High School Mathematics · Grades 10–12

26Differential Equations

A differential equation relates a function to its derivatives. 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 with constant coefficients.

26.1 The equation $y' = ay$

Definition 26.1 (Differential equation)

A differential equation is an equation whose unknown is a function $y$ , involving $y$ and its derivatives. A solution on an interval $I$ is a differentiable function satisfying the equation at every point of $I$ .

Theorem 26.2 (Solutions of $y' = ay$ )

Let $a \in \R$ . The solutions on $\R$ of the equation $y' = ay$ are exactly the functions

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: 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. — The equation $y' = y$ prescribes a slope 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.

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 for the half-life.

26.2 The equation $y' = ay + b$

Theorem 26.5 (Solutions of $y' = ay + b$ )

Let $a \neq 0$ and $b \in \R$ . The solutions on $\R$ of $y' = ay + b$ are exactly the functions

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

The constant 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, $z = C\eu^{ax}$ , 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 $y' = ay + b$ with an initial condition)

Find the equilibrium solution $y_p = -\frac{b}{a}$ (solve $y' = 0$ ).
Write the general solution $y = C\eu^{ax} + y_p$ .
Determine $C$ from the initial condition.
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 — extends to $y' = ay + f(x)$ : see Proposition 26.8.

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. — 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 on an 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 $y = C\eu^{ax} + y_p$ , $C \in \R$ .

Proof. As in Theorem 26.5: $y$ 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 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

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 satisfied by the population $N(t)$ and determine the proportionality constant.

Solution

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

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). By Proposition 26.8, 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

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 contains $60\%$ of the carbon-14 of a living organism. Estimate its age.

Solution

Solution of Exercise 26.5.

$N(t) = N_0\,\eu^{-\lambda t}$ with $\lambda = \frac{\ln 2}{5730}$ (Example 23.9). 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).

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

Solution

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.

Solve the equation with $v(0) = 0$ .
Compute the terminal velocity $\lim_{t\to+\infty} v(t)$ , and the time needed to reach $95\%$ of it.

Solution

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.) A population $y(t) \in \intoo{0}{1}$ (as a fraction of the maximal population) satisfies

y' = y(1 - y).

Let $z = \dfrac{1}{y}$ . Show that $z$ satisfies the linear equation $z' = -z + 1$ .
Solve for $z$ , then for $y$ , with $y(0) = \frac{1}{10}$ .
Show that $y(t) \to 1$ as $t \to +\infty$ and sketch the shape of the solution curve.

Solution

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?

Solution

Solution of Exercise 26.9.

By Theorem 26.5, $y(x) = C\eu^{ax} - \frac ba$ . 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 converge to the equilibrium $-\frac ba$ — and indeed the fixed point 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 is a law of change; solving it turns the law into a clock. The same tiny equation $y' = ay + b$ (Theorem 26.5) 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.

Solve $y' = 3y$ , $y(0) = 2$ ; then $y' = -2y + 6$ , $y(0) = 0$ (Method 26.6).
Re-derive the uniqueness trick behind Theorem 26.2: if $y' = ay$ , compute the derivative of $y(t)\,\eu^{-at}$ and conclude that every solution is $C\eu^{at}$ .
For $y' = -2y + 6$ : find the equilibrium solution, and describe the fate of every other solution as $t \to +\infty$ . (The fixed points of Problem 13.1’s recurrences, gone continuous.)
For the decay law $y' = -\frac{y}{\tau}$ : show that the half-life is $t_{1/2} = \tau \ln 2$ .
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.

Compute $\lambda$ (per year).
A bone retains $20\,\%$ of the living ratio: how old is it?
Charcoal from the painted caves of Lascaux retains about $15\,\%$ : date the paintings.
The Iceman Ötzi, found in an Alpine glacier, measured about $53\,\%$ : date him (the archaeologists say about $5\,300$ years — how did you do?).
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.
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 $T' = -k\,(T - T_a)$ .

Solve the equation (substitute $z = T - T_a$ ):

$T(t) = T_a + (T_0 - T_a)\,\eu^{-kt} .$
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.
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 (what does adding milk do to the gap $T - T_a$ , and how does the gap drive the loss?).
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.
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.

The logistic equation of Exercise 26.8, $y' = y(1 - y)$ 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 $y = \frac12$ — the epidemiologist’s date of Problem 22.1, now computable.
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.
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.
Finale — one 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 $\to$ solution $\to$ calibration $\to$ prediction $\to$ audit), and name the phenomenon that needs $y''$ — and which problem already met it.

Solution

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, the continuous twin of the recurrence fixed points of Problem 13.1.

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; and a body moved from elsewhere resets $T_a$ mid-decay, faking an earlier or later death. The 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 twin of the loan: constant inflow, proportional outflow — Exercise 26.9 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, four worlds. The loop: state the law of change; write the equation; solve (Method 26.6); 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.

26.1 The equation y′=ayy' = ayy′=ay

26.2 The equation y′=ay+by' = ay + by′=ay+b

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

26.4 Exercises

26.5 Problem: The clock inside things

26.1 The equation $y' = ay$

26.2 The equation $y' = ay + b$

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