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Module 9 · L16 of 20 ~40 min ⚡ +95 XP available

Modelling with Differential Equations

A drug is injected at a constant rate but is simultaneously cleared by the kidneys. A savings account earns interest but also has regular withdrawals. Both are rate-in minus rate-out problems — and both are solved by the same differential equation framework. In this lesson you'll build DEs from plain English, solve them with separation of variables or the integrating factor, and extract physical meaning from the solution.

Today's hook — Salt water flows into a 100 L tank at 2 L/min (concentration 0.5 kg/L) and the mixed solution drains out at 2 L/min. If the tank starts with pure water, write down a differential equation for the mass of salt $Q(t)$ in the tank. Don't solve it yet — just set it up.

0/5QUESTS

Recall — your gut answer first

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Salt water flows into a 100 L tank at 2 L/min (concentration 0.5 kg/L) and the well-mixed solution drains at 2 L/min. The tank starts with pure water. Without solving anything — write the differential equation for $Q(t)$, the mass of salt at time $t$. Also: what do you think $Q$ approaches as $t \to \infty$?

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The big idea

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Most applied DE problems follow a single pattern: identify what is changing, write its rate of change as rate in minus rate out, then solve.

The master template is:

$\dfrac{dQ}{dt} = (\text{rate in}) - (\text{rate out})$

Each term may depend on $t$, on $Q$ itself, or both. Once you have the DE, use separation of variables or an integrating factor to solve.

$\dfrac{dQ}{dt} = r_{\text{in}} - r_{\text{out}}$

Read the problem for rates

Identify units carefully. "Rate in" might be $c_{\text{in}} \times f_{\text{in}}$ (concentration × flow). "Rate out" uses the current concentration $Q/V$ times flow out.

Steady state

The long-run value $Q_\infty$ is found by setting $\tfrac{dQ}{dt} = 0$: rate in = rate out. This is the equilibrium — the solution approaches it as $t \to \infty$.

Separable or linear?

If the DE has the form $\tfrac{dQ}{dt} = a - bQ$ it is linear first-order. Separate variables: $\tfrac{dQ}{a - bQ} = dt$, or use the integrating factor $e^{bt}$.

What you'll master

Know

Key facts

Framework: $\tfrac{dQ}{dt} = \text{rate in} - \text{rate out}$
Mixing problem DE: $\tfrac{dQ}{dt} = c_{\text{in}} f_{\text{in}} - \tfrac{Q}{V} f_{\text{out}}$ (constant volume)
Steady state: set $\tfrac{dQ}{dt} = 0$ to find $Q_\infty$

Understand

Concepts

How to translate a word problem into a differential equation
Why the DE $\tfrac{dQ}{dt} = a - bQ$ has an exponential solution approaching $a/b$
The role of initial conditions in fixing the arbitrary constant

Can do

Skills

Set up DEs for mixing, pharmacokinetics, and simple financial models
Solve $\tfrac{dQ}{dt} = a - bQ$ by separation of variables
Find the steady-state value and interpret it in context

Key terms

Mathematical modelA differential equation (or system) that describes the evolution of a quantity in a real-world context.

Rate in / rate outThe rates at which a quantity enters and leaves a system. Units must be consistent (e.g.\ kg/min, \$/year).

Steady stateThe equilibrium value approached as $t \to \infty$; found by setting $\tfrac{dQ}{dt} = 0$.

Mixing problemA problem involving a substance flowing in and out of a container. Rate out depends on current concentration $Q/V$.

Initial conditionThe value $Q(0) = Q_0$ used to determine the arbitrary constant $C$ after integration.

Separable DEA DE of the form $\tfrac{dQ}{dt} = f(Q)g(t)$ that can be solved by writing $\tfrac{1}{f(Q)}\,dQ = g(t)\,dt$ and integrating both sides.

