Year: 2010
Paper: 3
Question Number: 8
Course: LFM Pure
Section: Differential equations
About 80% of candidates attempted at least five questions, and well less than 20% made genuine attempts at more than six. Those attempting more than six questions fell into three camps which were those weak candidates who made very little progress on any question, those with four or five fair solutions casting about for a sixth, and those strong candidates that either attempted 7th or even 8th questions as an "insurance policy" against a solution that seemed strong but wasn't, or else for entertainment!
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1531.5
Banger Comparisons: 4
Given that ${\rm P} (x) = {\rm Q} (x){\rm R}'(x) - {\rm Q}'(x){\rm R}(x)$, write down an expression for
\[
\int \frac{{\rm P} ( x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x\, .
\]
\begin{questionparts}
\item By choosing the function ${\rm R}(x)$ to be of the form
$a +bx+c x^2$, find
\[
\int \frac{5x^2 - 4x - 3} {(1 + 2x + 3x^2 )^2 } \, \d x
\,.\]
Show that the choice of ${\rm R}(x)$ is not unique and, by comparing the two functions ${\rm R}(x)$ corresponding to two different values of $a$, explain how the different
choices are related.
\item Find the general solution of
\[
(1+\cos x +2 \sin x) \frac {\d y}{\d x}
+(\sin x -2 \cos x)y = 5 - 3 \cos x + 4 \sin x\,.
\]
\end{questionparts}\begin{align*}
&& \int \frac{{\rm P} ( x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x &= \int \frac{{\rm Q} (x){\rm R}'(x) - {\rm Q}'(x){\rm R}(x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x \\
&&&= \int \frac{\d}{\d x} \left ( \frac{R(x)}{Q(x)} \right) \d x \\
&&&= \frac{R(x)}{Q(x)} + C
\end{align*}
\begin{questionparts}
\item Suppose $Q(x) = 1 + 2x + 3x^2, P(x) = 5x^2-4x-3$, and $R(x) = a +bx + cx^2$, then
\begin{align*}
&& 5x^2-4x-3 &= (1 + 2x + 3x^2)(2cx+b) - (6x+2)(a+bx+cx^2) \\
&&&= (6c-6c)x^3 + (3b+4c-6b-2c)x^2 + \\
&&&\quad+(2c+2b-6a-2b)x + (b-2a) \\
\Rightarrow && 2c-3b &= 5 \\
&& 2c-6a &= -4 \\
&& b - 2a &= -3 \\
\Rightarrow && b &= 2a - 3\\
&& c &= 3a-2
\end{align*}
So say $a = 0, b = -3, c = -2$ we will have
\begin{align*}
\int \frac{5x^2 - 4x - 3} {(1 + 2x + 3x^2 )^2 } \, \d x &= \frac{-3x-2x^2}{1+2x+3x^2} + C
\end{align*}
Suppose we have a different value of $a$, then we end up with:
\begin{align*}
\frac{a+(2a-3)x+(3a-2)x^2}{1+2x+3x^2} = a +\frac{-3x-2x^2}{1+2x+3x^2}
\end{align*}
So the different antiderivatives differ by a constant.
\item $\,$ \begin{align*}
&& \frac{\d }{\d x} \left ( \frac{1}{1+\cos x + 2 \sin x } y\right) &= \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \\
\end{align*}
We want to find $R(x) = A \cos x + B\sin x + C$ such that
\begin{align*}
&&5-3\cos x + 4 \sin x &= (1+\cos x + 2 \sin x)R'(x) - R(x)(-\sin x + 2 \cos x) \\
&&&= (1+\cos x + 2 \sin x)(-A\sin x + B \cos x) \\
&&&\quad- (A\cos x + B \sin x + C)(-\sin x + 2 \cos x) \\
&&&=(-A+C) \sin x + (B-2C)\cos x +\\
&&&\quad\quad (2B-A+A-2B)\sin x \cos x \\
&&&\quad\quad (-2A+B)\sin^2x+(B-2A)\cos^2x \\
&&&= (-A+C)\sin x + (B-2C)\cos x +(B-2A) \\
\Rightarrow && B-2A &= 5\\
&& C-A &= 4 \\
&& B-2C &= -3 \\
\end{align*}
There are many solutions so WLOG $C=4, A = 0, B = 5$ and so
\begin{align*}
&& \int \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \d x &= \frac{5\sin x +4}{1+\cos x + 2 \sin x} + K \\
\Rightarrow && y &= 5\sin x + 4 + K(1 + \cos x + 2 \sin x)
\end{align*}
\end{questionparts}
Three quarters of the candidates had a go at this, with moderate success. Most understood the method intended for part (i) and were aware of the method of using an integrating factor. Algebraic slips led to incorrect simultaneous equations in part (i), and few dealt with the non-uniqueness of R(x) satisfactorily. Having found the integrating factor for part (ii), most did not proceed further. Some candidates introduced a sign error into part (ii) which trivialized the left hand side to a differential of a product. A small number of candidates produced elegant solutions to part (ii) using the tan half angle substitution.