Question
The population, P, of a particular species of marsupial on a small remote island can be modelled by the logistic differential equation
\(\frac{dP}{dt}= kP\left ( 1-\frac{P}{N} \right )\)
where t is the time measured in years and k , N are positive constants.
The constant N represents the maximum population of this species of marsupial that the island can sustain indefinitely.
(a) In the context of the population model, interpret the meaning of \(\frac{dP}{dt}\).
(b) Show that \(\frac{d^{2}P}{dt^{2}}= k^{2}P\left ( 1-\frac{P}{N} \right )\left ( 1-\frac{2P}{N} \right )\)
(c) Hence show that the population of marsupials will increase at its maximum rate when \(P = \frac{N}{2}\). Justify your answer.
(d) Hence determine the maximum value of \(\frac{dP}{dt}\) in terms of k and N.
Let P0 be the initial population of marsupials.
(e) By solving the logistic differential equation, show that its solution can be expressed in the form
\(kt = In\frac{P}{P_{0}}\left ( \frac{N-P_{0}}{N-P} \right )\).
After 10 years, the population of marsupials is 3P0 . It is known that N = 4P0 .
(f) Find the value of k for this population model.
▶️Answer/Explanation
Ans:
(a) rate of growth (change) of the (marsupial) population (with respect to time)
Note: Do not accept growth (change) in the (marsupials) population per year.
OR
a correct and clearly labelled sign diagram (table) showing \(P = \frac{N}{2}\) corresponding to a local maximum point for \(\frac{dP}{dt}\)
Question
Consider the differential equation \(x^{2}\frac{dy}{dx}=y^{2}-2x^{2}\) for x > 0 and y > 2x . It is given that y = 3 when x = 1.
(a) Use Euler’s method, with a step length of 0.1, to find an approximate value of y when x = 1.5.
(b) Use the substitution y = vx to show that \(x\frac{dv}{dx}=v^{2}-v-2.\)
(c) (i) By solving the differential equation, show that \(y = \frac{8x+x^{4}}{4-x^{3}}.\)
(ii) Find the actual value of y when x = 1.5.
(iii) Using the graph of \(y = \frac{8x+x^{4}}{4-x^{3}}\) suggest a reason why the approximation given by Euler’s method in part (a) is not a good estimate to the actual value of y at x = 1.5.
▶️Answer/Explanation
Ans:
(a) attempt to use Euler’s method 
Note: A1 for any two correct y -values seen
y =10.6958…
y =10.7
Note: For the final A1, the value 10.7 must be the last value in a table or a list, or be given as a final answer, not just embedded in a table which has further lines.
(c) (i) attempt to separate variables v and x
Note: Condone absence of modulus signs throughout.
EITHER
attempt to find c using x = 1, y = 3, v= 3
OR
expressing both sides as a single logarithm
attempt to find A using x = 1, y = 3, v= 3
(ii) actual value at y(1.5) = 27.3
(iii) gradient changes rapidly (during the interval considered) OR the curve has a vertical asymptote at \(x = \sqrt[3]{4}(=1.5874…)\)
Question
The function \(f(x)\) is defined by the series \(f(x) = 1 + \frac{{(x + 2)}}{{3 \times 1}} + \frac{{{{(x + 2)}^2}}}{{{3^2} \times 2}} + \frac{{{{(x + 2)}^3}}}{{{3^3} \times 3}} + \ldots \) .
A.a.Write down the general term.[1]
A.b.Find the interval of convergence.[13]
B.Solve the differential equation \((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\) , giving your answer in the form \(u = f(v)\) .[8]
▶️Answer/Explanation
Markscheme
the general term is \(\frac{{{{(x + 2)}^n}}}{{{3^n}n}}\) A1
[1 mark]
\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{{(x + 2)}^{n + 1}}}}{{{3^{n + 1}}(n + 1)}} \times \frac{{{3^n}n}}{{{{(x + 2)}^n}}}} \right]\) M1A1A1
\( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{(x + 2){n^{}}}}{{3(n + 1)}}} \right]\) A1
\( = \frac{{(x + 2)}}{3}\) since \( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^{}}}}{{n + 1}}} \right] = 1\) A1R1
the series is convergent if \(\left| {\frac{{(x + 2)}}{3}} \right| < 1\) R1
then \( – 3 < x + 2 < 3 \Rightarrow – 5 < x < 1\) A1
if \(x = – 5\) , series is \(1 – 1 + \frac{1}{2} – \frac{1}{3} + \ldots + \frac{{{{( – 1)}^n}}}{n} + \ldots \) which converges M1A1
if \(x = 1\) , series is \(1 + 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} + \ldots \) which diverges M1A1
the interval of convergence is \( – 5 \le x < 1\) A1
[13 marks]
\((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\)
\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} = \frac{{(u + 3{v^3})}}{{2v}} = \frac{u}{{2v}} + \frac{{3{v^2}}}{2}\) M1A1
\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} – \frac{u}{{2v}} = \frac{{3{v^2}}}{2}\) A1
IF is \({{\rm{e}}^{\int {\frac{1}{{2v}}} {\rm{d}}v}} = {{\rm{e}}^{\frac{1}{2}\ln v}}\) M1
\( = {v^{\frac{1}{2}}}\) A1
\(\frac{u}{{\sqrt v }} = \int {\frac{{3{v^{\frac{3}{2}}}}}{2}} {\rm{d}}v\) M1
\( = \frac{3}{5}{v^{\frac{5}{2}}} + c\) A1
\(u = \frac{3}{5}{v^3} + c\sqrt v \) A1
[8 marks]
Question
Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\tan x = 2{\cos ^4}x\) given that \(y = 1\) when \(x = 0\).
