Differential equations

• The constant becomes a multiplier: \(\ln y = kt + c \Rightarrow y = e^c e^{kt} = Ae^{kt}\), never \(e^{kt} + c\).
• A decrease is \(-k\) with \(k > 0\); \(\displaystyle\int \frac{1}{g(y)}\,dy = \int f(x)\,dx\) with one constant; \(\ln|y| = kt + c\) gives \(y = Ae^{kt}\), never \(e^{kt} + c\).

• 3.8.1 I can formulate — formulate a simple statement involving a rate of change as a differential equation, introducing and evaluating a constant of proportionality where necessary
• 3.8.2 I can find — find by integration a general form of solution for a first order differential equation in which the variables are separable, using any integration technique from chapter 13
• 3.8.3 I can use — use an initial condition to find a particular solution
• 3.8.4 I can interpret — interpret the solution of a differential equation in the context of a problem being modelled by the equation

Accuracy for this chapter. Leave exact answers exact: \(k = \tfrac15\ln\tfrac43\), \(\tfrac{1}{10}\ln 2\), \(y = \ln\bigl((x - 1)e^x + 2\bigr)\), \(t = 10\ln 4\), \(\ln 2\). When a decimal is asked for, give it to 3 significant figures, and never round \(k\) before using it: carry the exact form or at least four significant figures marked “unrounded” (\(0.057\,536\)). Show the equation you solved for a time before the number: an unsupported answer earns nothing.

