Unit 31 Exam Practice
Differential Equations
These questions are written in the style of IB examination questions, and they are harder than the review exercises. The marks for each part are shown in square brackets on the right, for example [3]. Paper 1 means no calculator; Paper 2 means a graphic display calculator (GDC) is allowed. Try each question under exam conditions first, then open the markscheme.
This whole unit is HL only AHL 5.18 AHL 5.19, so there is no SL set.
| Mark | Meaning |
|---|---|
| M | Method mark: a correct method is used, even with a small slip |
| A | Accuracy mark: a correct answer or line of working (usually needs the M mark before it) |
| R | Reasoning mark: a clear reason or justification is given |
| N | Mark for a correct final answer with no working shown |
| AG | “Answer given” in the question, so no extra mark is awarded for writing it |
The number after the letter is the number of marks, so A2 is worth 2 marks. In the exam, give answers exactly or correct to 3 significant figures (s.f.) unless the question says otherwise.
HL questions
These questions cover the whole unit: homogeneous and linear equations, Maclaurin series from a differential equation, Euler’s method compared with an exact solution, and a logistic model.
HL 1 Paper 1
Consider the differential equation \[ x^{2}\frac{\mathrm{d}y}{\mathrm{d}x} = (x + y)^{2} + xy, \qquad x > 0, \] where \(y = 0\) when \(x = 1\).
(a) Show that the substitution \(y = vx\), where \(v\) is a function of \(x\), transforms the equation into \[ x\frac{\mathrm{d}v}{\mathrm{d}x} = (v + 1)^{2} . \] [2]
(b) Hence find \(y\) in terms of \(x\). [4]
(c) State the value of \(x\) for which this solution is not defined. [1]
Total 7 marks
(a)
\(y = vx \;\Rightarrow\; \dfrac{\mathrm{d}y}{\mathrm{d}x} = v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}\) M1
\(x^{2}\left(v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}\right) = x^{2}(1 + v)^{2} + x^{2}v\)
\(v + x\dfrac{\mathrm{d}v}{\mathrm{d}x} = (1 + v)^{2} + v\), so \(x\dfrac{\mathrm{d}v}{\mathrm{d}x} = (v + 1)^{2}\) A1 AG
[2 marks]
(b)
\(\displaystyle\int \frac{1}{(v + 1)^{2}}\,\mathrm{d}v = \int \frac{1}{x}\,\mathrm{d}x\) M1
\(-\dfrac{1}{v + 1} = \ln x + c\) A1
\(x = 1\), \(y = 0\) gives \(v = 0\), so \(-1 = 0 + c\) and \(c = -1\) A1
\(\dfrac{1}{v + 1} = 1 - \ln x \;\Rightarrow\; \dfrac{y}{x} = \dfrac{1}{1 - \ln x} - 1\)
\(y = \dfrac{x\ln x}{1 - \ln x}\) A1
Accept any equivalent form, for example \(y = \dfrac{x}{1 - \ln x} - x\).
[4 marks]
(c)
\(1 - \ln x = 0\), so \(x = \mathrm{e}\) A1
[1 mark]
See 31.4.
HL 2 Paper 1
Consider the differential equation \[ \left(x^{2} + 1\right)\frac{\mathrm{d}y}{\mathrm{d}x} + 4xy = \frac{12x}{x^{2} + 1}, \qquad x \in \mathbb{R}, \] where \(y = 1\) when \(x = 0\).
(a) Use an integrating factor to find \(y\) in terms of \(x\). [5]
(b) Find the maximum value of \(y\). [3]
Total 8 marks
(a)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} + \dfrac{4x}{x^{2} + 1}\,y = \dfrac{12x}{\left(x^{2} + 1\right)^{2}}\) M1
\(I(x) = \mathrm{e}^{\int \frac{4x}{x^{2} + 1}\,\mathrm{d}x} = \mathrm{e}^{2\ln\left(x^{2} + 1\right)} = \left(x^{2} + 1\right)^{2}\) A1
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\Big(\left(x^{2} + 1\right)^{2}y\Big) = 12x\) M1
\(\left(x^{2} + 1\right)^{2}y = 6x^{2} + c\) A1
\(x = 0\), \(y = 1\) gives \(c = 1\), so \(y = \dfrac{6x^{2} + 1}{\left(x^{2} + 1\right)^{2}}\) A1
[5 marks]
(b)
at a stationary point \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\), so \(4xy = \dfrac{12x}{x^{2} + 1}\), giving \(y = \dfrac{3}{x^{2} + 1}\) (or \(x = 0\)) M1
\(\dfrac{6x^{2} + 1}{\left(x^{2} + 1\right)^{2}} = \dfrac{3}{x^{2} + 1} \;\Rightarrow\; 6x^{2} + 1 = 3x^{2} + 3\), so \(x^{2} = \dfrac{2}{3}\) A1
maximum value \(y = \dfrac{3}{\frac{2}{3} + 1} = \dfrac{9}{5}\) A1
Accept the quotient rule: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4x\left(2 - 3x^{2}\right)}{\left(x^{2} + 1\right)^{3}}\) for M1. The stationary point at \(x = 0\) gives \(y = 1\), a minimum.
