Year: 2010
Paper: 1
Question Number: 6
Course: UFM Pure
Section: Second order differential equations
There were significantly more candidates attempting this paper than last year (just over 1000), and the scores were much higher than last year (presumably due to the easier first question): fewer than 2% of candidates scored less than 20 marks overall, and the median mark was 61. The pure questions were the most popular as usual, though there was much more variation than in some previous years: questions 1, 3, 4 and 6 were the most popular, while question 7 (on vectors) was intensely unpopular. About half of all candidates attempted at least one mechanics question, and 15% attempted at least one probability question. The marks were unsurprising: the pure questions generally gained the better marks, while the mechanics and probability questions generally had poorer marks. A sizeable number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have often restricted my attention below to those attempts which counted as one of the six highest-scoring answers, and referred to these as "significant attempts". The majority of candidates did begin with question 1 (presumably as it appeared to be the easiest), but some spent far longer on it than was wise. Some attempts ran to over eight pages in length, especially when they had made an algebraic slip early on, and used time which could have been far better spent tackling another question. It is important to balance the desire to finish the question with an appreciation of when to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, "Hence, or otherwise, show . . ." gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, but in Question 5 this year, the "otherwise" approach was very popular.) On some questions, some candidates tried to work forwards from the given question and backwards from the answer, hoping that they would meet somewhere in the middle. While this worked on occasion, it often required fudging. It is wise to remember that STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Show that, if $y=\e^x$, then
\[
(x-1) \frac{\d^2 y}{\d x^2} -x \frac{\d y}{\d x}
+y=0\,.
\tag{$*$}
\]
In order to find other solutions of this differential equation, now let $y=u\e^x$, where $u$ is a function of $x$.
By substituting this into $(*)$, show that
\[
(x-1) \frac{\d^2 u}{\d x^2} + (x-2) \frac{\d u}{\d x}
=0\,.
\tag{$**$}
\]
By setting $ \dfrac {\d u}{\d x}= v$ in $(**)$ and solving the resulting first order differential equation for $v$, find $u$ in terms of $x$.
Hence show that $y=Ax + B\e^x$ satisfies $(*)$, where $A$ and $B$ are any constants.\begin{align*}
&& y &= e^x \\
&& y' &= e^x \\
&& y'' &= e^x \\
\Rightarrow && (x-1)y'' - x y' + y &= (x-1)e^x - xe^x + e^x \\
&&&= 0
\end{align*}
Suppose $y = ue^x$ then
\begin{align*}
&& y' &= u'e^x + ue^x \\
&& y'' &= (u''+u')e^x + (u'+u)e^x \\
&&&= (u''+2u' +u)e^x \\
\\
&& 0 &= (x-1)y'' - x y' + y \\
&&&= [(x-1)(u''+2u'+u) - x(u'+u)+u]e^x \\
&&&= [(x-1)u'' +(x-2)u']e^x \\
\Rightarrow && 0 &= (x-1)u'' + (x-2)u' \\
v = u': && 0 &= (x-1)v' + (x-2) v \\
\Rightarrow && \frac{v'}{v} &= -\frac{x-2}{x-1} \\
&&&= -1-\frac{1}{x-1} \\
\Rightarrow && \ln v &= -x - \ln(x-1) + C \\
\Rightarrow && v &= A(x-1)e^{-x} \\
&& u &= \int Axe^{-x} - Ae^{-x} \d x \\
&&&= \left [-Axe^{-x} +Ae^{-x} \right] + \int Ae^{-x} \d x \\
&&&= -Axe^{-x} + D\\
\Rightarrow && y &= ue^x \\
&&&= -Ax + De^x
\end{align*}
This was another popular and well-answered question. The median mark was 13, and around one-fifth of attempts gained full marks. Almost all candidates were fine with the first step and showing that (∗) holds. A significant majority were also comfortable with deducing (∗∗), though there were some who had difficulties in applying the product rule twice or appreciating that u was a function of x rather than a constant. The final part of the question, however, caused numerous difficulties for the majority of candidates. First of all, some simply did not understand what was being asked of them, and thought of v as a constant (even though they had appreciated that u itself is a function of x). Then a significant number thought that d²u/dx² = v² rather than dv/dx, belying a lack of understanding of the meaning of a second derivative. Those who overcame these hurdles reached a correct first order ODE with no constant term, but many struggled to solve it; even those who correctly separated the variables could not figure out how to integrate (x−2)/(x−1), even though this is a standard A-level integration question. (Some tried integration by parts, with a predictable lack of success.) Those who managed to integrate to determine ln v then went on to exponentiate, mostly successfully, though a number forgot about the arbitrary constant or ended up with an expression of the form v = f(x) + c instead of v = cf(x). Those who reached this point generally appreciated that they now needed to write v = du/dx and integrate once more, and most realised that now was the time to integrate by parts. Unfortunately, many forgot to multiply their original arbitrary constant by the −∫(dv/dx)u dx term of the parts formula. For the last step, a number of candidates thought that it was sufficient to simply plug y = Ax + Bex into the differential equation; this, of course, gained no credit, as the question had explicitly said "Hence show that . . .".