The shape of the lesson
By the end of the lesson, learners can
- Separate, integrate, add one constant: Treat \(\dfrac{dy}{dx}\) as if it were a fraction: divide by the \(y\)-part and 'multiply' by \(dx\), so that each side has only one variable.
- Particular solutions in context: In a modelling question the initial condition is usually hidden in words ('initially', 'at the start', 'when \(t = 0\)'), and a second piece of data may fix the constant of proportionality.
- Using partial fractions and other integrals: The integral on the \(y\)-side may need any technique from section 3.5: partial fractions, \(\dfrac{1}{x^{2} + a^{2}}\), trigonometric identities.
- Separate: \(\displaystyle\int \frac{1}{g(y)}\,dy = \int f(x)\,dx\); split \(e^{a + b}\) into \(e^{a}e^{b}\) first.
- Integrate both sides, using partial fractions or standard integrals as needed.
- One constant, found from the initial condition straight away.
- Rearrange to the form asked for; exponentiating turns \(+C\) into a multiplier.
- Interpret: answer in context with units, and say what happens as \(t \to \infty\).
One picture every idea returns to
If \(\dfrac{dy}{dx} = f(x)\,g(y)\), then \(\displaystyle\int \frac{1}{g(y)}\,dy = \int f(x)\,dx\). Add ONE constant after integrating, use the initial condition to find it, and only then rearrange to make \(y\) the subject.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Sort the variables, then integrate.. Paper 3 only. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Separate, integrate, add one constant | Treat \(\dfrac{dy}{dx}\) as if it were a fraction: divide by the \(y\)-part and 'multiply' by \(dx\), so that each side has only one variable. Model the worked example A first separable equation on the board, one line at a time. Your turn: Solve \(\dfrac{dy}{dx} = y\cos x\), given that \(y = 2\) when \(x = 0\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: Particular solutions in context | In a modelling question the initial condition is usually hidden in words ('initially', 'at the start', 'when \(t = 0\)'), and a second piece of data may fix the constant of proportionality. Model the worked example Newton's law of cooling on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Using partial fractions and other integrals | The integral on the \(y\)-side may need any technique from section 3.5: partial fractions, \(\dfrac{1}{x^{2} + a^{2}}\), trigonometric identities. Model the worked example Logistic growth on the board, one line at a time. Your turn: Solve \(\dfrac{dy}{dx} = x(y^{2} + 1)\), given that \(y = 1\) when \(x = 0\). | Think-Pair-Share |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 29 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Using partial fractions and other integrals, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| \(\ln y = 2x + C\) gives \(y = e^{2x + C} = Ae^{2x}\), NOT \(y = e^{2x} + C\). | The constant becomes a multiplier when you exponentiate, so find \(C\) BEFORE removing the logarithm, or write \(A = e^{C}\). |
| Do not drop the 20: \(\theta = 80e^{-0.05t}\) solves a different equation, and it cools to zero instead of to room temperature. | Keep \(\theta - 20\) together throughout. |
| \(\displaystyle\int \frac{1}{20 - N}\,dN = -\ln(20 - N)\), not \(+\ln(20 - N)\): the inside has derivative \(-1\). | Losing this sign turns the logistic curve into nonsense. |
Solving by separating the variables: worksheet
6 questions, 29 marks. Written in the style of the papers; no past paper question is reproduced.
Find the general solution of \(\dfrac{dy}{dx} = \dfrac{2y}{x}\) for \(x > 0\), \(y > 0\), giving \(y\) in terms of \(x\).
Solve \(\dfrac{dx}{dt} = \cos^{2}x\), given that \(x = 0\) when \(t = 0\). Find the value of \(t\) when \(x = \tfrac{\pi}{4}\).
Solve \(\dfrac{dy}{dx} = x(y^{2} + 1)\), given that \(y = 1\) when \(x = 0\), giving \(y\) in terms of \(x\).
