Quick Recall
Quick check — pick the correct option in each question. These test the five steps of the modeling cycle.
Step 1 is to formulate the model: write the real situation as a differential equation, e.g. “rate of growth $\propto$ size” becomes $\dfrac{\text{d}y}{\text{d}x} = ky$.
We separate the variables, collecting all $y$ terms with $\text{d}y$ and all $x$ terms with $\text{d}x$: $\dfrac{\text{d}y}{g(y)} = f(x)\,\text{d}x$.
Integration introduces an arbitrary constant $C$, giving the general solution. The constant is fixed later using the initial condition.
We use the initial condition $y(x_0) = y_0$ — the known value at the start of the process — to determine $C$ and obtain the particular solution.
Step 5 is to interpret the solution back in the context of the problem: what prediction or meaning does it give for the real situation?
The Wall
A coffee shop's daily revenue $R$ (in dollars) grows at a rate proportional to $(1000 - R)$, where $1000$ is the market saturation level. Initially $R = 200$. When $R = 500$, the revenue grows at $50$ dollars per day.
You need to predict $R$ when $t = 10$ days. This requires all five steps chained together: formulate the DE, separate variables, integrate, apply the initial conditions, and interpret. No single skill from the past four lessons is enough — you need the complete workflow.
The Need
Today we put all five steps together into a single, complete workflow. Real modeling problems do not come pre-labeled as “separation” or “initial condition” problems — you must recognize which technique to use at each stage and chain them end-to-end. This is the comprehensive application of everything from W9-L1 through L4.
💡 The Bridge (click to reveal)
The coffee shop problem requires all five steps:
1. Formulate: $\dfrac{\text{d}R}{\text{d}t} = k(1000 - R)$ for some constant $k > 0$.
2. Separate: $\dfrac{\text{d}R}{1000 - R} = k\,\text{d}t$.
3. Integrate: $-\ln|1000 - R| = kt + C$, so $R = 1000 - Ae^{-kt}$.
4. Apply IC: $R(0) = 200 \Rightarrow A = 800$. Use $\dfrac{\text{d}R}{\text{d}t} = 50$ at $R = 500$: $50 = k(500) \Rightarrow k = 0.1$. So $R = 1000 - 800e^{-0.1t}$.
5. Interpret: $R(10) = 1000 - 800e^{-1} \approx 1000 - 294.2 \approx 705.8$. After 10 days the revenue is about $\$706$. As $t \to \infty$, $R \to 1000$ (the saturation level) — revenue approaches but never exceeds $\$1000$. This interpretation step confirms the model is sensible: revenue grows toward the market saturation level, matching the real-world expectation.
Today we integrate all five skills from W9-L1 through L4 into a single end-to-end workflow. Every real modeling problem follows the same cycle: Formulate → Separate → Integrate → Apply IC → Interpret.
Given a real situation, follow these five steps:
$\bullet$ Step 1 (Formulate): Translate “rate of change is proportional to …” into $\dfrac{\text{d}y}{\text{d}x} = k \cdot f(y)$.
$\bullet$ Step 2 (Separate): Rearrange to $\dfrac{\text{d}y}{f(y)} = k\,\text{d}x$.
$\bullet$ Step 3 (Integrate): Find the general solution $y = G(x) + C$.
$\bullet$ Step 4 (Apply IC): Substitute $y(x_0) = y_0$ to find $C$.
$\bullet$ Step 5 (Interpret): Answer the real question — predict, explain, or check validity.
Before integrating, identify the form of the DE. The form determines the technique:
$\bullet$ $\dfrac{\text{d}y}{\text{d}x} = f(x)$ → Direct integration: $y = \int f(x)\,\text{d}x$.
$\bullet$ $\dfrac{\text{d}y}{\text{d}x} = f(x)\,g(y)$ → Separation of variables: $\dfrac{\text{d}y}{g(y)} = f(x)\,\text{d}x$.
$\bullet$ $\dfrac{\text{d}y}{\text{d}x} = \dfrac{P(x)}{Q(y)}$ → Separation: $Q(y)\,\text{d}y = P(x)\,\text{d}x$.
$\bullet$ $\dfrac{\text{d}N}{\text{d}t} = kN(M - N)$ → Separation + partial fractions (logistic model).
After finding a particular solution, always check by substituting it back into both the DE and the initial condition:
$\bullet$ Check the DE: compute $\dfrac{\text{d}y}{\text{d}x}$ from your solution and verify it equals the right-hand side of the DE.
