growth proportional to size. This is why it appears everywhere in nature.

![The equation y' = y prescribes a slope at every point of the plane (gray segments). The solutions C x — here C = 1 (blue), C = 0.3 (red), C = -0.5 (orange) — are exactly the curves that follow this field of slopes.](https://one-course.com/images/onecourse/chapters/math-2/g12-diffeq/fig-4ca95fee37c9.svg)

*The [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $y' = y$ prescribes a [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) at every point of the plane (gray segments). The solutions $C\eu^{x}$ — here $C = 1$ (blue), $C = 0.3$ (red), $C = -0.5$ (orange) — are exactly the curves that follow this field of [slopes](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine).*

**Example 26.4 (Radioactive decay).**

A radioactive quantity satisfies $N' = -\lambda N$ with $\lambda > 0$, so $N(t) = N_0\,\eu^{-\lambda t}$; see [Example 23.9](https://one-course.com/books/math/2/en/chapter/23-exponential-and-logarithm#ex-g12-exp-halflife) for the half-life.

## 26.2 The equation $y' = ay + b$

**Theorem 26.5 (Solutions of y′=ay+by' = ay + by′=ay+b).**

Let $a \neq 0$ and $b \in \R$. The solutions on $\R$ of $y' = ay + b$ are exactly the [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function)

$$
y(x) = C\,\eu^{ax} - \frac{b}{a}, \qquad C \in \R .
$$

The constant [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $y_p = -\frac ba$ is the *equilibrium solution*. For every $(x_0, y_0)$ there is a unique solution with $y(x_0) = y_0$; if $a < 0$, every solution tends to the equilibrium $-\frac ba$ as $x \to +\infty$.

**Proof.** The constant $y_p = -\frac{b}{a}$ satisfies $y_p' = 0 = a y_p + b$. Now $y$ is a solution if and only if

$$
(y - y_p)' = y' = ay + b = a(y - y_p) + \underbrace{a y_p + b}_{=\,0}
= a (y - y_p),
$$

that is, if and only if $z = y - y_p$ solves $z' = az$. By [Theorem 26.2](#thm-g12-diffeq-homogeneous), $z = C\eu^{ax}$, whence the formula, the existence and the uniqueness. If $a < 0$, $\eu^{ax} \to 0$ as $x \to +\infty$, so $y(x) \to -\frac ba$. ∎

**Method 26.6 (Solving y′=ay+by' = ay + by′=ay+b with an initial condition).**

1. Find the equilibrium solution $y_p = -\frac{b}{a}$ (solve $y' = 0$ ).
2. Write the general solution $y = C\eu^{ax} + y_p$ .
3. Determine $C$ from the initial condition.
4. Check the long-term behavior against physical intuition (does the solution converge to equilibrium?).

The same strategy — *particular solution + general solution of the homogeneous [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation)* — extends to $y' = ay + f(x)$: see [Proposition 26.8](#prop-g12-diffeq-general).

**Example 26.7 (Newton’s law of cooling).**

A cup of coffee at $80\,^\circ$C sits in a room at $20\,^\circ$C. Newton’s law states that the temperature $T$ satisfies $T' = -k(T - 20)$ for some $k > 0$, *i.e.* $T' = -kT + 20k$. The equilibrium is $20$, and $T(t) = 20 + 60\,\eu^{-kt}$: the coffee cools exponentially fast to room temperature.

![Whatever their initial temperature, all solutions of T' = -k(T - 20) converge exponentially to the equilibrium T = 20.](https://one-course.com/images/onecourse/chapters/math-2/g12-diffeq/fig-3bb61fc635ba.svg)

*Whatever their initial temperature, all solutions of $T' = -k(T - 20)$ converge exponentially to the equilibrium $T = 20$.*

## 26.3 The equation $y' = ay + f(x)$

**Proposition 26.8 (Structure of the solution set).**

Let $f$ be [continuous](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $I$ and let $y_p$ be one particular solution of

$$
y' = ay + f(x)
$$

on $I$. Then the solutions on $I$ are exactly the [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $y = C\eu^{ax} + y_p$, $C \in \R$.