Setting up and solving a modelling DE

core concept

General strategy (4 steps):

Define the variable: choose $Q(t)$ with clear units.
Identify rates: write $\tfrac{dQ}{dt} = (\text{rate in}) - (\text{rate out})$.
Solve: separate variables (or use integrating factor) and integrate.
Apply initial condition to find $C$; interpret the solution in context.

Key form: Many applied DEs reduce to $\dfrac{dQ}{dt} = a - bQ$ where $a, b > 0$ are constants. Separate variables:

$$\int \frac{dQ}{a - bQ} = \int dt \quad \Rightarrow \quad -\frac{1}{b}\ln|a - bQ| = t + C$$

Exponentiating and applying $Q(0) = Q_0$:

$$\boxed{Q(t) = \frac{a}{b} + \left(Q_0 - \frac{a}{b}\right)e^{-bt}}$$

As $t \to \infty$, $Q \to \dfrac{a}{b}$ — the steady state. When $Q_0 < a/b$ the quantity grows to the steady state; when $Q_0 > a/b$ it decays down to it.

Steady-state shortcut. You don't need to solve the full DE to find $Q_\infty$. Just set $\tfrac{dQ}{dt} = 0$: $a - bQ_\infty = 0 \Rightarrow Q_\infty = a/b$. This is a powerful check — your solution should approach this value.

Key form: Many applied DEs reduce to $\dfrac{dQ}{dt} = a - bQ$ where $a, b > 0$ are constants. Separate variables:

Pause — copy the general form $\frac{dQ}{dt}=a-bQ$ and its solution $Q=\frac{a}{b}(1-e^{-bt})+Q_0 e^{-bt}$, identifying the long-run equilibrium $Q=a/b$ into your book.

Quick check: A tank holds 200 L of brine. Salt solution (0.3 kg/L) flows in at 4 L/min and the well-mixed solution drains at 4 L/min. Let $Q$ be the mass of salt (kg). Which DE is correct?

The salt tank — full solution

core concept

We just saw that many applied DEs reduce to $\frac{dQ}{dt}=a-bQ$, which is separable: $\int\frac{dQ}{a-bQ}=\int dt$ gives $Q=\frac{a}{b}(1-e^{-bt})+Q_0 e^{-bt}$. That raises a question: for the specific salt-tank hook (100 L tank, 0.5 kg/L inflow at 2 L/min), what are the explicit values of $a$, $b$, and $Q_0$? This card answers it → $a=0.5\times2=1$ kg/min (rate in); $b=2/100=0.02$ min$^{-1}$ (rate out per kg); $Q_0=0$.

Recall the hook problem: 100 L tank, pure water initially, salt solution (0.5 kg/L) flows in at 2 L/min, mixed solution drains at 2 L/min (volume stays constant at 100 L). Let $Q(t)$ = mass of salt (kg).

Set up:

Rate in: $0.5 \text{ kg/L} \times 2 \text{ L/min} = 1 \text{ kg/min}$
Rate out: $\dfrac{Q}{100} \text{ kg/L} \times 2 \text{ L/min} = \dfrac{Q}{50} \text{ kg/min}$

$$\frac{dQ}{dt} = 1 - \frac{Q}{50}$$

Here $a = 1$, $b = 1/50$, $Q_0 = 0$. The steady state is $Q_\infty = a/b = 50$ kg (sensible: $0.5 \text{ kg/L} \times 100 \text{ L}$). The solution is:

$$Q(t) = 50\left(1 - e^{-t/50}\right)$$

Units check. At $t = 0$: $Q = 50(1-1) = 0$ kg ✓. As $t \to \infty$: $Q \to 50$ kg ✓. The rate of approach is governed by the time constant $\tau = 1/b = 50$ min — after $50$ min the salt mass is $50(1 - e^{-1}) \approx 31.6$ kg.

Pause — copy the full salt-tank solution: values of $a=1$, $b=0.02$, $Q_0=0$; the particular solution $Q(t)=50(1-e^{-0.02t})$; and the equilibrium amount 50 kg into your book.

