IB Math AI HL Differential Equations — Complete Cheatsheet
Every differential-equations skill for IB Mathematics Applications & Interpretation HL — forming a DE from a described rate, exponential growth & decay, Newton's law of cooling, separable equations, logistic models, slope fields and Euler's method, all built into real modelling problems. Hand-built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Differential equations are where the IB Mathematics Applications & Interpretation HL calculus strand turns into genuine modelling. A differential equation (DE) ties a quantity to its own rate of change, so an AI question rarely asks you to solve one in the abstract — it asks you to build the equation from a described rate ("the rate of cooling is proportional to..."), then either solve it (separate the variables and integrate, or read off the standard growth / cooling / logistic template) or step it forward numerically with Euler's method when no neat formula exists. A GDC is always allowed, so the marks live in setting up the model, choosing the method and reading the answer back with the right units.
This cheatsheet condenses the whole range — from forming and solving DEs in AHL 5.14, through slope fields in AHL 5.15, to Euler's method in AHL 5.16 — onto one page, and flags the traps that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — Forming a differential equation AHL 5.14
A differential equation contains a derivative: it states the rule for how fast a quantity changes. Most AI questions hand you that rule in words, so your first job is to translate the sentence into calculus.
Rate of change:$\dfrac{dQ}{dt}$ is the phrase "the rate of change of $Q$" — this is always the left-hand side
Proportional to size:"rate is proportional to the amount present" $\Rightarrow \dfrac{dQ}{dt}=kQ$
Proportional to a gap:"rate is proportional to how far $Q$ is from $A$" $\Rightarrow \dfrac{dQ}{dt}=k(Q-A)$
Then solve it:separate the variables and integrate, or step forward with Euler $y_{n+1}=y_n+h\,f(x_n,y_n)$
NotePick the sign of $k$ from the context: growth uses $+k$, while decay or cooling uses $-k$ so the quantity falls. The constant of proportionality $k$ itself is found later, from the data.
TrickWrite "rate $=$" on the left first, then translate only the words that come after "proportional to" into the right-hand side. Almost every AI DE is one of the two templates above.
§2 — Exponential growth & decay AHL 5.14
The archetype DE: a quantity whose rate is proportional to its own size. One template covers app users, bacterial mass, radioactive decay and continuously compounded value.
Model:$\dfrac{dN}{dt}=kN$
Solution:$N=N_0e^{kt}$, where $N_0$ is the initial value at $t=0$
Growth vs decay:$k>0$ grows, $k<0$ decays; the doubling time or half-life comes from $e^{kt}=2$ or $\tfrac12$
Continuous rate:a continuous rate of $r\%$ per period means $k=\dfrac{r}{100}$
TrickWork out the exponent $kt$ first, then raise $e$ to that power on the GDC in one step — it avoids rounding a small $k$ too early. See Exponents & Logarithms for the log work behind it.
NoteRead $k$ straight off the equation whenever the DE is already written as $\dfrac{dN}{dt}=kN$. If instead you are given data, find $k$ first with the method in §3.
§3 — Finding the rate constant $k$ AHL 5.14
When $k$ is not given, two data points pin it down. Rearranging $N=N_0e^{kt}$ with logarithms isolates $k$.
From the start value:using $(t_1,N_1)$: $\;k=\dfrac{1}{t_1}\ln\!\dfrac{N_1}{N_0}$
From any two points:$k=\dfrac{1}{t_2-t_1}\ln\!\dfrac{N_2}{N_1}$
TrickDivide the two amounts first, take $\ln$ of that single number, then divide by the elapsed time — three clean calculator steps.
TrapFor decay the ratio $\dfrac{N_1}{N_0}$ is less than $1$, so its $\ln$ is negative and $k$ comes out negative. Keep that minus sign — it is what makes the model decay.
