Differential Equations (AA HL)
A differential equation describes how a quantity changes rather than giving its value directly - solving one means finding the function that satisfies it. This HL topic covers solving first order differential equations by separating variables, applying an initial condition to pin down a particular solution, and using Euler's method for a numerical approximation when an exact solution isn't required.
What the syllabus says
This topic is entirely HL content, covered by one point in the official IB Analysis & Approaches syllabus.
| Code | Syllabus content |
|---|---|
| AHL5.18 | First order differential equations. Numerical solution of \(\dfrac{dy}{dx}=f(x,y)\) using Euler's method. Variables separable, for example the logistic equation \(\dfrac{dn}{dt}=kn(a-n).\) Homogeneous differential equations of the form \(\dfrac{dy}{dx}=f\!\left(\dfrac{y}{x}\right)\) using the substitution \(y=vx.\) Solution of \(y'+P(x)y=Q(x)\) using an integrating factor. |
AHL5.18 is HL-only content; this page's questions focus on the variables-separable and Euler's-method parts of the syllabus point.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is a differential equation?
A differential equation is an equation involving a derivative, such as \(\dfrac{dy}{dx}\), rather than just \(x\) and \(y\) directly. Solving it means finding the original function \(y\) whose derivative satisfies the given equation.
e.g. \(\dfrac{dy}{dx}=2y\) is solved by \(y=Ae^{2x}\), since differentiating gives \(\dfrac{dy}{dx}=2Ae^{2x}=2y.\)
What does "variables separable" mean?
A differential equation has variables separable if it can be rearranged so every \(y\) term (with \(dy\)) sits on one side and every \(x\) term (with \(dx\)) sits on the other. Once separated, you integrate each side independently.
e.g. \(\dfrac{dy}{dx}=\dfrac{x^2}{y} \Rightarrow \int y\,dy=\int x^2\,dx \Rightarrow \dfrac{y^2}{2}=\dfrac{x^3}{3}+c.\)
What is a general solution vs a particular solution?
The general solution to a differential equation contains an unknown constant and represents a whole family of curves. Substituting a given initial condition finds that constant, producing the one particular solution that passes through the given point.
e.g. General solution \(y=Ae^{2x}\); with \(y(0)=5\), \(A=5\), giving the particular solution \(y=5e^{2x}.\)
What is Euler's method?
Euler's method estimates the solution to a differential equation numerically by taking small steps of size \(h\), using the approximation \(y_{n+1}=y_n+h\cdot f(x_n,y_n).\) It's an approximation, not an exact solution, and improves as \(h\) gets smaller.
e.g. For \(\dfrac{dy}{dx}=y,\ y(0)=1,\ h=0.5\): \(y_1=1+0.5(1)=1.5\), \(y_2=1.5+0.5(1.5)=2.25.\)
What is an initial condition?
An initial condition is a known pair of values, usually written like \(y(0)=5\), that the solution curve must pass through. It's substituted into the general solution to solve for the unknown constant of integration.
e.g. For \(y=20+Ae^{-kt}\) with \(y(0)=90\): \(90=20+A \Rightarrow A=70.\)
Key formulas
One core technique - separating variables - and one numerical method cover almost every question on this topic. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
Euler's method is on the official formula booklet; separating variables is a rearrangement technique built from integration you already know, rather than a single quoted formula.
| Formula | Used for | Booklet? |
|---|---|---|
| \(y_{n+1}=y_n+h\cdot f(x_n,y_n)\) | Euler's method | ✓ Yes |
| Rearrange to \(g(y)\,dy = f(x)\,dx\), then integrate both sides | Separating variables | Not a single listed formula - technique |
| \(\int\dfrac1x\,dx=\ln|x|+C\) | Common result when separating (e.g. Newton's cooling) | ✓ Yes |
Growth/decay models vs Newton's law of cooling
Both are solved the same way, by separating variables, but the two model families behave very differently in the long run.
| Feature | Growth / decay model | Newton's law of cooling |
|---|---|---|
| Equation | \(\dfrac{dP}{dt}=kP\) | \(\dfrac{dT}{dt}=-k(T-T_{\text{room}})\) |
| Solution shape | \(P=P_0e^{kt}\) | \(T=T_{\text{room}}+Ae^{-kt}\) |
| Long-run behaviour | Grows without bound (\(k>0\)) or decays to \(0\) (\(k<0\)) | Approaches the room temperature, never zero |
| Typical use | Population growth, radioactive decay | Cooling/warming objects, tank-mixing problems |
Separating and solving
The same three steps solve every variables-separable differential equation on this course.
1. Separate
Rearrange the equation so every \(y\) term is with \(dy\) on one side, and every \(x\) term is with \(dx\) on the other.
Not in the formula booklet - technique2. Integrate both sides
Integrate each side independently, remembering a single constant of integration for the whole equation (not one on each side).
Not in the formula booklet - technique3. Apply the initial condition
Substitute the given \(x\) and \(y\) values to solve for the constant, turning the general solution into the particular one.
Not in the formula booklet - techniqueNumerical and modelling methods
Not every differential equation has to be solved exactly - Euler's method gives an approximate numerical answer, and the same separating technique underlies several classic real-world models.