**Proof.** As in [Theorem 26.5](#thm-g12-diffeq-affine): $y$ is a solution if and only if $z = y - y_p$ satisfies $z' = (ay + f) - (ay_p + f) = az$, if and only if $z = C\eu^{ax}$. ∎

**Method 26.9 (Guessing a particular solution).**

Look for a particular solution *of the same shape* as $f$:

- $f$ polynomial of degree $n$ : try a polynomial of degree $n$ ;
- $f(x) = \alpha\,\eu^{kx}$ with $k \neq a$ : try $y_p = \beta\,\eu^{kx}$ ;
- $f(x) = \alpha\,\eu^{ax}$ (resonant case): try $y_p = \beta x\,\eu^{ax}$ .

Substitute into the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) and identify the coefficients.

**Example 26.10.**

Solve $y' = 2y + 4x$. Try $y_p = \alpha x + \beta$: $\alpha = 2(\alpha x + \beta) + 4x$ for all $x$ forces $2\alpha + 4 = 0$ and $\alpha = 2\beta$, so $\alpha = -2$, $\beta = -1$. General solution: $y = C\eu^{2x} - 2x - 1$.

## 26.4 Exercises

**Exercise 26.1 ★.**

Solve on $\R$: (a) $y' = 3y$ with $y(0) = 2$; (b) $2y' + y = 0$ with $y(0) = -1$; (c) $y' = -y + 5$ with $y(0) = 0$.

**Solution of Exercise 26.1.**

*(a)* $y = C\eu^{3x}$; $y(0) = 2$ gives $y = 2\eu^{3x}$.

*(b)* $y' = -\frac12 y$, so $y = C\eu^{-x/2}$; $y(0) = -1$ gives $y = -\eu^{-x/2}$.

*(c)* Equilibrium $y_p = 5$; $y = C\eu^{-x} + 5$; $y(0) = 0$ gives $C = -5$: $y = 5\left(1 - \eu^{-x}\right)$.

**Exercise 26.2 ★.**

A bacterial population grows at a rate proportional to its size, doubling every $3$ hours. Write the [differential equation](#def-g12-diffeq-ode) satisfied by the population $N(t)$ and determine the proportionality constant.

**Solution of Exercise 26.2.**

$N' = kN$, so $N(t) = N_0 \eu^{kt}$. Doubling in $3$ hours means $\eu^{3k} = 2$, *i.e.*

$$
k = \frac{\ln 2}{3} \approx 0.231\ \text{h}^{-1}.
$$

**Exercise 26.3 ★.**

Verify that $y_p(x) = x\,\eu^{x}$ is a solution of $y' = y + \eu^x$, and give all solutions on $\R$.

**Solution of Exercise 26.3.**

$y_p'(x) = \eu^x + x\eu^x = y_p(x) + \eu^x$: $y_p$ is a particular solution (this is the resonant case of [Method 26.9](#met-g12-diffeq-particular)). By [Proposition 26.8](#prop-g12-diffeq-general), the solutions are $y = C\eu^{x} + x\eu^{x} = (C + x)\,\eu^x$, $C \in \R$.

**Exercise 26.4 ★★.**

Solve $y' = -2y + \eu^{x}$ with $y(0) = 1$. (Hint: look for a particular solution of the form $\beta\,\eu^{x}$.)

**Solution of Exercise 26.4.**

Try $y_p = \beta\eu^x$: $\beta\eu^x = -2\beta\eu^x + \eu^x$ gives $3\beta = 1$, so $y_p = \frac13\eu^x$. General solution $y = C\eu^{-2x} + \frac13\eu^{x}$; the condition $y(0) = 1$ gives $C = \frac23$:

$$
y(x) = \frac{2}{3}\,\eu^{-2x} + \frac{1}{3}\,\eu^{x}.
$$

**Exercise 26.5 ★★.**

Carbon-14 decays with a half-life of $5730$ years. An archaeological [sample](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-sample) contains $60\%$ of the carbon-14 of a living organism. Estimate its age.

