| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Logistic/bounded growth |
| Difficulty | Standard +0.8 This is a substantial logistic differential equation problem requiring: (a) interpreting a rate statement to find the proportionality constant (non-trivial setup), and (b) separating variables, partial fractions decomposition, integration, and solving for x at a specific time. The multi-step nature, need to interpret the initial rate condition correctly, and algebraic complexity of solving the logistic equation place this above average difficulty. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) When \(x = \frac{1}{4}, \frac{dx}{dt} = \frac{3}{4} + 6 = \frac{7}{8}\) | M1 A1 | |
| \(\frac{dx}{dt} = k(1-x) \therefore \frac{7}{8} = k \times \frac{1}{4} \times \frac{3}{4}, k = \frac{7}{3} \therefore \frac{dx}{dt} = \frac{7}{3}x(1-x)\) | M1 A1 | |
| (b) \(\int\frac{1}{x(1-x)} \, dx = \int\frac{7}{3} \, dt\) | M1 | |
| \(\frac{1}{x(1-x)} = \frac{A}{x} + \frac{B}{1-x}, 1 \equiv A(1-x) + Bx\) | M1 | |
| \(x = 0 \Rightarrow A = 1\) | A1 | |
| \(x = 1 \Rightarrow B = 1\) | A1 | |
| \(\therefore \int(\frac{1}{x} + \frac{1}{1-x}) \, dx = \int\frac{7}{3} \, dt\) | M1 | |
| \(\ln | x | - \ln |
| \(t = 0, x = \frac{1}{4} \Rightarrow \ln\frac{1}{4} - \ln\frac{3}{4} = c, c = \ln\frac{1}{3}\) | M1 A1 | |
| \(t = 3 \Rightarrow \ln | x | - \ln |
| \(\ln\left | \frac{3x}{1-x}\right | = 2, \frac{3x}{1-x} = e^2\) |
| \(3x = e^2(1-x), x(e^2+3) = e^2\) | M1 | |
| \(x = \frac{e^2}{e^2+3} \therefore\) % destroyed \(= \frac{e^2}{e^2+3} \times 100\% = 71.1\%\) (3sf) | A1 | (15) |
**(a)** When $x = \frac{1}{4}, \frac{dx}{dt} = \frac{3}{4} + 6 = \frac{7}{8}$ | M1 A1 |
$\frac{dx}{dt} = k(1-x) \therefore \frac{7}{8} = k \times \frac{1}{4} \times \frac{3}{4}, k = \frac{7}{3} \therefore \frac{dx}{dt} = \frac{7}{3}x(1-x)$ | M1 A1 |
**(b)** $\int\frac{1}{x(1-x)} \, dx = \int\frac{7}{3} \, dt$ | M1 |
$\frac{1}{x(1-x)} = \frac{A}{x} + \frac{B}{1-x}, 1 \equiv A(1-x) + Bx$ | M1 |
$x = 0 \Rightarrow A = 1$ | A1 |
$x = 1 \Rightarrow B = 1$ | A1 |
$\therefore \int(\frac{1}{x} + \frac{1}{1-x}) \, dx = \int\frac{7}{3} \, dt$ | M1 |
$\ln|x| - \ln|1-x| = \frac{7}{3}t + c$ | M1 A1 |
$t = 0, x = \frac{1}{4} \Rightarrow \ln\frac{1}{4} - \ln\frac{3}{4} = c, c = \ln\frac{1}{3}$ | M1 A1 |
$t = 3 \Rightarrow \ln|x| - \ln|1-x| = 2 + \ln\frac{1}{3}$ | M1 |
$\ln\left|\frac{3x}{1-x}\right| = 2, \frac{3x}{1-x} = e^2$ | M1 |
$3x = e^2(1-x), x(e^2+3) = e^2$ | M1 |
$x = \frac{e^2}{e^2+3} \therefore$ % destroyed $= \frac{e^2}{e^2+3} \times 100\% = 71.1\%$ (3sf) | A1 | (15)
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**Total: (75)**7. When a plague of locusts attacks a wheat crop, the proportion of the crop destroyed after $t$ hours is denoted by $x$. In a model, it is assumed that the rate at which the crop is destroyed is proportional to $x ( 1 - x )$.
A plague of locusts is discovered in a wheat crop when one quarter of the crop has been destroyed.
Given that the rate of destruction at this instant is such that if it remained constant, the crop would be completely destroyed in a further six hours,
\begin{enumerate}[label=(\alph*)]
\item show that $\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 2 } { 3 } x ( 1 - x )$,
\item find the percentage of the crop destroyed three hours after the plague of locusts is first discovered.\\
7. continued\\
7. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [15]}}