Year: 2014
Paper: 2
Question Number: 5
Course: UFM Pure
Section: First order differential equations (integrating factor)
There were good solutions presented to all of the questions, although there was generally less success in those questions that required explanations of results or the use of diagrams and graphs to reach the solution. Algebraic manipulation was generally well done by many of the candidates although a range of common errors such as confusing differentiation and integration and simple arithmetic slips were evident. Candidates should also be advised to use the methods that are asked for in questions unless it is clear that other methods will be accepted (such as by the use of the phrase "or otherwise").
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1486.1
Banger Comparisons: 1
Given that $y=xu$, where $u$ is a function of $x$, write down an expression for $\dfrac {\d y}{\d x}$.
\begin{questionparts}
\item Use the substitution $y=xu$ to solve
\[
\frac {\d y}{\d x} = \frac {2y+x}{y-2x}
\]
given that the solution curve passes through the point $(1,1)$.
Give your answer in the form of a quadratic in $x$ and $y$.
\item Using the substitutions $x=X+a$ and $y=Y+b$ for appropriate values of $a$ and $b$, or otherwise, solve
\[
\frac {\d y}{\d x} = \frac {x-2y-4} {2x+y-3}\,,
\]
given that the solution curve passes through the point $(1,1)$.
\end{questionparts}\begin{align*}
\frac{\d y}{\d x} &= \frac{\d }{\d x} \l y \r \\
&= \frac{\d }{\d x} \l xu \r \\
&\underbrace{=}_{\text{product rule}} \frac{\d}{\d x} \l x \r u + x \frac{\d}{\d x} \l u \r \\
&= u + x \frac{\d u}{\d x}
\end{align*}
\begin{questionparts}
\item \begin{align*}
&& \frac{\d y}{\d x} &= \frac{2y + x}{y - 2x} \\
&& u + x \frac{\d u}{\d x} &= \frac{2u + 1}{u - 2} \\
&& x \frac{\d u}{\d x} &= \frac{2u-1-u^2+2u}{u-2} \\
\Rightarrow && \int \frac{2-u}{u^2-4u+1} \d u &= \int \frac{1}{x} \d x \\
&& \int \frac{2-u}{(u-2)^2-5} \d u &= \int \frac1x \d x \\
&& -\frac12\ln| (u-2)^2 - 5| &= \ln x + C \\
(x,y) = (1,1): && - \ln 2 &= C \\
\Rightarrow && \ln x^2 &= \ln 4 - \ln |5 - (u-2)^2| \\
\Rightarrow && x^2 &= \frac{4}{5- (u-2)^2} \\
\Rightarrow && 4 & = x^2(5 - (\frac{y}{x} - 2)^2) \\
&&&= 5x^2 - (y-2x)^2 \\
&&&= x^2+4xy-y^2
\end{align*}
\item It would be convienient if $x-2y -4 = X-2Y$ and $2x+y-3 = 2X+Y$, ie $a-2b = 4$ and $2a+b = 3$, ie $a = 2, b = -1$.
Now our differential equation is:
\begin{align*}
&& \frac{\d Y}{\d X} &= \frac{X - 2Y}{2X+Y} \\
&& \frac{\d X}{\d Y} &= \frac{2X + Y}{X-2Y}
\end{align*}
This is the same differential equation we have already solved, just with the roles of $x$ and $y$ interchanged with $Y$ and $X$ and with the point $(0,3)$ being on the curve, ie:
$Y^2 + 4XY-X^2 = c$ and $c = 9$, therefore our equation is:
\[ (y-1)^2 + 4(y-1)(x+2)-(x+2)^2 = 9\]
This was the most popular question on the paper and the question which had the highest average score. Most candidates correctly solved the differential equation in the first part of the question, but many then calculated the constant term incorrectly. In the second part of the question most candidates were able to find the appropriate values of a and b, but then did not see how to apply the result from part (i) and so did the integration again or just copied the answer from the first part. Some candidates again struggled to obtain the correct constant for the integration and others did not substitute the correct values for the point on the curve (taking (x, y) as (1,1) rather than (π, π)).