Year: 2018
Paper: 3
Question Number: 3
Course: UFM Pure
Section: Second order differential equations
The total entry was a record number, an increase of over 6% on 2017. Only question 1 was attempted by more than 90%, although question 2 was attempted by very nearly 90%. Every question was attempted by a significant number of candidates with even the two least popular questions being attempted by 9%. More than 92% restricted themselves to attempting no more than 7 questions with very few indeed attempting more than 8. As has been normal in the past, apart from a handful of very strong candidates, those attempting six questions scored better than those attempting more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
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the second-order differential equation
\[
x^2y''+(1-2p) x\, y' + (p^2-q^2) \, y= \f(x)
\,,
\]
where $p$ and $q$ are constants, can be written in the
form
\[
x^a
\big(x^b
(x^cy)'\big)' = \f(x)
\,,
\tag{$*$}
\]
where $a$, $b$ and $c$ are constants.
\begin{questionparts}
\item Use $(*)$ to derive the general solution of the equation
\[
x^{2}y''+(1-2p)xy'+(p^2-q^{2})y=0
\]
in the different cases that arise according to the values of $p$ and $q$.
\item Use $(*)$ to derive the general solution of the equation
\[
x^{2}y''+(1-2p)xy'+p^2y=x^{n}
\]
in the different cases that arise according to the values of $p$ and $n$.
\end{questionparts}Consider $x^a
\big(x^b
(x^cy)'\big)'$ then
\begin{align*}
x^a
\big(x^b
(x^cy)'\big)' &= x^a \big (bx^{b-1}(x^c y)'+x^b(x^cy)'' \big ) \\
&= x^a \big (bx^{b-1} (cx^{c-1}y + x^c y') + x^b(c(c-1)x^{c-2}y + 2cx^{c-1}y' + x^cy'') \\
&= x^{a+b+c}y'' + (2cx^{c-1+b+a}+bx^{c+b-1+a})y'+(c(b+c-1))x^{a+b+c-2} y
\end{align*}
So we need:
\begin{align*}
&&& \begin{cases}
a+b+c &= 2 \\
2c+b &= 1-2p \\
c(b+c-1) &= p^2-q^2
\end{cases} \\
\Rightarrow && c((1-2p)-2c+c-1) &=p^2-q^2 \\
\Rightarrow && c^2+2pc &= q^2-p^2
\end{align*}
Only 65% attempted this, and it was the second weakest of the Pure questions with a mean score of about 7/20. It was imperative for candidates to demonstrate high levels of algebraic accuracy to score highly. Most successfully differentiated the initial expression, and equated coefficients but then failed to solve explicitly for, and in terms of and (or demonstrate that such, and existed). Candidates without these explicit expressions then often failed to spot one of the main strands of the question, integration of in the two cases 1, and 1; some considered superfluous other cases. Some candidates fell at the final hurdle having used correct methods but then did not express their solutions in terms of and. Also, some were thrown by the two possible sets of solutions for, and, successful candidates realising that these gave the same solutions to the differential equations.