2018 Paper 2 Q8
Year: 2018
Paper: 2
Question Number: 8
Course: UFM Pure
Section: First order differential equations (integrating factor)
Difficulty: 1600.0
Banger: 1484.0
differential equation
first order
integrating factor
substitution
Bernoulli equation
curve sketching
comparison of solutions
Problem
- Use the substitution \(v= \sqrt y\) to solve the differential equation \[ \frac{\d y}{\d t} = \alpha y^{\frac12} - \beta y \ \ \ \ \ \ \ \ \ \ (y\ge0, \ \ t\ge0) \,, \] where \(\alpha\) and \(\beta\) are positive constants. Find the non-constant solution \(y_1(x)\) that satisfies \(y_1(0)=0\,\).
- Solve the differential equation \[ \frac{\d y}{\d t} = \alpha y^{\frac23} - \beta y \ \ \ \ \ \ \ \ \ \ (y\ge0, \ \ t\ge0) \,, \] where \(\alpha\) and \(\beta\) are positive constants. Find the non-constant solution \(y_2(x)\) that satisfies \(y_2(0)=0\,\).
- In the case \(\alpha=\beta\), sketch \(y_1(x)\) and \(y_2(x)\) on the same axes, indicating clearly which is \(y_1(x)\) and which is \(y_2(x)\). You should explain how you determined the positions of the curves relative to each other.
Solution
- Suppose \(v = \sqrt{y} \Rightarrow v^2 = y \Rightarrow 2v v' = y'\) so \begin{align*} && 2vv' &= \alpha v-\beta v^2 \\ \Rightarrow && 2v' &= \alpha - \beta v \\ \Rightarrow && v' + \frac{\beta}{2} v &= \frac{\alpha}{2} \\ \Rightarrow && \frac{\d}{\d t} \left (e^{\beta t/2} v \right) &= \frac{\alpha}{2} e^{\beta t/2} \\ \Rightarrow && e^{\beta t/2} v &=C+ \frac{\alpha}{\beta}e^{\beta t /2} \\ \Rightarrow && v &= Ce^{-\beta t/2} + \frac{\alpha}{\beta} \\ \Rightarrow && \sqrt{y} &= Ce^{-\beta t/2} + \frac{\alpha}{\beta} \\ y(0) = 0: && 0 &= C+\frac{\alpha}{\beta} \\ \Rightarrow && \sqrt{y} &= \frac{\alpha}{\beta} \left (1-e^{-\beta t/2} \right) \\ \Rightarrow && y &= \frac{ \alpha^2}{\beta^2} \left (1-e^{-\beta t/2} \right)^2 \end{align*}
- Try \(v = y^{1/3} \Rightarrow v^3 = y \Rightarrow 3v^2 v' = y'\) so \begin{align*} && y' &= \alpha v^2 - \beta y \\ \Rightarrow && 3v^2v' &= \alpha v^2 - \beta v^3 \\ \Rightarrow && v' +\frac{\beta}{3} v &= \frac{\alpha}{3} \\ \Rightarrow && (v e^{\beta t/3})' &= \frac{\alpha}{3}e^{\beta t/3} \\ \Rightarrow && v &= Ce^{-\beta t/3} + \frac{\alpha}{\beta} \\ v(0) = 0: && v &= \frac{\alpha}{\beta} \left (1 - e^{-\beta t/3} \right) \\ \Rightarrow && y &= \frac{\alpha^3}{\beta^3} \left (1 - e^{-\beta t/3} \right) ^3 \end{align*}
- \(y_1 = (1-e^{-\beta t/2})^2, y_2 = (1-e^{-\beta t/3})^3\)
By considering the differential equation, notice that \(0 < y_i < 1\) so \(y^{1/2} > y^{2/3}\) and therefore \(y_1' > y_2'\) and so \(y_1\) increases faster.
Examiner's report
— 2018 STEP 2, Question 8
From the paper introduction
The pure questions were again the most popular of the paper, with only two of those questions being attempted by fewer than half of the candidates (none of the other questions was attempted by more than half of the candidates). Good responses were seen to all of the questions, but in many cases, explanations lacked sufficient detail to be awarded full marks. Candidates should ensure that they are demonstrating that the results that they are attempting to apply are valid in the cases being considered. In several of the questions, later parts involve finding solutions to situations that are similar to earlier parts of the question. In general candidates struggled to recognise these similarities and therefore spent a lot of time repeating work that had already been done, rather than simply observing what the result must be.
Source: Cambridge STEP 2018 Examiner's Report
· 2018-p2.pdf
Rating Information
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
Show LaTeX source
Problem source
\begin{questionparts}
\item Use the substitution $v= \sqrt y$
to solve the differential equation
\[
\frac{\d y}{\d t} = \alpha y^{\frac12} - \beta y
\ \ \ \ \ \ \ \ \ \ (y\ge0, \ \ t\ge0)
\,,
\]
where $\alpha$ and $\beta$ are positive constants.
Find the non-constant solution
$y_1(x)$
that satisfies $y_1(0)=0\,$.
\item
Solve the differential equation
\[
\frac{\d y}{\d t} = \alpha y^{\frac23} - \beta y
\ \ \ \ \ \ \ \ \ \ (y\ge0, \ \ t\ge0)
\,,
\]
where $\alpha$ and $\beta$ are positive constants.