§4 — Newton's law of cooling AHL 5.14
An object cools at a rate proportional to how far its temperature is above the surroundings — the "proportional to a gap" template from §1. This is the DE behind forensic time-of-death estimates.
Model:$\dfrac{d\theta}{dt}=-k(\theta-\theta_s)$, where $\theta_s$ is the surrounding (ambient) temperature
Solution:$\theta=\theta_s+(\theta_0-\theta_s)e^{-kt}$ — the object cools toward $\theta_s$
Find $k$:a second reading gives $e^{-kt_1}=\dfrac{\theta_1-\theta_s}{\theta_0-\theta_s}$, so $k=-\dfrac{1}{t_1}\ln\!\dfrac{\theta_1-\theta_s}{\theta_0-\theta_s}$
Find the time:isolate $e^{-kt}$, take $\ln$, then divide by $-k$
NoteKeep the ambient term $\theta_s$ separate: only the difference $(\theta_0-\theta_s)$ is multiplied by $e^{-kt}$. Once $k$ is known you can predict any later temperature or the time to reach it.
TrapGet $e^{-kt}$ completely on its own before taking logs — you cannot $\ln$ a sum such as $\theta_s+(\dots)e^{-kt}$ term by term. Subtract $\theta_s$ first.
§5 — Separable first-order DEs AHL 5.14
The general method whenever the two variables can be split onto opposite sides. Gather each variable with its own differential, integrate both sides, then use the initial condition.
Separable form:$\dfrac{dy}{dx}=f(x)\,g(y)$
Method:$\displaystyle\int\frac{1}{g(y)}\,dy=\int f(x)\,dx$, then add a single constant $+C$
Example 1:$\dfrac{dy}{dx}=\dfrac{x}{y}\Rightarrow\displaystyle\int y\,dy=\int x\,dx\Rightarrow y^2=x^2+K$
Example 2:$\dfrac{dy}{dx}=xy\Rightarrow \ln y=\tfrac{x^2}{2}+C\Rightarrow y=Ae^{x^2/2}$
TrickSubstitute the given boundary point straight into the integrated equation to pin down the constant — do this before rearranging into an explicit $y$.
TrapThe $+C$ appears the moment you integrate, not at the end. And when a square root makes $y$ explicit (as in $y^2=x^2+K$), choose the sign that matches the initial condition — if $y_0>0$, take the positive root.
Solution:$P=\dfrac{K}{1+Ae^{-rt}}$, with $A=\dfrac{K-P_0}{P_0}$
Fastest growth:at $P=\dfrac{K}{2}$ (the steepest point); the initial rate is $rP_0\left(1-\dfrac{P_0}{K}\right)$
TrickGet $A$ from the initial value first, then substitute the required $t$. As $t\to\infty$, $e^{-rt}\to0$ and $P\to K$ — a quick sanity check on your answer.
NoteThe curve is S-shaped: near-exponential early on, then flattening toward $K$. It is the AI staple for disease spread, product adoption and constrained populations.
§7 — Approach to equilibrium: mixing & dosing AHL 5.14
Mixing tanks, drug doses and cooling all share one shape: a constant inflow against an outflow proportional to the current amount. The quantity relaxes exponentially to a steady state.
TrickFind the equilibrium $L=\dfrac{a}{b}$ first; then $e^{-bt}=\dfrac{y-L}{y_0-L}$ solves for the time to reach any level — exactly the Newton-cooling move from §4.
NoteThis is the same mathematics as cooling, with $L$ playing the role of the ambient temperature: whatever the start value, the system decays smoothly toward $L$.
§8 — Slope fields AHL 5.15
A slope field lets you read a DE geometrically without solving it. At each grid point the DE gives a gradient; drawing a short segment of that slope everywhere produces the field, and any solution curve simply threads through, staying tangent to the segments.
A slope field for a cooling / relaxation model $\frac{d\theta}{dt}=-k(\theta-L)$: each short segment is the gradient the DE prescribes there — steep far from equilibrium, flat close to it. The two solution curves (gold) flow tangent to the field, both relaxing toward $\theta=L$.