• “\(V\) decreases at a rate proportional to \(V\), so \(\dfrac{dV}{dt} = kV\).” Correct A decrease is \(-k\) with \(k > 0\). Write \(\dfrac{dV}{dt} = -kV\), \(k > 0\). The minus sign carries the word “decreases”; leaving it to a negative \(k\) is how the sign gets lost.
• “\(\ln V = -kt + c\), so \(V = e^{-kt} + c\).” Correct The constant becomes a multiplier. \(V = e^{-kt + c} = e^c e^{-kt} = Ae^{-kt}\). Differentiate \(e^{-kt} + c\) and you get \(-ke^{-kt}\), which is not \(-k(e^{-kt} + c)\): it is not a solution.
• “\(\displaystyle\int \frac1y\,dy = \int 2\,dx\) gives \(\ln y + c_1 = 2x + c_2\), and I will find both constants.” Correct One constant, fixed by one condition. Combine them into a single \(c\) on the side of the independent variable, at the moment of integrating: \(\ln y = 2x + c\).
• “\(V = 150\) when \(t = 5\), so \(-k(150) = \dots\)” Correct A value goes into the solution, never into the differential equation. Only a rate (“decreasing at 4 \(\mathrm{cm^3\,s^{-1}}\) when \(V = 50\)”) goes into \(\dfrac{dV}{dt} = -kV\).
• “\(\theta = 20 + 60e^{-kt}\), so in the long run \(\theta \to 0\).” Correct \(e^{-kt} \to 0\), so \(\theta \to 20\). Read the limit from the whole solution, not from the exponential alone.
• “\(h = (10 - 0.2t)^2\), so at \(t = 60\), \(h = 4\).” Correct The tank is empty at \(t = 50\) and the model ends there. A formula is only as good as the situation it describes.
• “\(V\) decreases at a rate proportional to \(V\), so \(\dfrac{dV}{dt} = kV\).” Repair A decrease is a negative rate: \(\dfrac{dV}{dt} = -kV\) with \(k > 0\).
• “\(V\) increases at a rate proportional to \(V\), so \(\dfrac{dV}{dt} = V\).” Repair Introduce the constant: \(\dfrac{dV}{dt} = kV\) with \(k > 0\), and find \(k\) from the data.
• “\(\ln V = -kt + c \Rightarrow V = e^{-kt} + c\).” Repair \(V = e^{-kt + c} = e^c e^{-kt} = Ae^{-kt}\). The constant multiplies.
• “\(\ln V = -kt + c \Rightarrow V = -kt + e^c\).” Repair The exponential of a sum is a product of exponentials; exponentiate the whole right side.
• Two constants, one on each side, carried to the end. Repair Combine them into one \(c\) at the integration step; one condition can fix only one constant.
• “\(\dfrac{dy}{dx} = x + y\), so \(\displaystyle\int dy = \int (x + y)\,dx\).” Repair That is not separated (\(y\) remains on the right) and cannot be: \(x + y\) is not a function of \(x\) times a function of \(y\). Every equation to be solved in 9709 can be written as \(f(x)g(y)\).
• “\(\dfrac{dy}{dx} = xy^2\), so \(\displaystyle\int y^2\,dy = \int x\,dx\).” Repair Divide by \(g(y)\): \(\displaystyle\int y^{-2}\,dy = \int x\,dx\), giving \(-\dfrac1y = \tfrac12x^2 + c\).
• Leaving out \(c\) and fitting the initial condition to \(\ln V = -kt\). Repair Without \(c\), \(V = 1\) at \(t = 0\) whatever the question says, and no other initial condition can be satisfied. Add \(c\) at the moment of integrating.
• Substituting “\(V = 150\) when \(t = 5\)” into \(\dfrac{dV}{dt} = -kV\). Repair That is a value, not a rate: substitute it into the solution. Only a rate (“decreasing at 4 \(\mathrm{cm^3}\) per second when \(V = 50\)”) goes into the differential equation.
• Rounding \(k\) to \(0.06\) and computing \(V(20) = 200e^{-0.06 \times 20} = 60.2\). Repair The true value is \(63.3\). Even \(k = 0.0575\) (3 s.f.) drifts: it gives \(V(60) = 6.35\) against the true \(6.34\). Keep \(k\) exact (\(\tfrac15\ln\tfrac43\)) or unrounded to the end.
• “In the long run \(\theta \to 0\).” Repair \(\theta - 20 \to 0\), so \(\theta \to 20\); read the limit from the solution, not from the exponential alone.
• Using \(h = (10 - 0.2t)^2\) at \(t = 60\) to give \(h = 4\). Repair The tank emptied at \(t = 50\); the model stops there, and \(h = 0\) afterwards.
• “\(\displaystyle\int \frac{1}{x(10 - x)}\,dx = \tfrac{1}{10}\ln\bigl(x(10 - x)\bigr)\).” Repair Split into partial fractions first: \(\tfrac{1}{10}\ln\left(\dfrac{x}{10 - x}\right)\). The integral of \(\dfrac{1}{10 - x}\) is \(-\ln(10 - x)\), which is why the logarithms subtract.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters this one stands on. No later chapter of 9709 builds on differential equations, so the useful interleaving runs backwards: this chapter is where chapter 13's integration and chapter 10's logarithms are used together. When you revise chapter 13, re-solve one separable equation from Retrieval practice for each integration technique it uses; when you revise chapter 10, redo the step from \(\ln|y| = kt + c\) to \(y = Ae^{kt}\); when you revise chapter 14, re-answer the question in section D whose time can only be found by iteration. Recalling a method inside a new problem is worth more than another pass over this chapter on its own, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 3 content, examined in Paper 3. Paper 3 (Pure Mathematics 3) is compulsory for the A Level and is 30% of it. This chapter's content is in Paper 3 only; none of it is in Paper 2. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• Two forms. A pure question gives the differential equation and a condition and asks for \(y\) in terms of \(x\), possibly as a “show that” with the particular solution printed. A modelling question gives a sentence about a rate, and its parts run in order: write the equation, solve it, fix the constants from the data, then find a value, a time or the long-term behaviour. The reasoning sits at the two ends — the translation and the interpretation.
• MF19's integration table, the by-parts formula and \(\displaystyle\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)|\) are printed, and this chapter uses them through chapter 13. Everything that is special to differential equations must be known: the translation of rate statements, the separation \(\displaystyle\int \frac{1}{g(y)}\,dy = \int f(x)\,dx\), and the step from \(\ln|y| = kt + c\) to \(y = Ae^{kt}\).
• Write one constant of integration, at the moment of integrating. Keep \(k\) exact (\(k = \tfrac15\ln\tfrac43\)) or unrounded until the final answer, then give values and times to 3 significant figures. Show the equation you solved for a time, \(\left(\tfrac34\right)^{t/5} = \tfrac14\), before the number: an unsupported calculator answer earns nothing.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.