Find the non-constant solution
$y_2(x)$
that satisfies
$y_2(0)=0\,$.
\item In the case $\alpha=\beta$, sketch
$y_1(x)$ and $y_2(x)$
on the same
axes, indicating clearly which is
$y_1(x)$ and which is $y_2(x)$.
You should explain how you determined the positions of the
curves relative to each other.
\end{questionparts}Solution source
\begin{questionparts}
\item Suppose $v = \sqrt{y} \Rightarrow v^2 = y \Rightarrow 2v v' = y'$ so
\begin{align*}
&& 2vv' &= \alpha v-\beta v^2 \\
\Rightarrow && 2v' &= \alpha - \beta v \\
\Rightarrow && v' + \frac{\beta}{2} v &= \frac{\alpha}{2} \\
\Rightarrow && \frac{\d}{\d t} \left (e^{\beta t/2} v \right) &= \frac{\alpha}{2} e^{\beta t/2} \\
\Rightarrow && e^{\beta t/2} v &=C+ \frac{\alpha}{\beta}e^{\beta t /2} \\
\Rightarrow && v &= Ce^{-\beta t/2} + \frac{\alpha}{\beta} \\
\Rightarrow && \sqrt{y} &= Ce^{-\beta t/2} + \frac{\alpha}{\beta} \\
y(0) = 0: && 0 &= C+\frac{\alpha}{\beta} \\
\Rightarrow && \sqrt{y} &= \frac{\alpha}{\beta} \left (1-e^{-\beta t/2} \right) \\
\Rightarrow && y &= \frac{ \alpha^2}{\beta^2} \left (1-e^{-\beta t/2} \right)^2
\end{align*}
\item Try $v = y^{1/3} \Rightarrow v^3 = y \Rightarrow 3v^2 v' = y'$ so
\begin{align*}
&& y' &= \alpha v^2 - \beta y \\
\Rightarrow && 3v^2v' &= \alpha v^2 - \beta v^3 \\
\Rightarrow && v' +\frac{\beta}{3} v &= \frac{\alpha}{3} \\
\Rightarrow && (v e^{\beta t/3})' &= \frac{\alpha}{3}e^{\beta t/3} \\
\Rightarrow && v &= Ce^{-\beta t/3} + \frac{\alpha}{\beta} \\
v(0) = 0: && v &= \frac{\alpha}{\beta} \left (1 - e^{-\beta t/3} \right) \\
\Rightarrow && y &= \frac{\alpha^3}{\beta^3} \left (1 - e^{-\beta t/3} \right) ^3
\end{align*}
\item $y_1 = (1-e^{-\beta t/2})^2, y_2 = (1-e^{-\beta t/3})^3$
\begin{center}
\begin{tikzpicture}
% Slightly expanded bounding limits for breathing room
\def\xl{-.3}; \def\xu{5};
\def\yl{-.3}; \def\yu{1.2};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the reusable styles to keep code clean
\tikzset{
x=\xscale cm, y=\yscale cm,
axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
grid/.style={thin, dashed, gray!30},
curveA/.style={very thick, color=cyan!70!black, smooth},
curveB/.style={very thick, color=orange!90!black, smooth},
dot/.style={circle, fill=black, inner sep=1.2pt},
labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
}
% Draw background grid
\draw[grid] (\xl,\yl) grid[step=0.5] (\xu,\yu);
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$y$};
% Define the bounding region with clip
\begin{scope}
\clip (\xl,\yl) rectangle (\xu,\yu);
\draw[curveA, domain=0:5, samples=150]
plot ({\x},{(1-exp(-\x))^2});
\draw[curveB, domain=0:5, samples=150]
plot ({\x},{(1-exp(-\x*2/3))^3});
% \draw[curveB, domain=-3:3, samples=150]
% plot ({(4*\x*\x+1)/3},{\x});
\end{scope}
% Annotate Function Names
\node[curveA, labelbox] at (1, {(1-exp(-1))^2}) {$y = (1-e^{-\beta t/2 })^2$};
\node[curveB, labelbox] at (3, {(1-exp(-4/3))^2}) {$y = (1-e^{-\beta t/3 })^3$};
\end{tikzpicture}
\end{center}
By considering the differential equation, notice that $0 < y_i < 1$ so $y^{1/2} > y^{2/3}$ and therefore $y_1' > y_2'$ and so $y_1$ increases faster.
\end{questionparts}
This was the third most popular question on the paper and a number of very good responses were seen from candidates. Most candidates were able to apply the given substitution to the differential equation and then separated the variables successfully. Candidates adopted a range of approaches to solving the differential equation, such as use of an integrating factor or a solution by finding a complementary function and then a particular integral. Success was seen with all of these methods. Some candidates omitted the constant of integration and then were unable to reach the correct answer. In the second part of the question most candidates were able to spot a suitable substitution and proceeded to solve the differential equation successfully. The best candidates spotted the similarity to part (i) and therefore saved some time on this part by modifying the answer to (i) rather than working through all of the steps again. The final part of the question was the least well answered. Although most candidates realised that ( ) > ( ), many did not justify this or substituted a value to check rather than demonstrating that it was true for all values. Many candidates also failed to recognise that the gradient should be 0 at the origin for both curves.