Year: 2013
Paper: 3
Question Number: 7
Course: UFM Pure
Section: Second order differential equations
With the number of candidates submitting scripts up by some 8% from last year, and whilst inevitably some questions were more popular than others, namely the first two, 7 then 4 and 5 to a lesser extent, all questions on the paper were attempted by a significant number of candidates. About a sixth of candidates gave in answers to more than six questions, but the extra questions were invariably scoring negligible marks. Two fifths of the candidates gave in answers to six questions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Let $y(x)$ be a solution of the differential equation
$ \dfrac {\d^2 y}{\d x^2}+y^3=0$ with $y = 1$ and $\dfrac{\d y}{\d x} =0$ at $x=0$, and let
\[
{\rm E} (x)=
\left ( \frac {\d y}{\d x}\right)^{\!\!2} + \tfrac 12 y^4\,.
\]
Show by differentiation that ${\rm E}$ is constant and deduce that $ \vert y(x) \vert \le 1$ for all $x$.
\item Let $v(x)$ be a solution of the differential equation
$ \dfrac{\d^2 v}{\d x^2} + x \dfrac {\d v}{\d x} +\sinh v =0$ with $v = \ln 3$ and $\dfrac{\d v}{\d x} =0$ at $x=0$, and let
\[
{\rm E} (x)=
\left ( \frac {\d v}{\d x}\right)^{\!\!2} + 2 \cosh v\,.
\]
Show that $\dfrac{\d{\rm E}}{\d x}\le 0$ for $x\ge0$ and deduce that $\cosh v(x) \le \frac53$ for $x\ge0$.
\item Let $w(x)$ be a solution of the differential equation
\[
\frac{\d^2 w}{\d x^2} + (5\cosh x - 4 \sinh x -3) \frac{\d w}{\d x} +
(w\cosh w + 2 \sinh w) =0
\]
with $\dfrac{\d w }{\d x}=\dfrac 1 { \sqrt 2 }$ and $w=0$ at $x=0$.
Show that $\cosh w(x) \le \frac54$ for $x\ge0$.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& E(x) &= \left ( \frac{\d y}{\d x} \right)^2 + \frac12 y^4 \\
\Rightarrow && E'(x) &= 2 \frac{\d y}{\d x} \frac{\d^2 y}{\d x^2} + 2y^3 \frac{\d y}{\d x} \\
&&&= 2\frac{\d y}{\d x} \left ( \frac{\d^2 y}{\d x^2} + y^3 \right) \\
&&&= 0
\end{align*}
Therefore $E$ is constant. $E(0) = \frac12$ and
\begin{align*}
&&\frac12 &= \left ( \frac{\d y}{\d x} \right)^2 + \frac12 y^4 \\
&&&\geq \frac12 y^4 \\
\Rightarrow && |y| &\leq 1
\end{align*}
\item $\,$ \begin{align*}
&& E(x) &= \left ( \frac{\d v}{\d x} \right)^2 + 2 \cosh v \\
\Rightarrow && E'(x) &= 2 \frac{\d v}{\d x}\frac{\d^2 v}{\d^2 x} + 2 \sinh v \frac{\d v}{\d x} \\
&&&= 2 \frac{\d v}{\d x} \left ( \frac{\d^2 v}{\d^2 x} + \sinh v\right) \\
&&&= 2 \frac{\d v}{\d x} \left ( -x \frac{\d v}{\d x}\right) \\
&&&= -2x \left ( \frac{\d v}{\d x} \right)^2 \leq 0 \tag{$x \geq 0$} \\
\\
&& E(0) &= 0^2 + 2 \cosh \ln 3 \\
&&&= 3 + \frac13 = \frac{10}{3} \\
\Rightarrow && \frac{10}{3} &\geq E(x) \\
&&&= \left ( \frac{\d v}{\d x} \right)^2 + 2 \cosh v \\
&&&\geq 2 \cosh v \\
\Rightarrow && \cosh v &\leq \frac53
\end{align*}
\item $\,$ \begin{align*}
&& E(x) &= \left ( \frac{\d w}{\d x} \right)^2 + 2(w \sinh w + \cosh w) \\
&& E'(x) &= 2 \frac{\d w}{\d x}\frac{\d^2 w}{\d^2 x} + 2(w \cosh w + 2 \sinh w) \frac{\d w}{\d x} \\
&&&= 2 \frac{\d w}{\d x} \left ( \frac{\d^2 w}{\d^2 x}+(w \cosh w + 2 \sinh w)\right) \\
&&&= -2 \left ( \frac{\d w}{\d x} \right)^2 \left (\underbrace{5\cosh x - 4 \sinh x -3}_{\geq0} \right) \\
&&&\leq 0 \\
&& E(0) &= \frac12 + 2 = \frac52 \\
\Rightarrow && \frac52 &\geq E(x) \\
&&&=\underbrace{ \left ( \frac{\d w}{\d x} \right)^2}_{\geq0} + 2(\underbrace{w \sinh w}_{\geq 0} + \cosh w) \\
&&&\geq2\cosh w \\
\Rightarrow && \cosh w &\leq \frac54
\end{align*}
\end{questionparts}
Three quarters attempted this with more success than question 6 but less than question 3. Sadly, it was not uncommon for candidates to fail to differentiate correctly. Many established that the expression equals 0 but then made errors, when considering the maximum which was not sufficient and missed the point of the squared term, with consequences for the rest of the question. Many followed the stationary points line of logic correctly by considering the maximum and minimum values in part (i). Having established the constant value, some candidates attempted to solve the differential equation, usually by incorrect methods. The errors of part (i) were largely replicated in part (ii). There were fewer attempts at part (iii), and a number fell at the first hurdle through not obtaining the correct expression. Further, numerous candidates assumed rather than proved that 5 cosh − 4 sinh − 3 ≥ 0.