Year: 2024
Paper: 3
Question Number: 6
Course: LFM Pure
Section: Differential equations
The total entry was an increase on that of 2023 by more than 10%. One question was attempted by more than 98% of candidates, another two by about 80%, and another five by between 50% and 70%. The remaining four questions were attempted by between 5% and 30% of candidates, these being from Section B: Mechanics, and Section C: Probability and Statistics, though the Statistics questions were in general attempted more often and more successfully. All questions were perfectly solved by some candidates. About 84% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item A particle moves in two-dimensional space. Its position is given by coordinates $(x, y)$ which satisfy
\[\frac{\mathrm{d}x}{\mathrm{d}t} = -x + 3y + u\]
\[\frac{\mathrm{d}y}{\mathrm{d}t} = x + y + u\]
where $t$ is the time and $u$ is a function of time. At time $t = 0$, the particle has position $(x_0,\, y_0)$.
\begin{enumerate}
\item By considering $\dfrac{\mathrm{d}x}{\mathrm{d}t} - \dfrac{\mathrm{d}y}{\mathrm{d}t}$, show that if the particle is at the origin $(0,0)$ at some time $t > 0$, then it is necessary that $x_0 = y_0$.
\item Given that $x_0 = y_0$, find a constant value of $u$ that ensures that the particle is at the origin at a time $t = T$, where $T > 0$.
\end{enumerate}
\item A particle whose position in three-dimensional space is given by co-ordinates $(x, y, z)$ moves with time $t$ such that
\[\frac{\mathrm{d}x}{\mathrm{d}t} = 4y - 5z + u\]
\[\frac{\mathrm{d}y}{\mathrm{d}t} = x - 2z + u\]
\[\frac{\mathrm{d}z}{\mathrm{d}t} = x - 2y + u\]
where $u$ is a function of time. At time $t = 0$, the particle has position $(x_0,\, y_0,\, z_0)$.
\begin{enumerate}
\item Show that, if the particle is at the origin $(0,0,0)$ at some time $t > 0$, it is necessary that $y_0$ is the mean of $x_0$ and $z_0$.
\item Show further that, if the particle is at the origin $(0,0,0)$ at some time $t > 0$, it is necessary that $x_0 = y_0 = z_0$.
\item Given that $x_0 = y_0 = z_0$, find a constant value of $u$ that ensures that the particle is at the origin at a time $t = T$, where $T > 0$.
\end{enumerate}
\end{questionparts}\begin{questionparts}
\item \begin{enumerate}
\item $\,$ \begin{align*}
&& \frac{\d x}{\d t} - \frac{\d y}{\d t} &= -2x + 2y \\
&&&= -2(x-y) \\
\Rightarrow && \dot{z} &= -2z \tag{$z = x-y$} \\
\Rightarrow && z &= Ae^{-2t} \\
z = 0, t : && A &= 0 \\
\Rightarrow && z &= 0 \quad \forall t \\
\Rightarrow && x &= y \quad \forall t \\
\Rightarrow && x_0 &= y_0
\end{align*}
\item Since $x = y$ for all $t$ our equation can we written as $\frac{\d x}{\d t} = 2x + u$. This has solution $x = Ae^{2t} - \frac{u}{2}$ we also have \begin{align*}
t = 0: && x_0 &= A - \frac{u}{2} \\
t = T: && 0 &= Ae^{2T} - \frac{u}{2} \\
\Rightarrow && x_0 &= \frac{u}{2} - e^{-2T}\frac{u}{2} \\
\Rightarrow && u &= \frac{2x_0}{1-e^{-2T}}
\end{align*}
\end{enumerate}
\item \begin{enumerate}
\item Let $w = y - \frac12(x+z)$ then \begin{align*}
&& \dot{w} &= \dot{y} - \tfrac12(\dot{x}+\dot{z}) \\
&&&= (x-2z+u) - \tfrac12(4y-5z+u+x-2y+u) \\
&&&= -y + \tfrac12(x+z) \\
&&&= -w \\
\Rightarrow && w &= Ae^{-t} \\
\text{at origin}: && w &= 0 \\
\Rightarrow && y &= \tfrac12(x+z)
\end{align*}
\item We now have $\dot{x} = 2x+2z-5z+u = 2x-3z+u$ and $\dot{z} = x - x- z +u = - z+u$ so in particular $(x - z)' = 2(x-z)$ or $x-z = Ae^{2t}$ and since we hit the origin $x = z$ for all $t$ and so $y = x = z$ for all $t$.
\item Notice we now have $\dot{x} = -x + u$ or $x = Ae^{-t} + u$
\begin{align*}
t = 0: && x_0 &= A + u \\
t = T: && 0 &= Ae^{-T} + u \\
\Rightarrow && x_0 &= u-e^{T} u \\
\Rightarrow && u &= \frac{x_0}{1-e^{T}}
\end{align*}
\end{enumerate}
\end{questionparts}
This was the least popular of the Pure Mathematics section, and by a large margin the least successful of the whole paper. Those candidates who were successful in part (i)(a) usually tackled the question by re-writing the differential equation as d(x−y)/dt = -2(x - y). There were also some candidates who rewrote the equation as dx/dt + 2x = dy/dt = 2y and used integrating factors effectively to solve this, although some integrated erroneously to achieve x + 2xt = y + 2yt. Some candidates correctly concluded that x = y but did not go on to say that this implied that x0 = y0. Most of the candidates gaining no credit for this question substituted x = y = 0 into their differential equation and then integrated that. In part (i)(b) those candidates who attempted it generally understood what was required, but some did not appreciate that the situation in this case had different initial conditions to that in part (a). Some candidates used the given differential equations to find a second order differential equation in x or y, which was a valid if inefficient method. Those attempting part (ii) generally performed in a similar way to part (i), either gaining most of the credit available or making the same mistakes they had made in the previous parts. There were some candidates who rather cleverly spotted that they could combine the last two differential equations to show that y = z, and then show that x = z and in so doing answer both parts (ii)(a) and (b) together.