Year: 2023
Paper: 3
Question Number: 8
Course: UFM Pure
Section: Second order differential equations
No solution available for this problem.
The total entry was a marginal increase on that of 2022 (by just over 1%). Two questions were attempted by more than 90% of candidates, another two by 80%, and another two by about two thirds. The least popular questions were attempted by more than a sixth of candidates. All the questions were perfectly answered by at least three candidates (but mostly more than this), with one being perfectly answered by eighty candidates. Very nearly 90% of candidates attempted no more than 7 questions. One general comment regarding all the questions is that candidates need to make sure that they read the question carefully, paying particular attention to command words such as "hence" and "show that".
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
If
\[y = \begin{cases} \mathrm{k}_1(x) & x \leqslant b \\ \mathrm{k}_2(x) & x \geqslant b \end{cases}\]
with $\mathrm{k}_1(b) = \mathrm{k}_2(b)$, then $y$ is said to be \emph{continuously differentiable} at $x = b$ if $\mathrm{k}_1'(b) = \mathrm{k}_2'(b)$.
\begin{questionparts}
\item Let $\mathrm{f}(x) = x\mathrm{e}^{-x}$. Verify that, for all real $x$, $y = \mathrm{f}(x)$ is a solution to the differential equation
\[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\frac{\mathrm{d}y}{\mathrm{d}x} + y = 0\]
and that $y = 0$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1$ when $x = 0$.
Show that $\mathrm{f}'(x) \geqslant 0$ for $x \leqslant 1$.
\item You are given the differential equation
\[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\left|\frac{\mathrm{d}y}{\mathrm{d}x}\right| + y = 0\]
where $y = 0$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1$ when $x = 0$. Let
\[y = \begin{cases} \mathrm{g}_1(x) & x \leqslant 1 \\ \mathrm{g}_2(x) & x \geqslant 1 \end{cases}\]
be a solution of the differential equation which is continuously differentiable at $x = 1$.
Write down an expression for $\mathrm{g}_1(x)$ and find an expression for $\mathrm{g}_2(x)$.
\item State the geometrical relationship between the curves $y = \mathrm{g}_1(x)$ and $y = \mathrm{g}_2(x)$.
\item Prove that if $y = \mathrm{k}(x)$ is a solution of the differential equation
\[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + p\frac{\mathrm{d}y}{\mathrm{d}x} + qy = 0\]
in the interval $r \leqslant x \leqslant s$, where $p$ and $q$ are constants, then, in a suitable interval which you should state, $y = \mathrm{k}(c - x)$ satisfies the differential equation
\[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - p\frac{\mathrm{d}y}{\mathrm{d}x} + qy = 0\,.\]
\item You are given the differential equation
\[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\left|\frac{\mathrm{d}y}{\mathrm{d}x}\right| + 2y = 0\]
where $y = 0$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1$ when $x = 0$.
Let $\mathrm{h}(x) = \mathrm{e}^{-x}\sin x$. Show that $\mathrm{h}'\!\left(\frac{1}{4}\pi\right) = 0$.
It is given that $y = \mathrm{h}(x)$ satisfies the differential equation in the interval $-\frac{3}{4}\pi \leqslant x \leqslant \frac{1}{4}\pi$ and that $\mathrm{h}'(x) \geqslant 0$ in this interval.
In a solution to the differential equation which is continuously differentiable at $(n + \frac{1}{4})\pi$ for all $n \in \mathbb{Z}$, find $y$ in terms of $x$ in the intervals
\begin{enumerate}
\item $\frac{1}{4}\pi \leqslant x \leqslant \frac{5}{4}\pi$,
\item $\frac{5}{4}\pi \leqslant x \leqslant \frac{9}{4}\pi$.
\end{enumerate}
\end{questionparts}
This was only a little less popular than question 2 and was only answered with moderate success having a mean score of 7.3/20. Part (i) was mostly well answered, although some candidates lost marks by not being thorough in demonstrating the inequality, or by using a characteristic equation and failing to verify their solution as required. Part (ii) was found difficult, for although g1 was almost always stated, many struggled to find g2, often just flipping the sign of g1. Even when a general solution was found, many candidates used the boundary conditions of (i) instead of appreciating the sign change of the derivative at x = 1. Part (iii) was done extremely poorly, even by candidates who had the right functions and the algebraic relationship between them. Candidates could often pick up marks on part (iv), even if they were less successful with the rest of the question, though notation of what derivatives were being taken was often ambiguous. In attempting to answer part (v), many candidates knew they needed to use the result of part (iv) for part (a), although a significant number lost marks for giving no explanation or working. Part (b) proved much harder than part (a), since few candidates realized they had to match their function at 5/4π, not at −1/4π again. Some candidates fully solved both (a) and (b) directly via the characteristic equation which led to a very lengthy solution.