$\bullet$ Check the IC: substitute $x = x_0$ into your solution and verify $y = y_0$.
If both checks pass, your solution is correct. This catches algebra errors, sign errors, and missed constants.
You are locked in the modelling lab. Three sealed doors each guard a word problem — build the model, solve it, and the door surrenders its number. The vault in the middle opens only when you enter the sum of the three door numbers. Work in teams of three: one member owns each door, and the vault is opened together. Every answer is a whole number.
Enter the sum of the numbers revealed by the three doors.
Team rule: the owner of each door explains the model to the group before the group types the number. A wrong number costs nothing but time — the clock is your opponent, not a failure.
On your own. Drag each differential equation — or tap it and then tap a box — into the family it belongs to. A card placed in the wrong box flashes red and returns to the pile, so read the shape of the right-hand side before you move. Six models, three families.
Hint: look at the right-hand side only. A single factor $y$ grows or decays exponentially; a factor like $y(M - y)$ levels off — that family is the logistic one; a power such as $\sqrt{y}$ or $\dfrac{1}{y^{2}}$ is the power family.
A poker, 80 cm long, has one end in a fire. The temperature of the poker decreases with respect to the distance from that end at a rate proportional to that distance. Halfway along the poker, the temperature is decreasing at a rate of $10^{\circ}\text{C cm}^{-1}$. Find a differential equation to model this situation.
👁 Reveal Solution
Let $T$ = temperature ($^{\circ}\text{C}$), $x$ = distance from the hot end (cm), $0 \leq x \leq 80$.
“Temperature decreases with respect to distance at a rate proportional to that distance”:
The rate of decrease of $T$ with respect to $x$ is $-\dfrac{\text{d}T}{\text{d}x}$ (negative because $T$ drops as $x$ grows).
This is proportional to $x$: $\;-\dfrac{\text{d}T}{\text{d}x} = kx\;$ for some constant $k > 0$.
Halfway along the poker: $x = 40$ cm. The rate of decrease there is $10^{\circ}\text{C cm}^{-1}$.
$-\dfrac{\text{d}T}{\text{d}x}\bigg|_{x=40} = 10$, and $-\dfrac{\text{d}T}{\text{d}x} = kx$, so $k \cdot 40 = 10$.
$k = \dfrac{10}{40} = \dfrac{1}{4}$.
$\boxed{\dfrac{\text{d}T}{\text{d}x} = -\dfrac{1}{4}x}$
Although the question only asks for the DE, the complete workflow continues:
Integrate: $T = -\dfrac{x^2}{8} + C$.
Apply IC: If $T = T_0$ at $x = 0$ (the hot end), then $C = T_0$, so $T = T_0 - \dfrac{x^2}{8}$.
Interpret: Temperature drops quadratically — slowly near the fire, rapidly at the far end. At $x = 80$: $T = T_0 - 800$. The far end is $800^{\circ}\text{C}$ cooler than the hot end.
A spherical balloon is allowed to deflate. The rate at which air is leaving the balloon is proportional to the volume $V$ of air left in the balloon. When the radius of the balloon is 15 cm, air is leaving at a rate of $8\,\text{cm}^3\text{s}^{-1}$. Find an expression for $\dfrac{\mathrm{d}V}{\mathrm{d}t}$.
👁 Reveal Solution
“Rate at which air is leaving the balloon is proportional to the volume $V$”:
Rate of leaving $= -\dfrac{\text{d}V}{\text{d}t} = kV$ for some $k > 0$.
So $\dfrac{\text{d}V}{\text{d}t} = -kV$.
When $r = 15$ cm: $V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi(15)^3 = \dfrac{4}{3}\pi \cdot 3375 = 4500\pi\,\text{cm}^3$.
Rate of leaving $= 8\,\text{cm}^3\text{s}^{-1}$: $kV = 8$, so $k \cdot 4500\pi = 8$.
$k = \dfrac{8}{4500\pi} = \dfrac{2}{1125\pi}$.
$\boxed{\dfrac{\text{d}V}{\text{d}t} = -\dfrac{2}{1125\pi}V}$
Separate: $\dfrac{\text{d}V}{V} = -\dfrac{2}{1125\pi}\,\text{d}t$.
Integrate: $\ln|V| = -\dfrac{2}{1125\pi}t + C$, so $V = V_0\,e^{-\frac{2}{1125\pi}t}$.