A metal bar cools according to \(\dfrac{d\theta}{dt} = -0.05(\theta - 18)\), where \(\theta\) is in °C and \(t\) in minutes. Initially \(\theta = 98\). (a) Find \(\theta\) in terms of \(t\). (b) Find the time taken for the bar to cool to \(38^\circ\)C. (c) State the temperature the bar approaches.
The number of people \(x\) (in thousands) who have heard a rumour satisfies \(\dfrac{dx}{dt} = \dfrac{x(6 - x)}{12}\), where \(t\) is in days. Initially \(x = 2\).
(a) Show that \(x = \dfrac{6}{1 + 2e^{-t/2}}\). (b) Find the time when 5000 people have heard the rumour. (c) State what happens to \(x\) in the long term.
The depth \(h\) metres of water in a draining tank satisfies \(\dfrac{dh}{dt} = -0.02\sqrt{h}\), where \(t\) is in minutes. Initially \(h = 0.81\).
(a) Solve the differential equation, giving \(h\) in terms of \(t\). (b) Find the depth after 30 minutes. (c) Find the time taken for the tank to empty.
Answer key and marking notes
- \(\displaystyle\int \frac{1}{y}\,dy = \int \frac{2}{x}\,dx\), so \(\ln y = 2\ln x + C\).
- \(y = e^{C}e^{2\ln x} = Ax^{2}\), where \(A = e^{C}\).
- \(\displaystyle\int \sec^{2}x\,dx = \int dt\), so \(\tan x = t + C\).
- \(x = 0\), \(t = 0\): \(C = 0\), so \(\tan x = t\), i.e. \(x = \tan^{-1} t\).
- \(x = \tfrac{\pi}{4}\): \(t = \tan\tfrac{\pi}{4} = 1\).
- \(\displaystyle\int \frac{1}{y^{2} + 1}\,dy = \int x\,dx\), so \(\tan^{-1} y = \tfrac{1}{2}x^{2} + C\).
- \(x = 0\), \(y = 1\): \(C = \tfrac{\pi}{4}\).
- \(y = \tan\left(\tfrac{1}{2}x^{2} + \tfrac{\pi}{4}\right)\).
- (a) \(\displaystyle\int \frac{1}{\theta - 18}\,d\theta = \int -0.05\,dt\): \(\ln(\theta - 18) = -0.05t + C\), \(C = \ln 80\).
- \(\theta = 18 + 80e^{-0.05t}\).
- (b) \(80e^{-0.05t} = 20\), so \(e^{-0.05t} = \tfrac{1}{4}\) and \(t = 20\ln 4 = 27.7\) minutes (3 s.f.).
- (c) \(e^{-0.05t} \to 0\), so \(\theta \to 18^\circ\)C, the room temperature.
- (a) \(\displaystyle\int \frac{12}{x(6 - x)}\,dx = \int dt\), with \(\dfrac{12}{x(6 - x)} = \dfrac{2}{x} + \dfrac{2}{6 - x}\).
- \(2\ln x - 2\ln(6 - x) = t + C\); \(t = 0\), \(x = 2\): \(C = 2\ln\tfrac{2}{4} = -2\ln 2\).
- \(2\ln\dfrac{2x}{6 - x} = t\), so \(\dfrac{2x}{6 - x} = e^{t/2}\), \(2x = 6e^{t/2} - xe^{t/2}\), \(x = \dfrac{6e^{t/2}}{2 + e^{t/2}} = \dfrac{6}{1 + 2e^{-t/2}}\).
- (b) \(x = 5\): \(\dfrac{10}{1} = e^{t/2}\), so \(t = 2\ln 10 = 4.61\) days (3 s.f.).
- (c) \(e^{-t/2} \to 0\), so \(x \to 6\): about 6000 people hear the rumour in the long run.
- (a) \(\displaystyle\int h^{-1/2}\,dh = \int -0.02\,dt\): \(2\sqrt{h} = -0.02t + C\).
