Year: 2012
Paper: 3
Question Number: 7
Course: UFM Pure
Section: Second order differential equations
No solution available for this problem.
The number of candidates attempting more than six questions was, as last year, about 25%, though most of these extra attempts achieved little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
A pain-killing drug is injected into the bloodstream. It
then diffuses into the brain, where it is absorbed.
The quantities at time $t$ of the drug in the blood
and the brain respectively are $y(t)$
and $z(t)$. These satisfy
\[
\dot y = - 2(y-z)\,,
\ \ \ \ \ \ \
\dot z = - \dot y -3z\, ,
\]
where the dot denotes differentiation with respect to $t$.
Obtain a second order differential equation for $y$ and hence derive
the solution
\[
y= A\e^{-t} + B\e ^{-6t}\,,
\ \ \ \ \ \ \
z= \tfrac12 A \e^{-t} - 2 B \e^{-6t}\,,
\]
where $A$ and $B$ are arbitrary constants.
\begin{questionparts}
\item Obtain the solution that satisfies
$z(0)=0$ and $y(0)= 5$. The quantity
of the drug in the brain for this solution is
denoted by $z_1(t)$.
\item Obtain the solution that satisfies
$
z(0)=z(1)= c$,
where $c$ is a given constant.
%\[
%C=2(1-\e^{-1})^{-1} - 2(1-\e^{-6})^{-1}\,.
%\]
The quantity
of the drug in the brain for this solution is
denoted by $z_2(t)$.
\item Show that for $0\le t \le 1$,
\[
z_2(t) = \sum _{n=-\infty}^{0} z_1(t-n)\,,
\]
provided $c$ takes a particular value that you should find.
\end {questionparts}
Two thirds attempted this too, with marginally greater success than question 2. Most did very well with the stem, though a few were unable to obtain a proper second order equation. Those that attempted part (i) were usually successful. The non‐trivial exponential calculations in part (ii) caused problems for some making computation mistakes whilst others were totally on top of this. Part (iii) tested the candidates on two levels, interpreting the sigma notation correctly, and recognising and using the geometric series. Some managed this excellently.