| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Separable variables |
| Difficulty | Standard +0.3 This is a straightforward separable variables question requiring routine separation, integration of standard forms (exponential and inverse tan), and application of initial conditions. Part (b) requires simple limit evaluation. While it involves multiple steps, all techniques are standard A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separate variables correctly | B1 | \(\int \dfrac{1}{4+9y^2}\,dy = \int e^{-(2x+1)}\,dx\). Condone missing integral signs or \(dx\) and \(dy\) missing. |
| Obtain term \(-\dfrac{1}{2}e^{-2x-1}\) | B1 | OE e.g. \(-\dfrac{1}{2e}e^{-2x}\) |
| Obtain term of the form \(a\tan^{-1}\!\left(\dfrac{3y}{2}\right)\) | M1 | |
| Obtain term \(\dfrac{1}{6}\tan^{-1}\!\left(\dfrac{3y}{2}\right)\) | A1 | OE e.g. \(\dfrac{1}{9}\times\dfrac{3}{2}\tan^{-1}\dfrac{3y}{2}\) |
| Use \(x=1,\, y=0\) to evaluate a constant or as limits in a solution containing or derived from terms of the form \(a\tan^{-1}(by)\) and \(ce^{\pm(2x+1)}\) | M1 | If they rearrange before evaluating the constant, the constant must be of the correct form. |
| Obtain correct answer in any form | A1 | e.g. \(\dfrac{1}{6}\tan^{-1}\dfrac{3y}{2} = \dfrac{1}{2}e^{-3} - \dfrac{1}{2}e^{-(2x+1)}\) |
| Obtain final answer \(y = \dfrac{2}{3}\tan\!\left(3e^{-3} - 3e^{-2x-1}\right)\) | A1 | OE. Allow with \(3e^{-3} = 0.149\ldots\) |
| 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State that \(y\) approaches \(\dfrac{2}{3}\tan(3e^{-3})\) | B1 FT | Or exact equivalent. The FT is on correct work on a solution containing \(e^{-2x-1}\). Condone \(y = \ldots\) Accept correct answer stated with minimal wording. \(0.10032\ldots\) is not exact so B0. |
| 1 |
## Question 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly | B1 | $\int \dfrac{1}{4+9y^2}\,dy = \int e^{-(2x+1)}\,dx$. Condone missing integral signs or $dx$ and $dy$ missing. |
| Obtain term $-\dfrac{1}{2}e^{-2x-1}$ | B1 | OE e.g. $-\dfrac{1}{2e}e^{-2x}$ |
| Obtain term of the form $a\tan^{-1}\!\left(\dfrac{3y}{2}\right)$ | M1 | |
| Obtain term $\dfrac{1}{6}\tan^{-1}\!\left(\dfrac{3y}{2}\right)$ | A1 | OE e.g. $\dfrac{1}{9}\times\dfrac{3}{2}\tan^{-1}\dfrac{3y}{2}$ |
| Use $x=1,\, y=0$ to evaluate a constant or as limits in a solution containing or derived from terms of the form $a\tan^{-1}(by)$ and $ce^{\pm(2x+1)}$ | M1 | If they rearrange before evaluating the constant, the constant must be of the correct form. |
| Obtain correct answer in any form | A1 | e.g. $\dfrac{1}{6}\tan^{-1}\dfrac{3y}{2} = \dfrac{1}{2}e^{-3} - \dfrac{1}{2}e^{-(2x+1)}$ |
| Obtain final answer $y = \dfrac{2}{3}\tan\!\left(3e^{-3} - 3e^{-2x-1}\right)$ | A1 | OE. Allow with $3e^{-3} = 0.149\ldots$ |
| | 7 | |
## Question 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| State that $y$ approaches $\dfrac{2}{3}\tan(3e^{-3})$ | B1 FT | Or exact equivalent. The FT is on correct work on a solution containing $e^{-2x-1}$. Condone $y = \ldots$ Accept correct answer stated with minimal wording. $0.10032\ldots$ is not exact so B0. |
| | 1 | |
---8
\begin{enumerate}[label=(\alph*)]
\item The variables $x$ and $y$ satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 + 9 y ^ { 2 } } { \mathrm { e } ^ { 2 x + 1 } } .$$
It is given that $y = 0$ when $x = 1$.\\
Solve the differential equation, obtaining an expression for $y$ in terms of $x$.
\item State what happens to the value of $y$ as $x$ tends to infinity. Give your answer in an exact form.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q8 [8]}}