Unit 31 Review Exercises
Differential Equations
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed. All of this unit is AHL content.
1 Paper 1 AHL Show that \(y = (x + c)\,\mathrm{e}^{2x}\) is a solution of \(\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y = \mathrm{e}^{2x}\) for any constant \(c\). Find the particular solution that passes through \((0, 3)\), and the gradient of this curve at that point.
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{2x} + 2(x + c)\mathrm{e}^{2x}\), so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y = \mathrm{e}^{2x}\) ✓.
\((0, 3)\): \(c = 3\), \(\;y = (x + 3)\mathrm{e}^{2x}\). Gradient: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2(3) + \mathrm{e}^0 = 7\). See 31.1.
2 Paper 2 AHL The function \(y\) satisfies \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \sin(x + y)\) and \(y(0) = 1\). Use Euler’s method with step size \(0.1\) to estimate \(y(0.3)\).
| \(x_n\) | 0 | 0.1 | 0.2 | 0.3 |
|---|---|---|---|---|
| \(y_n\) | 1 | 1.084,15 | 1.176,76 | 1.274,89 |
\(y(0.3) \approx 1.27\) (3 s.f.). See 31.2.
3 Paper 1 AHL Solve \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = y^2\sin x\), given that \(y(0) = 1\). Give \(y\) in terms of \(x\).
\(\displaystyle\int y^{-2}\,\mathrm{d}y = \int \sin x\,\mathrm{d}x \Rightarrow -\frac{1}{y} = -\cos x + c\). \(\;x = 0\): \(-1 = -1 + c\), \(c = 0\).
\(y = \dfrac{1}{\cos x} = \sec x\). See 31.3.
4 Paper 1 AHL Use the substitution \(y = vx\) to solve \(2xy\dfrac{\mathrm{d}y}{\mathrm{d}x} = x^2 + 3y^2\), \(x > 0\), given that \(y = 1\) when \(x = 1\). Give \(y^2\) in terms of \(x\).
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1 + 3v^2}{2v}\), so \(xv' = \dfrac{1 + 3v^2}{2v} - v = \dfrac{1 + v^2}{2v}\).
\(\displaystyle\int \frac{2v}{1 + v^2}\,\mathrm{d}v = \int \frac{1}{x}\,\mathrm{d}x \Rightarrow \ln(1 + v^2) = \ln x + c \Rightarrow 1 + v^2 = Ax\).
\(x^2 + y^2 = Ax^3\); at \((1, 1)\), \(A = 2\). \(\qquad y^2 = 2x^3 - x^2\). See 31.4.
5 Paper 1 AHL Solve \(\dfrac{\mathrm{d}y}{\mathrm{d}x} - \dfrac{3y}{x} = x^3\), \(x > 0\), given that \(y(1) = 2\).
\(I = \mathrm{e}^{-3\ln x} = x^{-3}\). \(\;\left(\dfrac{y}{x^3}\right)' = 1 \Rightarrow \dfrac{y}{x^3} = x + c\). \(\;y(1) = 2\): \(c = 1\).
\(y = x^4 + x^3\). See 31.5.
6 Paper 1 AHL The function \(y\) satisfies \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x + y\), \(y(0) = 1\).
(a) Find the Maclaurin series of \(y\) up to the term in \(x^3\).
(b) Solve the equation exactly, and check your answer to (a).
(a) \(y'(0) = 1\); \(y'' = 2 + y'\), \(y''(0) = 3\); \(y''' = y''\), \(y'''(0) = 3\). \(\qquad y \approx 1 + x + \dfrac{3}{2}x^2 + \dfrac{1}{2}x^3\).
(b) \(y' - y = 2x\), \(I = \mathrm{e}^{-x}\): \(\;y\,\mathrm{e}^{-x} = \displaystyle\int 2x\,\mathrm{e}^{-x}\,\mathrm{d}x = -2(x + 1)\mathrm{e}^{-x} + c\), so \(y = -2x - 2 + c\,\mathrm{e}^x\); \(y(0) = 1\) gives \(c = 3\). \(\;y = 3\mathrm{e}^x - 2x - 2\).
Series: \(3\left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right) - 2x - 2 = 1 + x + \frac{3}{2}x^2 + \frac{1}{2}x^3\) ✓. See 31.6 and 31.5.
7 Paper 2 AHL A population satisfies \(\dfrac{\mathrm{d}P}{\mathrm{d}t} = 0.4P\left(1 - \dfrac{P}{800}\right)\), \(P(0) = 100\), with \(t\) in years.
(a) Find \(P\) in terms of \(t\).
(b) Find when \(P = 600\).
(c) Find the largest rate of growth.
(a) \(C = \dfrac{800 - 100}{100} = 7\): \(\;P = \dfrac{800}{1 + 7\mathrm{e}^{-0.4t}}\).
(b) \(1 + 7\mathrm{e}^{-0.4t} = \dfrac{4}{3} \Rightarrow \mathrm{e}^{-0.4t} = \dfrac{1}{21} \Rightarrow t = 2.5\ln 21 = 7.61\) years.
(c) Fastest growth at \(P = 400\): \(\;\dfrac{\mathrm{d}P}{\mathrm{d}t} = 0.4(400)(0.5) = 80\) per year. See 31.3.
8 Paper 2 AHL A drink at \(15\ ^\circ\mathrm{C}\) is left in a room at \(35\ ^\circ\mathrm{C}\). After 10 minutes it is at \(20\ ^\circ\mathrm{C}\). Use Newton’s law of cooling to find when the drink reaches \(30\ ^\circ\mathrm{C}\).
\(\dfrac{\mathrm{d}T}{\mathrm{d}t} = -k(T - 35) \Rightarrow T = 35 - 20\mathrm{e}^{-kt}\). \(\;20 = 35 - 20\mathrm{e}^{-10k} \Rightarrow \mathrm{e}^{-10k} = \dfrac{3}{4}\).
\(30\): \(\mathrm{e}^{-kt} = \dfrac{1}{4} \Rightarrow t = \dfrac{10\ln 4}{\ln(4/3)} = 48.2\) minutes. See 31.3.
9 Paper 1 AHL Solve \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{x}{x^2 + 4}\), given that \(y(0) = \ln 2\).
\(y = \dfrac{1}{2}\ln(x^2 + 4) + c\). \(\;x = 0\): \(\ln 2 = \dfrac{1}{2}\ln 4 + c = \ln 2 + c\), so \(c = 0\).
\(y = \dfrac{1}{2}\ln(x^2 + 4)\). See 31.1.
10 Paper 2 AHL A medicine is given through a drip. The concentration \(C\) mg L−1 in the blood after \(t\) hours satisfies \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = 5 - 0.25C\), with \(C(0) = 0\).
(a) Solve the equation.
(b) State the long-term concentration.
(c) Find when \(C = 15\).
(a) \(C' + 0.25C = 5\), \(I = \mathrm{e}^{0.25t}\): \(\;C\,\mathrm{e}^{0.25t} = 20\mathrm{e}^{0.25t} + c\); \(C(0) = 0\) gives \(c = -20\). \(\;C = 20\left(1 - \mathrm{e}^{-0.25t}\right)\).
(b) \(C \to 20\) mg L−1.
(c) \(\mathrm{e}^{-0.25t} = 0.25 \Rightarrow t = 4\ln 4 = 5.55\) hours. See 31.5 and 31.6.