Did you get this? True or false: for the salt tank in card 06, the mass of salt approaches 50 kg as $t \to \infty$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · MIXING PROBLEM

A 500 L tank initially contains 10 kg of salt dissolved in water. Brine (0.2 kg/L) flows in at 5 L/min and the well-mixed solution drains at 5 L/min. Find $Q(t)$ and the steady-state mass.

Rate in $= 0.2 \times 5 = 1$ kg/min. Rate out $= \tfrac{Q}{500} \times 5 = \tfrac{Q}{100}$ kg/min. DE: $\tfrac{dQ}{dt} = 1 - \tfrac{Q}{100}$.

Apply the rate in − rate out template. Volume stays 500 L since in-rate = out-rate.

PROBLEM 2 · PHARMACOKINETICS

A drug is infused at a constant rate of 6 mg/hr. The body eliminates it at a rate proportional to the current amount, with constant $k = 0.3$ hr$^{-1}$. Initially $Q = 0$. Find $Q(t)$ and the steady-state drug level.

$\dfrac{dQ}{dt} = 6 - 0.3Q$, $Q(0) = 0$. Here $a = 6$, $b = 0.3$.

Rate in is constant (6 mg/hr); rate out is $kQ$ (proportional to amount present).

PROBLEM 3 · FINANCIAL MODEL

An account earns interest continuously at 5% p.a. and money is withdrawn at \$2000/year. Initially the balance is \$30 000. Find the balance $B(t)$ and determine whether the account is depleted.

Rate in (interest): $0.05B$. Rate out (withdrawal): 2000. DE: $\dfrac{dB}{dt} = 0.05B - 2000$.

Continuous interest accrues at rate $rB$; fixed withdrawals leave at rate 2000/year.

Fill the gap: For $\dfrac{dQ}{dt} = 8 - 0.4Q$ with $Q(0) = 0$, the steady-state value is $Q_\infty = $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Getting the rate out wrong

Rate out is $\dfrac{Q}{V} \times f_{\text{out}}$, not just $f_{\text{out}}$. The concentration $Q/V$ changes over time. Writing rate out $= f_{\text{out}}$ (a constant) ignores the feedback mechanism and gives a completely wrong DE.

Trap 02

Using the wrong sign

If $\tfrac{dQ}{dt} = a - bQ$ with $b > 0$, the solution decays to $a/b$ from above (if $Q_0 > a/b$) or grows to $a/b$ from below (if $Q_0 < a/b$). Writing $e^{+bt}$ instead of $e^{-bt}$ gives exponential blowup — always check the sign of the exponent using the physics.

Trap 03

Forgetting units

Every term in a DE must have the same units. Check: if $Q$ is in kg and $t$ in minutes, then $\tfrac{dQ}{dt}$ is in kg/min. The rate-in term must also be in kg/min. Unit errors usually signal a setup mistake.

Did you get this? True or false: for a tank where inflow rate equals outflow rate (volume stays constant), the rate of salt leaving equals $\dfrac{Q}{V} \times f_{\text{out}}$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A 200 L tank contains 4 kg of salt. Brine (0.1 kg/L) flows in at 3 L/min; mixed solution drains at 3 L/min. Write the DE for $Q(t)$ and state the steady-state mass.

Solve the DE from Activity 1 with initial condition $Q(0) = 4$. Give the full expression for $Q(t)$.

A drug is delivered at 10 mg/hr and eliminated at rate $0.2Q$ mg/hr. Find the steady-state drug level. How long does it take to reach 90% of this level? (Hint: solve $e^{-0.2t} = 0.1$.)

An account with balance $B$ earns 4% continuous interest and has \$3000/year withdrawn. If $B(0) = \$50\,000$, write and solve the DE. Does the account grow or decline? Why?

Verify that $Q(t) = 50(1 - e^{-t/50})$ satisfies $\tfrac{dQ}{dt} = 1 - \tfrac{Q}{50}$ with $Q(0) = 0$. Show all substitutions.

