| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - partial fractions |
| Difficulty | Standard +0.3 This is a straightforward two-part question combining partial fractions (routine A-level technique) with a separable differential equation. Part (i) is standard bookwork, and part (ii) requires separation of variables followed by integration using the partial fractions result—all well-practiced techniques with no novel insight required. The 'no logarithms' instruction adds minor algebraic manipulation but nothing conceptually challenging. Slightly easier than average due to its predictable structure. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use any relevant method to determine a constant | M1 | |
| Obtain one of the values \(A = 1, B = -2, C = 4\) | A1 | |
| Obtain a second value | A1 | |
| Obtain the third value | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) Separate variables and obtain one term by integrating \(\frac{1}{y}\) or a partial fraction | M1 | |
| Obtain \(\ln y = -\frac{1}{2} - 2\ln(2x + 1) + c\), or equivalent | A3 | |
| Evaluate a constant, or use limits \(x = 1, y = 1\), in a solution containing at least three terms of the form \(k\ln y\), \(l/x\), \(m\ln x\) and \(n\ln(2x + 1)\), or equivalent | M1 | |
| Obtain solution \(\ln y = -\frac{1}{2} - 2\ln x + 2\ln(2x + 1) + c\), or equivalent | A1 | |
| Substitute \(x = 2\) and obtain \(y = \frac{25}{36}e^2\), or exact equivalent free of logarithms | A1 | [7] |
**(i)** Use any relevant method to determine a constant | M1 |
Obtain one of the values $A = 1, B = -2, C = 4$ | A1 |
Obtain a second value | A1 |
Obtain the third value | A1 | [4]
[If $A$ and $C$ are found by the cover up rule, give B1 + B1 then M1A1 for finding $B$. If only one is found by the rule, give B1M1A1A1.]
**(ii)** Separate variables and obtain one term by integrating $\frac{1}{y}$ or a partial fraction | M1 |
Obtain $\ln y = -\frac{1}{2} - 2\ln(2x + 1) + c$, or equivalent | A3 |
Evaluate a constant, or use limits $x = 1, y = 1$, in a solution containing at least three terms of the form $k\ln y$, $l/x$, $m\ln x$ and $n\ln(2x + 1)$, or equivalent | M1 |
Obtain solution $\ln y = -\frac{1}{2} - 2\ln x + 2\ln(2x + 1) + c$, or equivalent | A1 |
Substitute $x = 2$ and obtain $y = \frac{25}{36}e^2$, or exact equivalent free of logarithms | A1 | [7]
(The f.t. is on $A, B, C$. Give A2 if there is only one error or omission in the integration; A1 if two.)\begin{enumerate}[label=(\roman*)]
\item Express $\frac{1}{x^2(2x + 1)}$ in the form $\frac{A}{x^2} + \frac{B}{x} + \frac{C}{2x + 1}$. [4]
\item The variables $x$ and $y$ satisfy the differential equation
$$y = x^2(2x + 1)\frac{dy}{dx},$$
and $y = 1$ when $x = 1$. Solve the differential equation and find the exact value of $y$ when $x = 2$. Give your value of $y$ in a form not involving logarithms. [7]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2013 Q8 [11]}}