Euler's method
\[y_{n+1}=y_n+h\cdot f(x_n,y_n)\]
Step forward from a known starting point using the gradient at each step - smaller \(h\) gives a more accurate estimate, at the cost of more steps.
✓ In the formula bookletExponential growth & decay
\(\dfrac{dP}{dt}=kP \Rightarrow P=P_0e^{kt}.\) Used for population growth, radioactive decay and doubling/half-life problems.
Not a listed formula - modelling applicationNewton's law of cooling
\(\dfrac{dT}{dt}=-k(T-T_{\text{room}}) \Rightarrow T=T_{\text{room}}+Ae^{-kt}.\) The temperature approaches, but never reaches, the surrounding temperature.
Not a listed formula - modelling applicationWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
\(\dfrac{dy}{dx}=\dfrac{x^2}{y}.\)
(a) Find the general solution.
(b) Given that \(y=3\) when \(x=0,\) find \(C.\)
Worked solution
(a) Separate: \(\int y\,dy=\int x^2\,dx.\) M1
\(\dfrac{y^2}{2}=\dfrac{x^3}{3}+c.\) A1
So \(y^2=\dfrac{2x^3}{3}+C.\) A1
(b) \(9=0+C\Rightarrow C=9.\) A1
A body cools so that \(\dfrac{dT}{dt} = -k(T - 20)\), with \(T\) in °C. Initially \(T = 90\)°C and after 10 min \(T = 60\)°C.
(a) Solve for \(T\) in terms of \(t\) and \(k.\)
(b) Find \(k.\)
Worked solution
(a) \(\int \dfrac{dT}{T - 20} = -k\,dt \Rightarrow \ln|T - 20| = -kt + c \Rightarrow T\) M1
\(= 20 + Ae^{-kt}.\) A1
\(T(0)=90 \Rightarrow A=70\), so \(T = 20 + 70e^{-kt}.\) A1
(b) \(60 = 20 + 70e^{-10k} \Rightarrow e^{-10k}\) M1
\(= \tfrac47\) A1
\(k = -\tfrac{1}{10}\ln\tfrac47 \approx 0.0560.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Forgetting the constant of integration when separating variables. Only one constant \(c\) is needed for the whole equation, added once you've integrated both sides - not one on each side.
- Losing the modulus inside a logarithm. \(\displaystyle\int\dfrac{dT}{T-20}\) integrates to \(\ln|T-20|\), not \(\ln(T-20)\) - dropping the modulus can make later algebra (like rearranging to \(T-20=Ae^{-kt}\)) look wrong when it isn't.
- Mixing up the general and particular solution. The general solution still has an unknown constant in it - always apply the initial condition to find that constant before treating the equation as the final answer.
- Using too large a step size in Euler's method without saying so. A bigger \(h\) means fewer steps but a less accurate estimate - if the question asks for a specific \(h,\) use exactly that value, not a bigger one to save arithmetic.
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
Faster and safer than algebra for messy equations - useful once you've substituted an initial condition and just need to isolate a constant like \(k\).
- Graph \(f(x)\) first so you can see how many solutions exist and roughly where they are.
- Rearrange so everything is on one side: \(f(x) = 0\) - or graph both sides as separate functions and find intersections.
- MATH → Solver: enter the expression, type a starting guess close to one root, press ALPHA + ENTER. Move the guess to near a different root and repeat for each solution.TI-84
- Type nSolve(f(x)=0, x, guess) - include a guess or interval e.g. nSolve(f(x)=0, x, 2) or nSolve(f(x)=0, x, {1,5}) to target a specific root.Nspire
- Run-Matrix → SolveN(f(x), x) returns all real roots at once; or use the Equation app for a visual approach.Casio
- For transcendental equations (e.g. \(e^{-10k}=\tfrac47\)), graph both sides, count crossings, then use the intersection tool for each one.
- Always verify each solution by substituting back into the original equation.
Tip: The solver finds ONE root near your starting guess - change the guess to find others. The graph shows you how many to expect.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Differential equations questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What does "variables separable" mean?
It means a differential equation can be rearranged so every \(y\) term (including \(dy\)) is on one side and every \(x\) term (including \(dx\)) is on the other. Once separated, you integrate each side independently to reach the general solution.
How do I find the particular solution instead of the general one?
Solve the differential equation as normal to get the general solution, which includes an unknown constant. Then substitute the given initial condition - a known \(x\) and \(y\) pair - to work out the value of that constant.
What is Euler's method used for?
Euler's method estimates the solution to a differential equation numerically, using small steps of size \(h\) and the approximation that the curve is locally straight: each new \(y\) value is the previous one plus \(h\) times the gradient there. It's an approximation, and gets less accurate as \(h\) grows.
Are Newton's law of cooling and population growth the same type of equation?
They're both first order differential equations solved by separating variables, but they behave differently. Growth and decay models like \(\dfrac{dP}{dt} = kP\) grow or shrink without bound, while Newton's law of cooling, \(\dfrac{dT}{dt} = -k(T - T_{\text{room}})\), approaches a fixed limiting value rather than zero or infinity.
Sub-topics
Differential Equations broken down into its individual skills, each with its own focused page.
Related topics
More Calculus topics from the same AA HL syllabus unit, in case you want to keep going.