- \(t = 0\): \(C = 2\sqrt{0.81} = 1.8\), so \(\sqrt{h} = 0.9 - 0.01t\) and \(h = (0.9 - 0.01t)^{2}\).
- (b) \(t = 30\): \(h = 0.6^{2} = 0.36\) m.
- (c) Empty when \(\sqrt{h} = 0\): \(0.9 - 0.01t = 0\), so \(t = 90\) minutes.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Find the general solution of \(\dfrac{dy}{dx} = 3x^{2}y\), for \(y > 0\).
- A\(y = Ae^{x^{3}}\)
- B\(y = e^{x^{3}} + c\)
- C\(y = Ax^{3}\)
- D\(y = x^{3} + c\)
Solve \(\dfrac{dy}{dx} = \dfrac{x}{y}\), given that \(y = 3\) when \(x = 4\). Give \(y^{2}\) in terms of \(x\).
- A\(y^{2} = x^{2} - 7\)
- B\(y^{2} = x^{2} + 7\)
- C\(y^{2} = x^{2} - 14\)
- D\(y^{2} = x^{2} - \dfrac{7}{2}\)
After separating the variables in \(\dfrac{dy}{dx} = e^{-y}\cos x\) and integrating, which equation is correct?
- A\(e^{y} = \sin x + c\)
- B\(-e^{-y} = \sin x + c\)
- C\(e^{-y} = \sin x + c\)
- D\(e^{y} = -\sin x + c\)
Given \(\dfrac{dy}{dx} = 2y\) and \(y = 5\) when \(x = 0\), find the exact value of \(y\) when \(x = 1\).
- A\(5e^{2}\)
- B\(e^{2} + 4\)
- C\(15\)
- D\(5e^{\frac{1}{2}}\)
A quantity \(x\) satisfies \(\dfrac{dx}{dt} = k(10 - x)\), where \(k\) is a positive constant and \(x = 0\) when \(t = 0\). Find \(x\) in terms of \(k\) and \(t\).
- A\(x = 10(1 - e^{-kt})\)
- B\(x = 10e^{-kt}\)
- C\(x = 10(1 - e^{kt})\)
- D\(x = 10 - e^{-kt}\)
Solve \(\dfrac{dy}{dx} = xy^{2}\), given that \(y = 1\) when \(x = 0\). Express \(y\) in terms of \(x\).
- A\(y = \dfrac{2}{2 - x^{2}}\)
- B\(y = \dfrac{2}{2 + x^{2}}\)
- C\(y = 1 - \dfrac{x^{2}}{2}\)
- D\(y = e^{\frac{x^{2}}{2}}\)
Solve \(\dfrac{dy}{dx} = y\tan x\), for \(-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}\), given that \(y = 2\) when \(x = 0\).
- A\(y = 2\sec x\)
- B\(y = 2\cos x\)
- C\(y = \sec x + 1\)
- D\(y = 2 + \ln(\sec x)\)
A population \(N\) satisfies \(\dfrac{dN}{dt} = kN\). When \(t = 0\), \(N = 100\), and when \(t = 2\), \(N = 150\). Find the time at which \(N = 400\), to 3 significant figures.
- A\(6.84\)
- B\(3.42\)
- C\(29.6\)
- D\(12\)
Quiz key
\(\ln y = x^{3} + c\), so \(y = e^{x^{3} + c} = e^{c}e^{x^{3}} = Ae^{x^{3}}\). The tempting \(e^{x^{3}} + c\) adds the constant after exponentiating, but \(e^{x^{3} + c}\) is \(e^{c}\) TIMES \(e^{x^{3}}\). \(Ax^{3}\) treats \(\ln y = \ln x^{3}\), and \(x^{3} + c\) integrates \(3x^{2}\) but ignores the \(y\) on the right.
