Year: 2013
Paper: 1
Question Number: 7
Course: LFM Pure
Section: Differential equations
Around 1500 candidates sat this paper, a significant increase on last year. Overall, responses were good with candidates finding much to occupy them profitably during the three hours of the examination. In hindsight, two or three of the questions lacked sufficient 'punch' in their later parts, but at least most candidates showed sufficient skill to identify them and work on them as part of their chosen selection of questions. On the whole, nearly all candidates managed to attempt 4-6 questions – although there is always a significant minority who attempt 7, 8, 9, … bit and pieces of questions – and most scored well on at least two. Indeed, there were many scripts with 6 question-attempts, most or all of which were fantastically accomplished mathematically, and such excellence is very heart-warming.
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item Use the substitution $y=ux$, where $u$ is a function of $x$, to show that the solution of the differential equation
\[
\frac{\d y}{\d x} = \frac x y + \frac y x
\quad \quad (x > 0, y> 0)
\] that satisfies $y=2$ when $x=1$
is
\[
y= x\, \sqrt{4+2\ln x \, }
( x > \e^{-2}).
\]
\item Use a substitution to find the solution of the differential equation
\[
\frac{\d y}{\d x} = \frac x y + \frac {2y} x
\quad \quad (x > 0, y > 0)
\] that satisfies $y=2$ when $x=1$.
\item Find the solution of the differential equation
\[
\frac{\d y}{\d x} = \frac {x^2} y + \frac {2y} x
\quad \quad (x> 0, \ y> 0)
\] that satisfies $y=2$ when $x=1$.
\end{questionparts}\begin{questionparts}
\item Let $y = ux$, then $\frac{\d y}{\d x} = x\frac{\d u}{\d x} = u$ and the differential equation becomes,
\begin{align*}
&& xu' + u &= \frac{1}{u} +u \\
\Rightarrow && u' &= \frac{1}{ux} \\
\Rightarrow && u u' &= \frac1{x} \\
\Rightarrow && \frac12 u^2 &= \ln x + C \\
(x,y) = (1,2): && \frac12 4 &= C \\
\Rightarrow && \frac12 \frac{y^2}{x^2} &= \ln x + 2 \\
\Rightarrow && y^2 &= x^2 \l 2\ln x + 4 \r \\
\Rightarrow && y &= x \sqrt{4 + 2 \ln x} \quad (x > e^{-2})
\end{align*}
\item Let $y = ux^2$ then
\begin{align*}
&& \frac{\d y}{\d x} &= \frac{x^2}{y} + \frac{2y}{x} \\
\Rightarrow && u' x^2 + 2x u &= \frac{1}{u} + 2x u \\
\Rightarrow && u' u &= \frac{1}{x^2} \\
\Rightarrow && \frac12 u^2 &= -\frac{1}{x} + C \\
(x,y) = (1,2): && 2 &= C - 1 \\
\Rightarrow && \frac12 \frac{y^2}{x^4} &= 3 - \frac{1}{x} \\
\Rightarrow && y &= x\sqrt{2(3x^2-x)}, \quad (x > \frac13)
\end{align*}
\end{questionparts}
Around 1300 candidates attempted this question, making it the second most popular question on the paper. It was also the second highest-scoring question on average which, if nothing else, pays tribute to the candidates' ability to spot the right questions to attempt. In hindsight, this was possibly a little too straightforward; this was undoubtedly partly due to the appearance of similar questions (on what are known as homogeneous differential equations) in recent years' STEPs, but also to the fact that part (ii) could be solved by the use of the given approach for part (i). It was part (iii) that required of candidates a stretch of the imagination – the use of y = ux2 – but even this helped make the question more approachable, as this substitution could also be used to solve part (ii) if it turned out that candidates got imaginative a bit earlier than anticipated. For those making essentially correct attempts at parts (ii) and (iii), the only final hurdle to complete success lay in the hoped-for statement of a domain for the functions which had been found as solutions. We allowed as obvious the taking of non-negative square-roots (since the given "initial" values of y are positive – though, in general, candidates should be encouraged to state that they recognise they are doing this) but expected candidates to indicate a suitable interval for the x values in each case: the hint lay in the given answer to part (i).