**Solution of Exercise 26.5.**

$N(t) = N_0\,\eu^{-\lambda t}$ with $\lambda = \frac{\ln 2}{5730}$ ([Example 23.9](https://one-course.com/books/math/2/en/chapter/23-exponential-and-logarithm#ex-g12-exp-halflife)). We solve $\eu^{-\lambda t} = 0.6$:

$$
t = \frac{\ln(1/0.6)}{\lambda} = 5730\,\frac{\ln(5/3)}{\ln 2}
\approx 5730 \times \frac{0.5108}{0.6931} \approx 4220 \text{ years}.
$$

**Exercise 26.6 ★★.**

A tank contains $100$ L of pure water. Brine containing $0.2$ kg of salt per liter flows in at $5$ L/min, the mixture (kept uniform) flows out at the same rate. Let $m(t)$ be the mass of salt in the tank at time $t$ (in minutes).

1. Justify that $m' = 1 - \dfrac{m}{20}$ .
2. Solve, and determine the limit of $m(t)$ as $t \to +\infty$ . Interpret.

**Solution of Exercise 26.6.**

*1.* Salt flows in at $0.2 \times 5 = 1$ kg/min. The outflow carries concentration $\frac{m}{100}$ kg/L at $5$ L/min, *i.e.* $\frac{m}{20}$ kg/min. Hence $m' = 1 - \frac{m}{20}$.

*2.* Equilibrium $m_p = 20$; $m(t) = 20 + C\eu^{-t/20}$, and $m(0) = 0$ gives $C = -20$:

$$
m(t) = 20\left(1 - \eu^{-t/20}\right) \xrightarrow[t\to+\infty]{} 20 \text{ kg}.
$$

In the long run the tank’s concentration equals that of the incoming brine: $0.2$ kg/L $\times$ $100$ L $= 20$ kg.

**Exercise 26.7 ★★.**

A skydiver of mass $80$ kg falls subject to gravity ($g = 9.8\ \text{m/s}^2$) and air resistance proportional to speed, so that her speed satisfies $v' = g - \frac{k}{m}v$ with $k = 16$ kg/s.

1. Solve the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) with $v(0) = 0$ .
2. Compute the *terminal velocity* $\lim_{t\to+\infty} v(t)$ , and the time needed to reach $95\%$ of it.

**Solution of Exercise 26.7.**

*1.* $v' = g - \frac{k}{m} v$ with $\frac km = \frac{16}{80} = 0.2$. Equilibrium $v_\infty = \frac{mg}{k} = \frac{9.8}{0.2} = 49$ m/s; $v(t) = 49 + C\eu^{-0.2t}$, and $v(0) = 0$ gives

$$
v(t) = 49\left(1 - \eu^{-0.2 t}\right).
$$

*2.* Terminal velocity $49$ m/s ($\approx 176$ km/h). We want $1 - \eu^{-0.2t} = 0.95$, *i.e.* $\eu^{-0.2t} = 0.05$:

$$
t = \frac{\ln 20}{0.2} \approx \frac{3.00}{0.2} \approx 15 \text{ s}.
$$

**Exercise 26.8 ★★★.**

*(Logistic [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation).)* A population $y(t) \in \intoo{0}{1}$ (as a fraction of the maximal population) satisfies

$$
y' = y(1 - y).
$$

1. Let $z = \dfrac{1}{y}$ . Show that $z$ satisfies the linear [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $z' = -z + 1$ .
2. Solve for $z$ , then for $y$ , with $y(0) = \frac{1}{10}$ .
3. Show that $y(t) \to 1$ as $t \to +\infty$ and sketch the shape of the solution curve.

**Solution of Exercise 26.8.**

*1.* $z = \frac1y$ gives $z' = -\frac{y'}{y^2}
= -\frac{y(1-y)}{y^2} = -\frac{1-y}{y} = -\frac1y + 1 = -z + 1$.

*2.* Equilibrium $z_p = 1$, so $z(t) = 1 + C\eu^{-t}$. From $y(0) = \frac{1}{10}$, $z(0) = 10$, so $C = 9$ and

$$
y(t) = \frac{1}{1 + 9\,\eu^{-t}} .
$$

*3.* As $t \to +\infty$, $9\eu^{-t} \to 0$ and $y(t) \to 1$. The curve is the classic S-shaped (*sigmoid*) logistic curve: slow growth at first ($y$ small, $y' \approx y$), fastest growth when $y = \frac12$ (where $y' = y(1-y)$ is maximal), then saturation towards the carrying capacity $1$.

