Year: 2011
Paper: 3
Question Number: 1
Course: UFM Pure
Section: Second order differential equations
The percentages attempting larger numbers of questions were higher this year than formerly. More than 90% attempted at least five questions and there were 30% that didn't attempt at least six questions. About 25% made substantive attempts at more than six questions, of which a very small number indeed were high scoring candidates that had perhaps done extra questions (well) for fun, but mostly these were cases of candidates not being able to complete six good solutions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item
Find the general solution of the differential equation
\[
\frac{\d u}{\d x} - \left(\frac { x +2}{x+1}\right)u =0\,.
\]
\item Show that substituting$y=z\e^{-x}$ (where $z$ is a function of $x$) into the second order differential equation
\[
(x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y = 0
\tag{$*$}
\]
leads to a first order differential equation for $\dfrac{\d z}{\d x}\,$.
Find $z$ and hence show that the general solution of $(*)$ is
\[
y= Ax + B\e^{-x}\,,
\]
where $A$ and $B$ are arbitrary constants.
\item Find the general solution of the differential equation
\[
(x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y =
(x+1)^2 .
\]
\end{questionparts}\begin{questionparts}
\item
\begin{align*}
&& 0 &= \frac{\d u}{\d x} - \left ( \frac{x+2}{x+1} \right)u \\
\Rightarrow && \int \frac1u \d u &= \int 1 + \frac1{x+1} \d x \\
\Rightarrow && \ln |u| &= x + \ln |x+1| + C \\
\Rightarrow && u &= A(x+1)e^x
\end{align*}
\item If $y = ze^{-x}$, $y' = (z'-z)e^{-x}$, $y'' = (z''-2z'+z)e^{-x}$
\begin{align*}
&& 0 &= (x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y \\
y = ze^{-x}: && 0 &= (x+1) \left ( \frac{\d^2 z}{\d x^2} - 2\frac{\d z}{\d x} +z\right)e^{-x} +x \left ( \frac{\d z}{\d x} -z\right)e^{-x} - ze^{-x} \\
&&&= (x+1) \frac{\d^2 z}{\d x^2} -(x+2)\frac{\d z}{\d x} \\
\Rightarrow && \frac{\d}{\d x} \left ( \frac{\d z}{\d x}\right) &= \left ( \frac{x+2}{x+1}\right) \frac{\d z}{\d x}
\end{align*}
Therefore $\frac{\d z}{\d x} = A(x+1)e^x $ and so
\begin{align*}
z &= A \int (x+1)e^{x} \d x \\
&= A \left ( \left [ (x+1)e^x\right] - \int e^x \d x \right) \\
&= A(x+1)e^x - Ae^x + B \\
y &= Ax + Be^{-x}
\end{align*}
\item We have found the complementary solution. To find a particular integral consider $y = ax^2 + bx + c$, then $y' = 2ax+b, y'' = 2a$ and we have
\begin{align*}
&& x^2+2x+1 &= 2a(x+1) + x(2ax+b) - (ax^2+bx+c) \\
\Rightarrow && x^2+2x+1 &= ax^2+ 2ax + 2a-c \\
\Rightarrow && a = 1, &c=1
\end{align*}
so the general solution should be
\[ y = Ax + Be^{-x} + x^2+1 \]
\end{questionparts}
As might be expected, this was a very popular question, in fact the most popular being attempted by very nearly all the candidates. Fortunately, it was also generally well-attempted, with scores well above those for other questions. Apart from frequent algebraic errors and overlooking terms, especially when using results from a previous part that required adaptation, the main difficulties were in showing that (*) in part (ii) did indeed lead to a first order differential equation in the required variable, and the consequent solution of that equation. Part (iii) was generally well done. At the other end of the scale, some candidates did leave their answers to part (i) in the form ln.