\(\dfrac{1}{2}y^{2} = \dfrac{1}{2}x^{2} + c\). At \((4, 3)\): \(\dfrac{9}{2} = 8 + c\), \(c = -\dfrac{7}{2}\). So \(y^{2} = x^{2} - 7\). Check: \(16 - 7 = 9\). The tempting \(x^{2} + 7\) gets the sign of the constant wrong (\(9 = 16 + C\) gives \(C = -7\)). \(x^{2} - 14\) doubles the halves but not the constant, using \(c = -7\) in \(\dfrac{1}{2}y^{2} = \dfrac{1}{2}x^{2} + c\), and \(x^{2} - \dfrac{7}{2}\) doubles the \(\dfrac{1}{2}y^{2}\) and \(\dfrac{1}{2}x^{2}\) but forgets to double \(c\).
\(\dfrac{1}{e^{-y}} = e^{y}\), so \(\displaystyle\int e^{y}\,dy = \int \cos x\,dx\) gives \(e^{y} = \sin x + c\). The tempting \(-e^{-y}\) integrates \(e^{-y}\) without separating it to the other side first (it is on the right, multiplying \(\cos x\)). \(e^{-y} = \sin x + c\) also fails to move it, and \(-\sin x\) is the integral of \(-\cos x\), not of \(\cos x\).
\(\ln y = 2x + c\), so \(y = Ae^{2x}\); \(y(0) = 5\) gives \(A = 5\), so \(y(1) = 5e^{2}\). The tempting \(e^{2} + 4\) writes \(y = e^{2x} + c\) (adding the constant after exponentiating). \(15\) treats the rate as constant at its starting value \(2 \times 5 = 10\) for one unit, and \(5e^{\frac{1}{2}}\) divides by the 2 instead of keeping it as a factor, getting \(\ln y = \dfrac{x}{2} + c\).
\(-\ln(10 - x) = kt + c\). At \(t = 0\), \(x = 0\): \(c = -\ln 10\). So \(\ln\dfrac{10 - x}{10} = -kt\), \(10 - x = 10e^{-kt}\) and \(x = 10(1 - e^{-kt})\), which rises from 0 towards 10. The tempting \(10e^{-kt}\) is \(10 - x\), not \(x\), and starts at 10 rather than 0. \(10(1 - e^{kt})\) loses the minus sign from integrating \(\dfrac{1}{10 - x}\) (it goes negative), and \(10 - e^{-kt}\) has the constant in the wrong place, so it does not give \(x = 0\) at \(t = 0\).
\(-\dfrac{1}{y} = \dfrac{x^{2}}{2} + c\); at \((0, 1)\), \(c = -1\). So \(\dfrac{1}{y} = 1 - \dfrac{x^{2}}{2} = \dfrac{2 - x^{2}}{2}\) and \(y = \dfrac{2}{2 - x^{2}}\). The tempting \(\dfrac{2}{2 + x^{2}}\) integrates \(y^{-2}\) as \(+\dfrac{1}{y}\). \(1 - \dfrac{x^{2}}{2}\) is \(\dfrac{1}{y}\), not \(y\), and \(e^{\frac{x^{2}}{2}}\) integrates \(\dfrac{1}{y^{2}}\) as \(\ln y\).
\(\ln y = \ln(\sec x) + c\), so \(y = A\sec x\); \(y(0) = 2\) gives \(A = 2\), so \(y = 2\sec x\). The tempting \(2\cos x\) takes \(\displaystyle\int \tan x\,dx = \ln\cos x\), missing the minus sign. \(\sec x + 1\) adds the constant after exponentiating, and \(2 + \ln(\sec x)\) forgets that the left side was \(\ln y\), not \(y\).
\(150 = 100e^{2k}\) gives \(k = \dfrac{1}{2}\ln 1.5\). Then \(e^{kt} = 4\), so \(t = \dfrac{\ln 4}{k} = \dfrac{2\ln 4}{\ln 1.5} = 6.84\). The tempting \(3.42\) uses \(k = \ln 1.5\), forgetting to divide by the time 2. \(29.6\) solves \(e^{kt} = 400\), forgetting the initial value 100, and \(12\) assumes linear growth of 50 every 2 time units.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.