Apply IC: If $V(0) = 4500\pi$ (when $r = 15$), then $V_0 = 4500\pi$.
$V = 4500\pi\,e^{-\frac{2}{1125\pi}t}$.
Interpret: Volume decays exponentially. As $t \to \infty$, $V \to 0$ — the balloon never fully empties (asymptotic). The half-life is $t_{1/2} = \dfrac{1125\pi \ln 2}{2} \approx 1224$ seconds — about 20 minutes to lose half the air.
Test your understanding of all four lesson objectives. Each question targets one objective — try each before moving on!
A population grows at a rate proportional to its size $N$. Which differential equation models this?
"Rate proportional to size $N$" means $\dfrac{dN}{dt} \propto N$, so $\dfrac{dN}{dt} = kN$ where $k$ is the constant of proportionality.
The DE $\dfrac{dy}{dx} = \dfrac{x^2}{y}$ is best described as:
The DE can be written as $y\,dy = x^2\,dx$, separating the $y$ terms on one side and $x$ terms on the other. This is the hallmark of a separable DE.
Solve $\dfrac{dy}{dx} = 2xy$ with $y(0) = 3$. What is the particular solution?
Separate: $\dfrac{dy}{y} = 2x\,dx$. Integrate: $\ln|y| = x^2 + C$. So $y = Ae^{x^2}$. Apply $y(0)=3$: $3 = Ae^0 = A$, so $A = 3$. Solution: $y = 3e^{x^2}$.
The solution to a logistic DE is $N(t) = \dfrac{M}{1 + Be^{-kt}}$ where $M, B, k > 0$. What happens as $t \to \infty$?
As $t \to \infty$, $e^{-kt} \to 0$ (since $k > 0$), so the denominator $\to 1$ and $N \to M$. $M$ is the carrying capacity of the logistic model.
A tank is shaped as a cuboid with a square base of side 10 cm. Water runs out through a hole in the base at a rate proportional to the square root of the height, $h$ cm, of water in the tank. At the same time, water is pumped into the tank at a constant rate of $2\,\text{cm}^3\,\text{s}^{-1}$. Find an expression for $\dfrac{\text{d}h}{\text{d}t}$.
💡 Hint
Let the outflow rate be $k\sqrt{h}$ for some $k > 0$. The net rate of change of volume is $\dfrac{\text{d}V}{\text{d}t} = (\text{inflow}) - (\text{outflow}) = 2 - k\sqrt{h}$.
The volume is $V = (\text{base area}) \times h = 100h$.
Use the chain rule: $\dfrac{\text{d}V}{\text{d}t} = \dfrac{\text{d}V}{\text{d}h} \cdot \dfrac{\text{d}h}{\text{d}t}$.
👁 Reveal Solution
Base area $= 10 \times 10 = 100\,\text{cm}^2$.
$V = 100h$, so $\dfrac{\text{d}V}{\text{d}h} = 100$.
Inflow $= 2\,\text{cm}^3\text{s}^{-1}$ (constant).
Outflow $= k\sqrt{h}$ (proportional to $\sqrt{h}$, $k > 0$).
$\dfrac{\text{d}V}{\text{d}t} = 2 - k\sqrt{h}$.
$\dfrac{\text{d}V}{\text{d}t} = \dfrac{\text{d}V}{\text{d}h} \cdot \dfrac{\text{d}h}{\text{d}t} = 100\,\dfrac{\text{d}h}{\text{d}t}$.
So $100\,\dfrac{\text{d}h}{\text{d}t} = 2 - k\sqrt{h}$.
$\boxed{\dfrac{\text{d}h}{\text{d}t} = \dfrac{2 - k\sqrt{h}}{100}}$
where $k$ is a positive constant of proportionality.
Interpretation: When $h$ is large, outflow dominates and $h$ decreases. When $h$ is small, inflow dominates and $h$ increases. The equilibrium height (where $\dfrac{\text{d}h}{\text{d}t} = 0$) is $h = \left(\dfrac{2}{k}\right)^2$.
The variables $x$ and $y$ are related by the differential equation $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{5}xy^{\frac{1}{2}}\sin\left(\dfrac{1}{3}x\right)$.
(i) Find the general solution, giving $y$ in terms of $x$.
(ii) Given that $y = 100$ when $x = 0$, find the value of $y$ when $x = 25$.
💡 Hint
For (i): separate as $y^{-1/2}\,\text{d}y = \dfrac{1}{5}x\sin\!\left(\dfrac{x}{3}\right)\text{d}x$.
