2007 Paper 3 Q8

Year: 2007
Paper: 3
Question Number: 8

Course: UFM Pure
Section: Second order differential equations

Difficulty: 1700.0 Banger: 1487.5

Problem

  1. Find functions \({\rm a}(x)\) and \({\rm b}(x)\) such that \(u=x\) and \(u=\e^{-x}\) both satisfy the equation $$\dfrac{\d^2u}{\d x^2} +{\rm a}(x) \dfrac{\d u}{\d x} + {\rm b} (x)u=0\,.$$ For these functions \({\rm a}(x)\) and \({\rm b}(x)\), write down the general solution of the equation. Show that the substitution \(y= \dfrac 1 {3u} \dfrac {\d u}{\d x}\) transforms the equation \[ \frac{\d y}{\d x} +3y^2 + \frac {x} {1+x} y = \frac 1 {3(1+x)} \tag{\(*\)} \] into \[ \frac{\d^2 u}{\d x^2} +\frac x{1+x} \frac{\d u}{\d x} - \frac 1 {1+x} u=0 \] and hence show that the solution of equation (\(*\)) that satisfies \(y=0\) at \(x=0\) is given by \(y = \dfrac{1-\e^{-x}}{3(x+\e^{-x})}\).
  2. Find the solution of the equation $$ \frac{\d y}{\d x} +y^2 + \frac x {1-x} y = \frac 1 {1-x} $$ that satisfies \(y=2\) at \(x=0\).

No solution available for this problem.

Examiner's report
— 2007 STEP 3, Question 8
Above Average Ranked alongside Q5 in popularity and success

This ranked alongside question 5 in popularity and success. Frequently, it was calculation errors that obscured the path through part (i) and the two differences between part (i) and part (ii) were enough to put most off the track for part (ii), even if they had completed (i) successfully.

Source: Cambridge STEP 2007 Examiner's Report · 2007-full.pdf
Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1487.5

Banger Comparisons: 1

Show LaTeX source
Problem source
\begin{questionparts}
\item Find functions ${\rm a}(x)$ and ${\rm b}(x)$ such that $u=x$ and
  $u=\e^{-x}$
both satisfy the equation 
$$\dfrac{\d^2u}{\d x^2} +{\rm a}(x) \dfrac{\d u}{\d x} + {\rm b} (x)u=0\,.$$ 
For these functions ${\rm a}(x)$ and ${\rm b}(x)$, write 
down the  general solution of the equation.
Show that  the substitution $y= \dfrac 1 {3u} \dfrac {\d u}{\d x}$
transforms the equation
\[
\frac{\d y}{\d x} +3y^2 + \frac {x} {1+x} y = \frac 1 {3(1+x)}
\tag{$*$}
\]
into
\[
\frac{\d^2 u}{\d x^2} +\frac x{1+x} \frac{\d u}{\d x} - \frac 1 {1+x}
u=0
\]
and hence show that the solution of equation ($*$) that satisfies
$y=0$ at $x=0$ is given by
 $y = \dfrac{1-\e^{-x}}{3(x+\e^{-x})}$.
\item
Find the solution of the equation
$$
\frac{\d y}{\d x} +y^2 + \frac x {1-x} y = \frac 1 {1-x}
$$
that satisfies $y=2$ at $x=0$.

\end{questionparts}