CAIE P3 2010 June — Question 4 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2010
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeSeparable variables - standard (polynomial/exponential x-side)
DifficultyStandard +0.3 This is a separable differential equation requiring standard techniques: separation of variables, partial fractions to integrate, and applying initial conditions. While it involves multiple steps (7 marks), each step follows routine procedures taught in P3 with no novel insight required, making it slightly easier than average.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)

Given that \(x = 1\) when \(t = 0\), solve the differential equation $$\frac{dx}{dt} = \frac{1}{x} - \frac{x}{4},$$ obtaining an expression for \(x^2\) in terms of \(t\). [7]

AnswerMarks Guidance
Separate variables correctlyB1
Obtain term \(k\ln(4-x^2)\), or terms \(k_1\ln(2-x) + k_2\ln(2+x)\)B1
Obtain term \(-2\ln(4-x^2)\), or \(-2\ln(2-x) - 2\ln(2+x)\), or equivalentB1
Obtain term \(t\), or equivalentB1
Evaluate a constant or use limits \(x = 1\), \(t = 0\) in a solution containing terms \(c\ln(4-x^2)\) and \(bt\) or terms \(c\ln(2-x)\), \(d\ln(2+x)\) and \(bt\)M1
Obtain correct solution in any form, e.g. \(c = -2 \ln(4-x^2) = t - 2\ln 3\)A1
Rearrange and obtain \(x^2 = 4 - 3\exp(-\frac{1}{4}t)\), or equivalent (allow use of \(2\ln 3 = 2.20\))A1 [7]
Separate variables correctly | B1 | —

Obtain term $k\ln(4-x^2)$, or terms $k_1\ln(2-x) + k_2\ln(2+x)$ | B1 | —

Obtain term $-2\ln(4-x^2)$, or $-2\ln(2-x) - 2\ln(2+x)$, or equivalent | B1 | —

Obtain term $t$, or equivalent | B1 | —

Evaluate a constant or use limits $x = 1$, $t = 0$ in a solution containing terms $c\ln(4-x^2)$ and $bt$ or terms $c\ln(2-x)$, $d\ln(2+x)$ and $bt$ | M1 | —

Obtain correct solution in any form, e.g. $c = -2 \ln(4-x^2) = t - 2\ln 3$ | A1 | —

Rearrange and obtain $x^2 = 4 - 3\exp(-\frac{1}{4}t)$, or equivalent (allow use of $2\ln 3 = 2.20$) | A1 | [7]

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Given that $x = 1$ when $t = 0$, solve the differential equation
$$\frac{dx}{dt} = \frac{1}{x} - \frac{x}{4},$$
obtaining an expression for $x^2$ in terms of $t$. [7]

\hfill \mbox{\textit{CAIE P3 2010 Q4 [7]}}