[3 marks]
See 31.5.
HL 3 Paper 1
The function \(y = f(x)\) satisfies the differential equation \[ \frac{\mathrm{d}y}{\mathrm{d}x} = \cos x - y^{2}, \] and \(y = 1\) when \(x = 0\).
(a) Find the first three non-zero terms of the Maclaurin series for \(y\). [6]
(b) Hence find \(\displaystyle\lim_{x \to 0} \frac{1 - y}{x^{3}}\). [2]
Total 8 marks
(a)
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -\sin x - 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) M1 A1
\(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = -\cos x - 2\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^{2} - 2y\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\) A1
\(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = \sin x - 6\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} - 2y\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\) A1
at \(x = 0\): \(y' = 1 - 1 = 0\), \(\;y'' = 0\), \(\;y''' = -1\), \(\;y^{(4)} = 2\) A1
\(y = 1 - \dfrac{x^{3}}{6} + \dfrac{x^{4}}{12} + \cdots\) A1
Award M1 for using the chain rule on \(y^{2}\). Follow through from their derivatives for the last two A1 marks.
[6 marks]
(b)
\(\dfrac{1 - y}{x^{3}} = \dfrac{1}{6} - \dfrac{x}{12} + \cdots\) M1
limit \(= \dfrac{1}{6}\) A1
[2 marks]
HL 4 Paper 2
Consider the differential equation \[ \frac{\mathrm{d}y}{\mathrm{d}x} = y\,(2 - x), \] where \(y = 1\) when \(x = 0\).
(a) Show that Euler’s method with step length \(h = 0.25\) gives \[ y_{n+1} = y_n\left(1.5 - 0.25x_n\right) . \] [2]
(b) Complete the table, and hence write down an estimate for \(y(1)\). [3]
| \(n\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(x_n\) | 0 | 0.25 | 0.5 | 0.75 | 1 |
| \(y_n\) | 1 | 1.5 |
(c) By solving the differential equation, find \(y\) in terms of \(x\). [3]
(d) Find the percentage error in your estimate in part (b). [2]
(e) Show that \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = y\left((2 - x)^{2} - 1\right)\). Hence explain why Euler’s method gives an underestimate of \(y(1)\). [3]
(f) Use Euler’s method with step length \(h = 0.05\) to find a new estimate for \(y(1)\). [2]
Total 15 marks
(a)
\(y_{n+1} = y_n + 0.25\,y_n\left(2 - x_n\right)\) M1
\(= y_n + 0.5y_n - 0.25x_n y_n = y_n\left(1.5 - 0.25x_n\right)\) A1 AG
[2 marks]
(b)
\(y_2 = 1.5 \times 1.4375 = 2.15625\) (for example, using a Lists & Spreadsheet table on a GDC) A1
\(y_3 = 2.15625 \times 1.375 = 2.96484\ldots\) A1
\(y(1) \approx y_4 = 2.96484\ldots \times 1.3125 = 3.89\) \((3.89135\ldots)\) A1
[3 marks]
(c)
\(\displaystyle\int \frac{1}{y}\,\mathrm{d}y = \int (2 - x)\,\mathrm{d}x\) M1
\(\ln y = 2x - \dfrac{x^{2}}{2} + c\) A1
\(x = 0\), \(y = 1\) gives \(c = 0\), so \(y = \mathrm{e}^{2x - \frac{x^{2}}{2}}\) A1
[3 marks]
(d)
\(y(1) = \mathrm{e}^{1.5} = 4.48168\ldots\), so the error is \(\dfrac{4.48168\ldots - 3.89135\ldots}{4.48168\ldots} \times 100\) M1
\(= 13.2\%\) \((13.1720\ldots)\) A1
[2 marks]
(e)
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = (2 - x)\dfrac{\mathrm{d}y}{\mathrm{d}x} - y\) M1
\(= (2 - x)^{2}y - y = y\left((2 - x)^{2} - 1\right)\) A1 AG
for \(0 \le x < 1\), \((2 - x)^{2} > 1\) and \(y > 0\), so \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} > 0\): the curve is concave up, so each tangent step lies below the curve R1
[3 marks]
(f)
\(20\) steps of \(y_{n+1} = y_n + 0.05\,y_n\left(2 - x_n\right)\) (for example, using a Lists & Spreadsheet table on a GDC) M1
\(y(1) \approx 4.34\) \((4.33994\ldots)\) A1
[2 marks]
HL 5 Paper 1
A lake has a surface area of \(12\) hectares (ha). A water plant starts to spread across the lake. At time \(t\) months after it is first measured, the plant covers \(x\) ha of the lake, where \(0 < x < 12\). The rate of increase of \(x\) is proportional to the product of the area covered by the plant and the area of the lake not yet covered.