Segment slope:substitute the point $(x,y)$ into $\dfrac{dy}{dx}$ to get the gradient drawn there
TrickA solution curve is tangent to the field everywhere. Start at the initial point and "flow" with the segments — you can sketch the behaviour of $y$ without any integration.
NoteEquilibrium solutions show up as rows of horizontal segments, where $\dfrac{dy}{dx}=0$ — for example $\theta=\theta_s$ in cooling, or $P=K$ in the logistic model.
§9 — Euler's method: numerical solution AHL 5.16
When a DE has no tidy formula, Euler's method marches the solution forward along the tangent slope, one small step $h$ at a time.
Update rule:$y_{n+1}=y_n+h\,f(x_n,y_n)$, and $x_{n+1}=x_n+h$
One step:take the gradient $f(x_n,y_n)$ at the old point, multiply by $h$, add to $y_n$
Step length:smaller $h$ means more steps but a closer estimate
Euler's method steps along straight tangent pieces (navy). Because the exact solution (gold) curves upward — it is convex — each step lands slightly low, so the running estimate drifts below the truth. The gap at $x_4$ is the error, and it shrinks as the step $h$ is reduced.
TrapUpdate $y$ using the old $x$ and $y$, then advance $x$ — do it one step at a time. Feeding the new $x$ into the same step is the classic Euler error.
NoteEuler walks straight tangent pieces, so on a convex (upward-curving) solution it steadily drifts below the true value — an underestimate. When you judge the accuracy, always put the exact value in the denominator of the percentage error, never the estimate.
§10 — Exam attack plan All sections
Question cue
What to do
Watch for
"the rate of ... is proportional to ..."
Form $\dfrac{dQ}{dt}=kQ$ or $k(Q-A)$; name the variable first
Sign of $k$: $+$ growth, $-$ decay / cooling
"grows / decays exponentially; find the amount"
$N=N_0e^{kt}$; a continuous $r\%$ gives $k=r/100$
Compute the exponent $kt$ before pressing $e^x$
"find the growth / decay constant"
Two points: $k=\dfrac{1}{t_2-t_1}\ln\dfrac{N_2}{N_1}$
Decay makes $k$ negative — keep the sign
"temperature after $t$ / when does it cool to..."
$\theta=\theta_s+(\theta_0-\theta_s)e^{-kt}$; isolate $e^{-kt}$, then $\ln$
Keep $\theta_s$ separate; subtract it before logs
"solve $\dfrac{dy}{dx}=f(x)g(y)$"
Separate: $\int\frac{1}{g}\,dy=\int f\,dx$, add $+C$, apply the condition
Exact in the denominator; Euler underestimates a convex curve
Worked Example — Cooling Coffee by Euler's Method
Question (AI HL Paper 2 style — 8 marks)
A cup of coffee at $90^\circ$C is left to cool in a room held at a constant $20^\circ$C. Its temperature $\theta$ (in $^\circ$C) after $t$ minutes is modelled by $$\frac{d\theta}{dt}=-0.1(\theta-20),\qquad \theta(0)=90.$$ (a) Using Euler's method with step length $h=5$, estimate the temperature after $15$ minutes. (b) Solve the differential equation and hence find the exact temperature after $15$ minutes. (c) Find the percentage error of the Euler estimate. (d) Explain why Euler's method underestimates the true temperature here.