**Exercise 26.9 ★★★.**

Let $y$ be a solution of $y' = ay + b$ and let $u_n = y(n)$ for $n \in \N$. Show that $(u_n)$ satisfies an arithmetico-geometric recurrence $u_{n+1} = q u_n + r$, and express $q$ and $r$ in terms of $a$ and $b$. What does the condition $\abs{q} < 1$ correspond to for the [differential equation](#def-g12-diffeq-ode)?

**Solution of Exercise 26.9.**

By [Theorem 26.5](#thm-g12-diffeq-affine), $y(x) = C\eu^{ax} - \frac ba$. Hence

$$
u_{n+1} = C\eu^{a(n+1)} - \frac ba
= \eu^{a}\left(C\eu^{an} - \frac ba\right) + \frac ba\left(\eu^a - 1\right)
= q\,u_n + r
$$

with $q = \eu^{a}$ and $r = \frac{b}{a}\left(\eu^{a} - 1\right)$. The condition $\abs q < 1$ means $\eu^a < 1$, *i.e.* $a < 0$: exactly the condition under which the solutions of the [differential equation](#def-g12-diffeq-ode) converge to the equilibrium $-\frac ba$ — and indeed the [fixed point](https://one-course.com/books/math/2/en/chapter/3-functions#pb-g10-functions-1) of the recurrence is $\frac{r}{1 - q} = -\frac{b}{a}$.

## 26.5 Problem: The clock inside things

**Problem 26.1.**

Weekend problem — carbon-14 dates the caves, Newton’s cooling law times a crime, and one small equation wears four costumes

A [differential equation](#def-g12-diffeq-ode) is a law of change; solving it turns the law into a *clock*. The same tiny [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $y' = ay + b$ ([Theorem 26.5](#thm-g12-diffeq-affine)) ticks inside prehistoric charcoal, cooling coffee, falling raindrops and hospital drips — and reading those clocks is this problem’s business: it dates the Lascaux paintings, fixes a time of death, and audits its own assumptions like a good scientist.

**Part I — Fluency.**

1. Solve $y' = 3y$ , $y(0) = 2$ ; then $y' = -2y + 6$ , $y(0) = 0$ ( [Method 26.6](#met-g12-diffeq-solve) ).
2. Re-derive the uniqueness trick behind [Theorem 26.2](#thm-g12-diffeq-homogeneous) : if $y' = ay$ , compute the [derivative](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) of $y(t)\,\eu^{-at}$ and conclude that *every* solution is $C\eu^{at}$ .
3. For $y' = -2y + 6$ : find the equilibrium solution, and describe the fate of every other solution as $t \to +\infty$ . (The [fixed points](https://one-course.com/books/math/2/en/chapter/3-functions#pb-g10-functions-1) of [Problem 13.1](https://one-course.com/books/math/2/en/chapter/13-sequences-a-first-course#pb-g11-seq-1) ’s recurrences, gone [continuous](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) .)
4. For the decay law $y' = -\frac{y}{\tau}$ : show that the half-life is $t_{1/2} = \tau \ln 2$ .
5. Sketch (or describe) the family of solutions of $y' = -2y + 6$ : what do solutions starting above $3$ do? Below $3$ ? On $3$ ?

**Part II — Carbon-14.** Living tissue maintains a constant ratio of radioactive carbon-14; at death the intake stops and the stock decays: $N' = -\lambda N$, with half-life $5\,730$ years.

6. Compute $\lambda$ (per year).
7. A bone retains $20\,\%$ of the living ratio: how old is it?
8. Charcoal from the painted caves of Lascaux retains about $15\,\%$ : date the paintings.
9. The Iceman Ötzi, found in an Alpine glacier, measured about $53\,\%$ : date him (the archaeologists say about $5\,300$ years — how did you do?).
10. Why can carbon-14 not date dinosaurs? Compute the fraction remaining after ten half-lives, state the fraction after $65$ million years as a power of $2$ , and conclude.
11. Error bars: if Ötzi’s $53\,\%$ is only known to $\pm 1\,\%$ , compute the age range. What lab precision buys what dating precision?