Odd one out: A tank (100 L, pure water) has brine (0.4 kg/L) flowing in at 2 L/min and draining at 2 L/min. Three of these statements are correct. Which one is NOT?

Revisit your thinking

Earlier you wrote the DE for the salt tank (100 L, 0.5 kg/L brine, 2 L/min each way).

The correct DE is $\tfrac{dQ}{dt} = 1 - \tfrac{Q}{50}$. Rate in $= 0.5 \times 2 = 1$ kg/min; rate out $= \tfrac{Q}{100} \times 2 = \tfrac{Q}{50}$ kg/min. The steady state is $Q_\infty = 50$ kg (0.5 kg/L $\times$ 100 L). Did you get the rate-out expression right? That is the most common error in mixing problems.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. A 400 L tank holds 8 kg of salt. Brine (0.25 kg/L) flows in at 4 L/min and drains at 4 L/min. Write the DE for $Q(t)$ and state the steady-state salt mass. (2 marks)

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ApplyBand 43 marks

Q2. Solve the DE from Q1 with $Q(0) = 8$. Find the time at which $Q = 90$ kg. Give the time in exact form. (3 marks)

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AnalyseBand 53 marks

Q3. An investor's portfolio $B(t)$ (in thousands of dollars) satisfies $\dfrac{dB}{dt} = 0.06B - 5$, with $B(0) = 70$. Find $B(t)$ and determine whether the portfolio grows indefinitely or is eventually depleted. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers: 1. Rate in $=0.3$ kg/min, rate out $=\tfrac{3Q}{200}$ kg/min, DE: $\tfrac{dQ}{dt}=0.3-\tfrac{3Q}{200}$, $Q_\infty=20$ kg · 2. $Q(t)=20-16e^{-3t/200}$ · 3. $Q_\infty=50$ mg; 90% at $t=-\tfrac{\ln(0.1)}{0.2}=\tfrac{\ln 10}{0.2}=5\ln 10\approx11.5$ hr · 4. DE: $\tfrac{dB}{dt}=0.04B-3000$; $B^*=75\,000$; since $B_0=50\,000 < 75\,000$ the account is declining · 5. $\tfrac{dQ}{dt}=e^{-t/50}$; RHS: $1-\tfrac{50(1-e^{-t/50})}{50}=1-(1-e^{-t/50})=e^{-t/50}$ ✓; $Q(0)=0$ ✓.

Q1 (2 marks): Rate in $=0.25\times4=1$ kg/min; rate out $=\tfrac{Q}{400}\times4=\tfrac{Q}{100}$ kg/min [1]. DE: $\tfrac{dQ}{dt}=1-\tfrac{Q}{100}$; $Q_\infty=100$ kg [1].

Q2 (3 marks): $a=1$, $b=1/100$, $Q_0=8$; $Q(t)=100-92e^{-t/100}$ [1]. Set $100-92e^{-t/100}=90$: $e^{-t/100}=\tfrac{10}{92}=\tfrac{5}{46}$ [1]. $t=100\ln\!\tfrac{46}{5}$ [1].

Q3 (3 marks): Rewrite as $\tfrac{dB}{dt}=0.06(B-\tfrac{250}{3})$; equilibrium $B^*=5/0.06=\tfrac{250}{3}\approx83.3$ [\$'000] [1]. Separate: $B-\tfrac{250}{3}=Ce^{0.06t}$; $B(0)=70$ gives $C=70-\tfrac{250}{3}=-\tfrac{40}{3}$ [1]. $B(t)=\tfrac{250}{3}-\tfrac{40}{3}e^{0.06t}$. Since $B_0=70

Boss battle · The DE Modeller

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Science Jump · platform challenge

Climb platforms by answering DE modelling questions. Lighter alternative to the boss.

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Module overview · Extension 1 · Checkpoint 4