Integrate the LHS: $\int y^{-1/2}\,\text{d}y = 2\sqrt{y}$.
For the RHS, use integration by parts: $\int x\sin(ax)\,\text{d}x$ with $a = \dfrac{1}{3}$.
For (ii): substitute $x = 0$, $y = 100$ to find the constant, then evaluate at $x = 25$.
👁 Reveal Solution
Separate: $y^{-1/2}\,\text{d}y = \dfrac{1}{5}x\sin\!\left(\dfrac{x}{3}\right)\text{d}x$.
Integrate LHS: $\int y^{-1/2}\,\text{d}y = 2y^{1/2} = 2\sqrt{y}$.
Integrate RHS using integration by parts ($u = x$, $dv = \sin\!\left(\dfrac{x}{3}\right)\text{d}x$):
$\int x\sin\!\left(\dfrac{x}{3}\right)\text{d}x = -3x\cos\!\left(\dfrac{x}{3}\right) + 3\int\cos\!\left(\dfrac{x}{3}\right)\text{d}x = -3x\cos\!\left(\dfrac{x}{3}\right) + 9\sin\!\left(\dfrac{x}{3}\right)$.
So: $2\sqrt{y} = \dfrac{1}{5}\!\left[-3x\cos\!\left(\dfrac{x}{3}\right) + 9\sin\!\left(\dfrac{x}{3}\right)\right] + C$.
$\sqrt{y} = \dfrac{9}{10}\sin\!\left(\dfrac{x}{3}\right) - \dfrac{3x}{10}\cos\!\left(\dfrac{x}{3}\right) + A$ where $A = \dfrac{C}{2}$.
$\boxed{y = \left[\dfrac{9}{10}\sin\!\left(\dfrac{x}{3}\right) - \dfrac{3x}{10}\cos\!\left(\dfrac{x}{3}\right) + A\right]^2}$
Substitute $x = 0$, $y = 100$: $\sqrt{100} = 0 - 0 + A$, so $A = 10$.
$\sqrt{y} = \dfrac{9}{10}\sin\!\left(\dfrac{x}{3}\right) - \dfrac{3x}{10}\cos\!\left(\dfrac{x}{3}\right) + 10$.
At $x = 25$: $\sqrt{y} = \dfrac{9}{10}\sin\!\left(\dfrac{25}{3}\right) - \dfrac{75}{10}\cos\!\left(\dfrac{25}{3}\right) + 10$.
Since $\dfrac{25}{3} \approx 8.333$ rad, and $\dfrac{25}{3} - 2\pi \approx 2.050$ rad:
$\sin\!\left(\dfrac{25}{3}\right) \approx 0.8873$, $\;\cos\!\left(\dfrac{25}{3}\right) \approx -0.4612$.
$\sqrt{y} \approx 0.9(0.8873) - 7.5(-0.4612) + 10 \approx 0.7986 + 3.4590 + 10 \approx 14.258$.
$\boxed{y \approx 203.3}$ (to 1 d.p.)
Verification: At $x = 0$: $\sqrt{y} = 0 - 0 + 10 = 10$, so $y = 100$. ✓
Comprehensive problem: this question requires formulating the approach (separation + partial fractions), solving, applying an initial condition, and expressing $y$ explicitly — the complete workflow.
Given that $y = 1$ when $x = 0$, solve the differential equation $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x(3y^2 + 10y + 3)$, obtaining an expression for $y$ in terms of $x$.
💡 Hint
Factorise $3y^2 + 10y + 3 = (3y + 1)(y + 3)$, then use partial fractions on $\dfrac{1}{(3y+1)(y+3)}$.
Separate, integrate, exponentiate, then substitute $x = 0$, $y = 1$ to find the constant.
Finally, solve for $y$ explicitly by collecting $y$-terms.
👁 Reveal Solution
$3y^2 + 10y + 3 = (3y + 1)(y + 3)$.
Separate: $\dfrac{\text{d}y}{(3y+1)(y+3)} = 4x\,\text{d}x$.
$\dfrac{1}{(3y+1)(y+3)} = \dfrac{3/8}{3y+1} - \dfrac{1/8}{y+3}$.