**Part III — Newton’s cooling, and a crime.** A body at temperature $T$ in surroundings at constant $T_a$ cools by $T' = -k\,(T - T_a)$.

12. Solve the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) (substitute $z = T - T_a$): $$T(t) = T_a + (T_0 - T_a)\,\eu^{-kt} .$$
13. Coffee poured at $90\,^\circ$ C in a $20\,^\circ$ C room reads $70\,^\circ$ C after $5$ minutes. Determine $k$ , then the waiting time until the drinkable $55\,^\circ$ C.
14. The milk question: to drink the warmest possible coffee in ten minutes, should the cold milk go in now or at the last moment? Answer with the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) (what does adding milk do to the gap $T - T_a$ , and how does the gap drive the loss?).
15. Forensics: a body is found at midnight at $30\,^\circ$ C in a $20\,^\circ$ C room; one hour later it reads $28\,^\circ$ C. Assuming $37\,^\circ$ C at the time of death, recover $k$ from the two measurements, then compute the time of death.
16. Audit the coroner’s clock: name three real-world violations of the model’s assumptions and the direction in which each would bias the estimated time of death.

**Part IV — Four costumes and an S-curve.**

17. The logistic [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of [Exercise 26.8](#exo-g12-diffeq-8) , $y' = y(1 - y)$ with $y(0) = 0.1$ : using the exercise’s substitution, derive $y(t) = \dfrac{1}{1 + 9\eu^{-t}}$ , and find the time of the [inflection point](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-inflection) $y = \frac12$ — the epidemiologist’s date of [Problem 22.1](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#pb-g12-deriv-1) , now computable.
18. A raindrop obeys $v' = 10 - \frac v2$ (gravity minus drag), $v(0) = 0$ . Find the terminal velocity, solve for $v(t)$ , and compute when the drop reaches $95\,\%$ of terminal speed.
19. A hospital drip delivers a drug at constant rate while the body eliminates it proportionally: $c' = 4 - 0.5c$ , $c(0) = 0$ . Find the steady-state concentration and the time to reach half of it. State in one sentence the parallel with the loan recurrences of [Exercise 26.9](#exo-g12-diffeq-9) .
20. Finale — one [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) , four costumes: decay, cooling, falling, dosing, all $y' = ay + b$ with different signs and names. Recite the modeling loop this problem ran four times (law $\to$ [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $\to$ solution $\to$ calibration $\to$ prediction $\to$ audit), and name the phenomenon that needs $y''$ — and which problem already met it.

**Solution of Problem 26.1.**

**1.** $y = 2\eu^{3t}$. For the second: equilibrium $3$, so $y = 3 + C\eu^{-2t}$ with $y(0) = 0$: $C = -3$: $y = 3\left(1 - \eu^{-2t}\right)$.

**2.** $\left(y\eu^{-at}\right)' = y'\eu^{-at} -
ay\eu^{-at} = (y' - ay)\eu^{-at} = 0$: the product is a constant $C$, so $y = C\eu^{at}$ — no solution escapes.

**3.** Equilibrium: $y \equiv 3$. Every other solution is $3 + C\eu^{-2t}$ with $C \neq 0$: the exponential dies and the solution glides to $3$ — a stable [fixed point](https://one-course.com/books/math/2/en/chapter/3-functions#pb-g10-functions-1), the [continuous](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) twin of the recurrence [fixed points](https://one-course.com/books/math/2/en/chapter/3-functions#pb-g10-functions-1) of [Problem 13.1](https://one-course.com/books/math/2/en/chapter/13-sequences-a-first-course#pb-g11-seq-1).

**4.** $y = y_0\eu^{-t/\tau}$ halves when $\eu^{-t/\tau} = \frac12$: $t = \tau\ln 2$.

**5.** All solutions are vertical translates of the decay towards $3$: those starting above fall to $3$, those below rise to $3$, and the constant solution $3$ sits still — a funnel around the equilibrium.