Solution
(a) The update rule is $\theta_{n+1}=\theta_n+h\,f(t_n,\theta_n)=\theta_n+5\big[-0.1(\theta_n-20)\big]$. (M1)
$n$
$t_n$
$\theta_n$
$f=-0.1(\theta_n-20)$
$\theta_{n+1}=\theta_n+5f$
0
0
90
$-7$
$90+5(-7)=55$
1
5
55
$-3.5$
$55+5(-3.5)=37.5$
2
10
37.5
$-1.75$
$37.5+5(-1.75)=28.75$
3
15
28.75
—
—
So after $15$ minutes, $\theta\approx28.75\approx28.8^\circ$C. (A1)(A1)
(b) Separate the variables and integrate: $\displaystyle\int\frac{d\theta}{\theta-20}=\int-0.1\,dt\Rightarrow\ln|\theta-20|=-0.1t+C$, so $\theta-20=Ae^{-0.1t}$. The condition $\theta(0)=90$ gives $A=70$, hence $\theta=20+70e^{-0.1t}$. At $t=15$: $\theta=20+70e^{-1.5}=35.6^\circ$C (3 s.f.). (M1)(A1)
(d) The exact curve is decreasing but concave up — its slope flattens as $\theta\to20$. Euler follows the steeper starting tangent across each whole step, so it drops too fast and lands below the true curve, giving an underestimate. (R1)
Examiner's note: keep the sign of $f$ (negative, because the coffee is cooling) and read $e^{-1.5}$ straight off the GDC. The step $h=5$ is deliberately coarse, which is why the error is as large as $19\%$; halving $h$ would roughly halve it. Always put the exact value in the denominator of the percentage error — using the estimate there is a common slip.
Common Student Questions
How do I turn a worded "rate is proportional to" sentence into a differential equation?
Put the derivative on the left and the proportional quantity (times $k$) on the right. "The rate of change of $N$ is proportional to $N$" becomes $\dfrac{dN}{dt}=kN$; "the rate of cooling is proportional to the temperature above the room" becomes $\dfrac{d\theta}{dt}=-k(\theta-\theta_s)$. Use $+k$ for growth and $-k$ for decay or cooling, and find the actual value of $k$ later from the data.
When do I separate the variables and when do I use Euler's method?
If the DE is separable — you can get all the $y$ terms on one side and all the $x$ terms on the other — and both integrals are doable, solve it exactly by integrating. Use Euler's method when the equation cannot be separated or integrated neatly, or whenever the question explicitly asks for a numerical estimate with a given step $h$. In AI, exponential, cooling and simple separable models are solved exactly; awkward ones are stepped with Euler.
Why does Euler's method usually underestimate the answer?
Euler replaces the curved solution with straight tangent steps. If the true solution is convex (curving upward, concave up), each straight step cuts underneath the curve, so the running total falls below the exact value. Making the step length $h$ smaller reduces this gap. When you report the percentage error, the exact value always goes in the denominator, never the Euler estimate.
In Newton's law of cooling, what is $k$ and how do I find it?
$k$ is a positive cooling constant that sets how fast the object loses heat. Rearrange the solution $\theta=\theta_s+(\theta_0-\theta_s)e^{-kt}$ so the exponential is on its own: $e^{-kt_1}=\dfrac{\theta_1-\theta_s}{\theta_0-\theta_s}$ using a second temperature reading $(\,t_1,\theta_1)$. Take natural logs and divide by $-t_1$ to get $k$. Keep $k>0$; the minus sign in the model already produces the decay.
Do I need the "$+C$" when I integrate a differential equation?
Yes — always. Integrating gives the general solution, which contains an arbitrary constant ($+C$, or the $A$ inside $Ae^{kt}$). The initial or boundary condition then fixes that constant to give the one particular solution the question wants. Add the constant the moment you integrate and substitute the given point straight away; forgetting it leaves you with a family of curves instead of the answer.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — every model from exponential growth to logistic and cooling, with fully worked separation-of-variables derivations and edge cases.
Tutorial booklet — 25+ AI HL-style questions from forming a DE to Newton cooling, logistic modelling and Euler's method with error analysis.
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations, Euler tables and GDC screenshots.
Predicted-paper questions — the exact differential-equation and Euler's-method question types most likely in the next session.
