| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2010 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Deriving the differential equation |
| Difficulty | Moderate -0.3 This is a straightforward first-order differential equation question that follows a standard template. Part (i) simply requires setting up a proportionality relationship and using given initial conditions to find the constant—pure substitution with no problem-solving. Parts (ii)-(iv) involve routine separation of variables or integrating factor method, standard integration, and limit behavior. While it requires multiple steps, each is a textbook procedure with no novel insight required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts4.10c Integrating factor: first order equations |
| Answer | Marks |
|---|---|
| (i) State or imply \(\frac{dx}{dt} = k(20 - x)\) | B1 |
| Show that \(k = 0.05\) | B1 |
| Answer | Marks |
|---|---|
| (ii) Separate variables correctly and integrate both sides | B1 |
| Obtain term \(-\ln(20 - x)\), or equivalent | B1 |
| Obtain term \(-\frac{1}{20}t\), or equivalent | B1 |
| Evaluate a constant or use limits \(t = 0, x = 0\) in a solution containing terms \(a \ln(20 - x)\) and \(bt\) | M1* |
| Obtain correct answer in any form, e.g. \(\ln 20 - \ln(20 - x) = \frac{1}{20}t\) | A1 |
| Answer | Marks |
|---|---|
| (iii) Substitute \(t = 10\) and calculate \(x\) | M1(dep*) |
| Obtain answer \(x = 7.9\) | A1 |
| Answer | Marks |
|---|---|
| (iv) State that \(x\) approaches 20 | B1 |
**(i)** State or imply $\frac{dx}{dt} = k(20 - x)$ | B1 |
Show that $k = 0.05$ | B1 |
[2 marks total]
**(ii)** Separate variables correctly and integrate both sides | B1 |
Obtain term $-\ln(20 - x)$, or equivalent | B1 |
Obtain term $-\frac{1}{20}t$, or equivalent | B1 |
Evaluate a constant or use limits $t = 0, x = 0$ in a solution containing terms $a \ln(20 - x)$ and $bt$ | M1* |
Obtain correct answer in any form, e.g. $\ln 20 - \ln(20 - x) = \frac{1}{20}t$ | A1 |
[5 marks total]
**(iii)** Substitute $t = 10$ and calculate $x$ | M1(dep*) |
Obtain answer $x = 7.9$ | A1 |
[2 marks total]
**(iv)** State that $x$ approaches 20 | B1 |
[1 mark total]10 A certain substance is formed in a chemical reaction. The mass of substance formed $t$ seconds after the start of the reaction is $x$ grams. At any time the rate of formation of the substance is proportional to $( 20 - x )$. When $t = 0 , x = 0$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 1$.\\
(i) Show that $x$ and $t$ satisfy the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.05 ( 20 - x ) .$$
(ii) Find, in any form, the solution of this differential equation.\\
(iii) Find $x$ when $t = 10$, giving your answer correct to 1 decimal place.\\
(iv) State what happens to the value of $x$ as $t$ becomes very large.
\hfill \mbox{\textit{CAIE P3 2010 Q10 [10]}}