(a) Write down a differential equation for \(x\), using \(k\) as the constant of proportionality. [1]
(b) At \(t = 0\) the plant covers \(2\) ha. By solving your differential equation, show that \[ x = \frac{12}{1 + 5\,\mathrm{e}^{-12kt}} . \] [6]
(c) After \(3\) months the plant covers \(4\) ha. Find the exact value of \(k\). [3]
(d) Find the exact time at which the plant covers \(10\) ha. [3]
(e) Find the exact rate at which the area covered is increasing when the plant is spreading fastest. [3]
Total 16 marks
(a)
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = kx(12 - x)\), where \(k > 0\) A1
[1 mark]
(b)
\(\displaystyle\int \frac{1}{x(12 - x)}\,\mathrm{d}x = \int k\,\mathrm{d}t\) M1
\(\dfrac{1}{x(12 - x)} = \dfrac{1}{12}\left(\dfrac{1}{x} + \dfrac{1}{12 - x}\right)\) A1
\(\dfrac{1}{12}\big(\ln x - \ln(12 - x)\big) = kt + c\) A1
\(\ln\dfrac{x}{12 - x} = 12kt + C\); at \(t = 0\), \(x = 2\): \(C = \ln\dfrac{2}{10} = -\ln 5\) A1
\(\dfrac{x}{12 - x} = \dfrac{\mathrm{e}^{12kt}}{5} \;\Rightarrow\; 5x = \mathrm{e}^{12kt}(12 - x)\) M1
\(x\left(5 + \mathrm{e}^{12kt}\right) = 12\,\mathrm{e}^{12kt}\), so \(x = \dfrac{12}{1 + 5\,\mathrm{e}^{-12kt}}\) A1 AG
No modulus signs are needed because \(0 < x < 12\).
[6 marks]
(c)
\(4 = \dfrac{12}{1 + 5\,\mathrm{e}^{-36k}} \;\Rightarrow\; 1 + 5\,\mathrm{e}^{-36k} = 3\) M1
\(\mathrm{e}^{-36k} = \dfrac{2}{5}\) A1
\(k = \dfrac{1}{36}\ln\dfrac{5}{2}\) A1
[3 marks]
(d)
\(1 + 5\,\mathrm{e}^{-12kt} = \dfrac{6}{5} \;\Rightarrow\; \mathrm{e}^{-12kt} = \dfrac{1}{25}\) M1 A1
\(12kt = \ln 25\), so \(t = \dfrac{\ln 25}{\frac{1}{3}\ln\frac{5}{2}} = \dfrac{6\ln 5}{\ln\frac{5}{2}}\) months A1
Accept \(\dfrac{3\ln 25}{\ln 2.5}\) or any equivalent exact form.
[3 marks]
(e)
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = k\left(12x - x^{2}\right)\) is a quadratic in \(x\); \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(12x - x^{2}\right) = 12 - 2x = 0\) M1
\(x = 6\) (a maximum, since the coefficient of \(x^{2}\) is negative) A1
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 36k = \ln\dfrac{5}{2}\) ha per month A1
[3 marks]