Year: 2025
Paper: 2
Question Number: 7
Course: UFM Pure
Section: Second order differential equations
As is commonly the case, the vast majority of candidates focused on the Pure questions in Section A of the paper, with a good number of attempts made on all of those questions. Candidates that attempted the Mechanics questions in Section B generally answered both questions. More candidates attempted Question 11 in Section C than either Mechanics question, but very few attempted Question 12 in that section. There were a large number of good responses seen for all the questions, but a significant number of responses lacked sufficient detail in the presentation, particularly when asked to prove a given result or provide an explanation. Candidates who did well on this paper generally: gave careful explanations of each step within their solutions; indicated all points of interest on graphs and other diagrams clearly; made clear comments about the approach that needed to be taken, particularly when having to explore a number of cases as part of the solution to a question; used mathematical terminology accurately within their solutions. Candidates who did less well on this paper generally: made errors with basic algebraic manipulation, such as incorrect processing of indices; produced sketches of graphs in which significant points were difficult to see clearly because of the chosen scale; skipped important lines within lengthy sections of algebraic reasoning.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The differential equation
\[\frac{d^2x}{dt^2} = 2x\frac{dx}{dt}\]
describes the motion of a particle with position $x(t)$ at time $t$. At $t = 0$, $x = a$, where $a > 0$.
\begin{questionparts}
\item Solve the differential equation in the case where $\frac{dx}{dt} = a^2$ when $t = 0$.
What happens to the particle as $t$ increases from 0?
\item Solve the differential equation in the case where $\frac{dx}{dt} = a^2 + p$ when $t = 0$, where $p > 0$.
What happens to the particle as $t$ increases from 0?
\item Solve the differential equation in the case where $\frac{dx}{dt} = a^2 - q^2$ when $t = 0$, where $q > 0$.
What happens to the particle as $t$ increases from 0? Give conditions on $a$ and $q$ for the different cases which arise.
\end{questionparts}Let $v = \frac{\d x}{\d t}$ and notice that $\frac{\d}{\d t} \left ( \frac{\d x}{\d t} \right) = \frac{\d }{\d x} \left ( v \right) \frac{\d x}{\d t} = v \frac{\d v}{\d x}$. Also notice that:
\begin{align*}
&& v \frac{\d v}{\d x} &= 2x v \\
\Rightarrow && \frac{\d v}{\d x} &= 2x \\
\Rightarrow && v &= x^2 + C \\
\Rightarrow && \frac{\d x}{\d t} &= x^2 + C \\
\end{align*}
\begin{questionparts}
\item When $t = 0, \frac{\d x}{\d t} = a^2$ so $C = 0$, therefore $\frac{\d x}{\d t} = x^2 \Rightarrow t = -x^{-1} + k$ and so $k = a^{-1}$ and $x = \frac{a}{1-at}$. As $t$ increases from $0$ the particle heads to infinity at an increasing rate, `reaching' infinity around $t=\frac{1}{a}$
\item When $t = 0, \frac{\d x}{\d t} = a^2 + p$ so $C = p$. Therefore $\frac{\d x}{\d t} = x^2 + p \Rightarrow t = \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) + c$. When $c = - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right)$, so
\begin{align*}
&& t &= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right) \\
&&&= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} \right) \\
\Rightarrow && \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} &= \tan (\sqrt{p} t) \\
\Leftrightarrow && \sqrt{p}(x-a) &= \tan (\sqrt{p} t)(\sqrt{p}-ax) \\
\Leftrightarrow && x(\sqrt{p}+a\tan (\sqrt{p} t)) &= \sqrt{p} (\tan(\sqrt{p}t) + a) \\
\Leftrightarrow && x &= \frac{\sqrt{p} (\tan(\sqrt{p}t) + a)}{\sqrt{p}+a\tan (\sqrt{p} t)}
\end{align*}
The particle heads to $\frac{\sqrt{p}}{a}$.
\item When $t = 0, \frac{\d x}{\d t} = a^2-q^2$ so $C = -q^2$. Therefore
\begin{align*}
&& \frac{\d x}{\d t} &= x^2 -q^2 \\
\Rightarrow && \int \d t &= \int \frac{1}{(x-q)(x+q)} \d x \\
&&&= \frac{1}{2q} \int \left ( \frac{1}{x-q}- \frac{1}{x+q} \right )\d x \\
&&&= \frac{1}{2q} \left ( \ln (x-q) - \ln(x+q) \right) \\
&&&= \frac{1}{2q} \ln \left ( \frac{x-q}{x+q} \right)\\
\Rightarrow && \frac{x-q}{x+q} &= Ae^{2qt} \\
\underbrace{\Rightarrow}_{t= 0} && A &= \frac{a-q}{a+q} \\
\Rightarrow && x-q &= \frac{a-q}{a+q}e^{2qt}(x+q) \\
\Leftrightarrow && x\left (1-\frac{a-q}{a+q}e^{2qt} \right) &= q\left (1 + \frac{a-q}{a+q}e^{2qt} \right) \\
\Leftrightarrow && x &= q \frac{1 + \frac{a-q}{a+q}e^{2qt}}{1-\frac{a-q}{a+q}e^{2qt}}
\end{align*}
\end{questionparts}
While there were several very good responses to this question, there were also a significant number of candidates who did not recognise that the first differential equation could easily be turned into a first-order differential equation. Where the correct solution method was not identified, responses often did not make any further progress with the question, and many responses were very brief before the candidate opted to move on to a different question. Those who recognised the method that was needed were often able to solve part (i) well although many candidates appeared to assume that the information dx/dt = a² at t = 0 would mean that dx/dt = x² for all t. In part (ii) many candidates appeared to be familiar with the form of the required integral and so were able to reach a solution of the differential equation, although many struggled to explain the initial motion sufficiently clearly. Part (iii) was generally answered well by those that attempted it, but in many cases the absolute value signs within the integrals was not dealt with sufficiently clearly within the solution. Many responses to this part of the question did not consider all of the possible cases, with the case where a = q being the most commonly omitted. Throughout the question, responses often attempted to explain the motion of the particle as t → ∞, rather than the motion as t increases from 0.