**6.** $\lambda = \frac{\ln 2}{5730} \approx 1.21 \times
10^{-4}$ per year.

**7.** $\eu^{-\lambda t} = 0.2$: $t = \frac{\ln 5}{\lambda} \approx 13\,300$ years.

**8.** $t = \frac{\ln(1/0.15)}{\lambda} \approx 15\,700$ years: the bulls of Lascaux are late Ice Age.

**9.** $t = \frac{\ln(1/0.53)}{\lambda} \approx 5\,250$ years: within a lifetime of the archaeologists’ value — the clock works.

**10.** Ten half-lives leave $2^{-10} \approx 0.1\,\%$: at the edge of measurement. After $65$ million years the fraction is $2^{-65\,000\,000/5\,730} \approx 2^{-11\,344}$: no atom of the original stock remains in any fossil — dinosaurs are dated by slower clocks (potassium–argon and kin).

**11.** $52\,\%$ gives $\approx 5\,405$ years, $54\,\%$ gives $\approx 5\,095$: the $\pm 1\,\%$ of chemistry becomes roughly $\pm 155$ years of history — precision in the lab is precision in the museum label.

**12.** $z = T - T_a$ satisfies $z' = -kz$: $z = (T_0 - T_a)\eu^{-kt}$, and $T = T_a + z$.

**13.** The initial gap is $T_0 - T_a = 70$; after five minutes the gap is $70 - 20 = 50$, so $50 = 70\eu^{-5k}$: $k = \frac{\ln(70/50)}{5} \approx 0.067$ per minute. Drinkable: $55 - 20 = 35 = 70\eu^{-kt}$: $t = \frac{\ln 2}{k} \approx 10.3$ minutes.

**14.** The loss rate is proportional to the gap $T - T_a$: piping-hot coffee bleeds heat fastest. Adding the milk *now* slashes the gap immediately, so less heat is lost over the ten minutes; adding it at the end lets the coffee cool at full speed first. For the warmest cup: milk first. (Impatient hosts have the thermodynamics backwards.)

**15.** Gaps from ambient: at midnight $10$, an hour later $8$: $\eu^{-k} = 0.8$: $k = \ln\frac{10}{8} \approx 0.223$ per hour. At death the gap was $17$; it decayed to $10$ by midnight: $s = \frac{\ln(17/10)}{k} \approx 2.4$ hours. Time of death: about $21{:}40$.

**16.** A heated or draughty room ($T_a$ not constant) bends the clock either way; clothing or body mass changes $k$ (the coroner’s tables adjust for it) — a wrong $k$ scales the whole [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval); and a body moved from elsewhere resets $T_a$ mid-decay, faking an earlier or later death. The [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) is honest; the assumptions carry the risk.

**17.** $z = \frac1y$ obeys $z' = 1 - z$: $z = 1 + 9\eu^{-t}$ (from $z(0) = 10$), so $y = \frac{1}{1 + 9\eu^{-t}}$. Inflection at $y = \frac12$: $9\eu^{-t} = 1$: $t = \ln 9 \approx 2.2$ — the epidemic’s turning date, straight from the formula.

**18.** Terminal velocity: $v' = 0$ at $v = 20$ m/s. Solution: $v(t) = 20\left(1 - \eu^{-t/2}\right)$. Then $0.95$: $\eu^{-t/2} = 0.05$: $t = 2\ln 20 \approx 6$ s — raindrops hit their cruising speed within seconds, which is why rain does not kill.

**19.** Steady state: $c = 8$. Half of it: $4 = 8\left(1 - \eu^{-t/2}\right)$: $\eu^{-t/2} = \frac12$: $t = 2\ln 2 \approx 1.4$ hours. The drip is the [continuous](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) twin of the loan: constant inflow, proportional outflow — [Exercise 26.9](#exo-g12-diffeq-9) makes the dictionary exact.

**20.** Decay ($a < 0$, $b = 0$), cooling ($y = T - T_a$), falling ($a < 0$, $b > 0$: approach to equilibrium from below), dosing (same, in a vein): one [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation), four worlds. The loop: state the law of change; write the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation); solve ([Method 26.6](#met-g12-diffeq-solve)); calibrate constants on measurements; predict; then audit the assumptions (question 16). Oscillation needs $y'' = -\omega^2 y$ — met, solved and set swinging in [Problem 24.1](https://one-course.com/books/math/2/en/chapter/24-trigonometric-functions#pb-g12-trigo-1).