2008 Paper 3 Q6

Year: 2008
Paper: 3
Question Number: 6

Course: UFM Pure
Section: Second order differential equations

Difficulty: 1700.0 Banger: 1500.0

Problem

In this question, \(p\) denotes \(\dfrac{\d y}{\d x}\,\).
  1. Given that \[ y=p^2 +2 xp\,, \] show by differentiating with respect to \(x\) that \[ \frac{\d x}{\d p} = -2 - \frac {2x} p . \] Hence show that \(x = -\frac23p +Ap^{-2}\,,\) where \(A\) is an arbitrary constant. Find \(y\) in terms of \(x\) if \(p=-3\) when \(x=2\).
  2. Given instead that \[ y=2xp +p \ln p\,,\] and that \(p=1\) when \(x=-\frac14\), show that \(x=-\frac12 \ln p - \frac14\,\) and find \(y\) in terms of \(x\).

No solution available for this problem.

Examiner's report
— 2008 STEP 3, Question 6
Mean: ~14 / 20 (inferred) 80% attempted Inferred ~14/20: 'more success than any other question' must exceed Q8 (~13/20). Conservatively set to 14.

More than 80% attempted this, and with more success than any other question. Having obtained the relation between x and p in each part, quite a few attempts then treated these as differential equations rather than merely substituting back to find expressions for y, and consequent inaccuracies lost marks.

Most candidates attempted five, six or seven questions, and scored the majority of their total score on their best three or four. Those attempting seven or more tended not to do well, pursuing no single solution far enough to earn substantial marks.

Source: Cambridge STEP 2008 Examiner's Report · 2008-full.pdf
Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0

Show LaTeX source
Problem source
In this question, $p$ denotes $\dfrac{\d y}{\d x}\,$.
\begin{questionparts}
\item Given that 
\[
y=p^2 +2 xp\,,
\]
show by differentiating with respect to $x$ that 
\[
\frac{\d x}{\d p} = -2 -  \frac {2x} p .
\]
Hence show that $x = -\frac23p +Ap^{-2}\,,$ where $A$ is an arbitrary
 constant.
Find $y$ in terms of $x$ if $p=-3$ when $x=2$.
\item Given instead that 
\[ y=2xp +p \ln p\,,\]
and that $p=1$ when $x=-\frac14$, show that 
$x=-\frac12 \ln p - \frac14\,$ and find $y$ in terms of $x$.
\end{questionparts}