$\displaystyle\int \!\left(\dfrac{3/8}{3y+1} - \dfrac{1/8}{y+3}\right)\text{d}y = \int 4x\,\text{d}x$
$\dfrac{1}{8}\ln|3y+1| - \dfrac{1}{8}\ln|y+3| = 2x^2 + C$
$\dfrac{1}{8}\ln\!\left|\dfrac{3y+1}{y+3}\right| = 2x^2 + C$
$\ln\!\left|\dfrac{3y+1}{y+3}\right| = 16x^2 + 8C$
$\dfrac{3y+1}{y+3} = Ae^{16x^2}$ where $A = e^{8C}$.
$\dfrac{3(1)+1}{1+3} = Ae^{0} \implies \dfrac{4}{4} = A \implies A = 1$.
So $\dfrac{3y+1}{y+3} = e^{16x^2}$.
$3y + 1 = (y+3)e^{16x^2}$
$3y + 1 = ye^{16x^2} + 3e^{16x^2}$
$y\!\left(3 - e^{16x^2}\right) = 3e^{16x^2} - 1$
$\boxed{y = \dfrac{3e^{16x^2} - 1}{3 - e^{16x^2}}}$
Verification: At $x = 0$: $y = \dfrac{3 - 1}{3 - 1} = 1$. ✓
Comprehensive problem: this is a full logistic model — formulate, separate, use partial fractions, integrate, apply IC, solve for $N$, and interpret the long-term behaviour.
The number of birds of a certain species in a forested region is recorded over several years. At time $t$ years, the number of birds is $N$, where $N$ is treated as a continuous variable. The variation in the number of birds is modelled by $\dfrac{\mathrm{d}N}{\mathrm{d}t} = \dfrac{N(1800 - N)}{3600}$. It is given that $N = 300$ when $t = 0$.
(i) Find an expression for $N$ in terms of $t$.
(ii) According to the model, how many birds will there be after a long time?
💡 Hint 1
This is a logistic differential equation. Separate as $\dfrac{3600}{N(1800 - N)}\,\text{d}N = \text{d}t$.
Use partial fractions: $\dfrac{3600}{N(1800 - N)} = \dfrac{A}{N} + \dfrac{B}{1800 - N}$. Multiply through by the full denominator $N(1800 - N)$ to get $3600 \equiv A(1800 - N) + BN$, then substitute $N = 0$ and $N = 1800$ to find $A$ and $B$.
💡 Hint 2
After integrating: $2\ln|N| - 2\ln|1800 - N| = t + C$, i.e. $\ln\!\left|\dfrac{N}{1800 - N}\right| = \dfrac{t}{2} + C'$.
Exponentiate and apply $N = 300$ at $t = 0$ to find the constant.
For part (ii): take $t \to \infty$ and see what $N$ approaches.
👁 Reveal Solution
Separate: $\dfrac{3600}{N(1800 - N)}\,\text{d}N = \text{d}t$.
Partial fractions: $3600 = A(1800 - N) + BN$. Set $N = 0$: $A = 2$. Set $N = 1800$: $B = 2$.
So $\dfrac{3600}{N(1800 - N)} = \dfrac{2}{N} + \dfrac{2}{1800 - N}$.
Integrate: $\int\!\left(\dfrac{2}{N} + \dfrac{2}{1800 - N}\right)\text{d}N = \int \text{d}t$.
$2\ln|N| - 2\ln|1800 - N| = t + C$.
$\ln\!\left|\dfrac{N}{1800 - N}\right| = \dfrac{t}{2} + \dfrac{C}{2}$. Let $K = \dfrac{C}{2}$: $\dfrac{N}{1800 - N} = Ae^{t/2}$ where $A = e^K$.
Apply $N = 300$ at $t = 0$: $\dfrac{300}{1500} = A \implies A = \dfrac{1}{5}$.
So $\dfrac{N}{1800 - N} = \dfrac{1}{5}e^{t/2}$.
Solve: $5N = (1800 - N)e^{t/2} \implies N(5 + e^{t/2}) = 1800e^{t/2}$.
$\boxed{N = \dfrac{1800}{1 + 5e^{-t/2}}}$
Verification: At $t = 0$: $N = \dfrac{1800}{1 + 5} = 300$. ✓
As $t \to \infty$, $e^{-t/2} \to 0$, so $N \to \dfrac{1800}{1 + 0} = 1800$.
$\boxed{N \to 1800 \text{ birds}}$
Interpretation: 1800 is the carrying capacity — the maximum sustainable population. The population grows fastest when $N = 900$ (half the carrying capacity), then slows as it approaches 1800. This is the characteristic S-curve of logistic growth.
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides future review.