(a) Solve the differential equation, giving your answer in the form \(y = f(x)\).
(b) (i) By differentiating both sides of the differential equation, show that
\[\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = – 10\sin x{\cos ^3}x\]
(ii) Hence find the first four terms of the Maclaurin series for \(y\).
▶️Answer/Explanation
Markscheme
(a) integrating factor \( = {e^{\int {\tan x{\text{d}}x} }}\) M1
\( = {{\text{e}}^{\ln \sec x}}\) A1
\( = \sec x\) A1
\(\sec x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\sec x\tan x = 2{\cos ^3}x\) (M1)
integrating,
\(y\sec x = 2\int {{{\cos }^3}x{\text{d}}x} \) A1
\( = 2\int {\cos x(1 – {{\sin }^2}x){\text{d}}x} \) A1
\( = 2\left( {\sin x – \frac{{{{\sin }^3}x}}{3}} \right) + C\) A1
Note: Condone the absence of \(C\).
(substituting \(x = 0,{\text{ }}y = 1\))
\(1 = C\) M1
the solution is
\(y = 2\cos x\left( {\sin x – \frac{{{{\sin }^3}x}}{3}} \right) + \cos x\) A1
[9 marks]
(b) (i) differentiating the equation,
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y{\sec ^2}x + \tan x\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 8{\cos ^3}x\sin x\) A1A1
Note: A1 for each side.
substituting for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\),
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y{\sec ^2}x + \tan x\left( {2{{\cos }^4}x – y\tan x} \right) = – 8{\cos ^3}x\sin x\) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y({\sec ^2}x – {\tan ^2}x) = – 8{\cos ^3}x\sin x – 2\tan x{\cos ^4}x\) (or equivalent) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = – 10\sin x{\cos ^3}x\) AG
(ii) differentiating again,
\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} + \frac{{{\text{d}}y}}{{{\text{d}}x}} = – 10{\cos ^4}x + {\text{term involving }} \sin x\) A1
it follows that
\(y(0) = 1,{\text{ }}y'(0) = 2\) A1
\(y”(0) = – 1,{\text{ }}y”'(0) = – 12\) A1
attempting to use \(y = y(0) + xy'(0) + \frac{{{x^2}}}{2}y”(0) + \frac{{{x^3}}}{6}y”'(0) + \ldots \) (M1)
\(y = 1 + 2x – \frac{{{x^2}}}{2} – 2{x^3}\) A1
[9 marks]
Question
Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x + y – 1\) with boundary condition \(y = 1\) when \(x = 0\).
a.Using Euler’s method with increments of \(0.2\), find an approximate value for \(y\) when \(x = 1\).[5]
b.Explain how Euler’s method could be improved to provide a better approximation.[1]
c.Solve the differential equation to find an exact value for \(y\) when \(x = 1\).[9]
d.(i) Find the first three non-zero terms of the Maclaurin series for \(y\).
(ii) Hence find an approximate value for \(y\) when \(x = 1\).[5]
▶️Answer/Explanation
Markscheme
(M1)(A1)(A1)(A1)
Note: Award M1 for equivalent of setting up first row of table, A1 for each of row 2, 3 and 5.
approximate solution \(y = 1.98\) A1
make the increments smaller or any specific correct instruction – for example change increment from \(0.2\) to \(0.1\) A1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} – y = 2x – 1\)
integrating factor is \({{\text{e}}^{\int { – 1{\text{d}}x} }} = {{\text{e}}^{ – x}}\) (M1)(A1)
\(\frac{{\text{d}}}{{{\text{d}}x}}(y{{\text{e}}^{ – x}}) = {{\text{e}}^{ – x}}(2x – 1)\) M1
attempt at integration by parts of \(\int {{{\text{e}}^{ – x}}(2x – 1){\text{d}}x} \) (M1)
\( = – (2x – 1){{\text{e}}^{ – x}} + \int {2{{\text{e}}^{ – x}}{\text{d}}x} \) A1
\( = – (2x – 1){{\text{e}}^{ – x}} – 2{{\text{e}}^{ – x}}( + c)\) A1
\(y{{\text{e}}^{ – x}} = – (1 + 2x){{\text{e}}^{ – x}} + c\)
\(y = – (1 + 2x) + c{{\text{e}}^x}\)
when \(x = 0,{\text{ }}y = 1 \Rightarrow c = 2\) M1
\(y = – (1 + 2x) + 2{{\text{e}}^{ – x}}\) A1
when \(x = 1,{\text{ }}y = – 3 + 2{\text{e}}\) A1
(i) METHOD 1
\(f(0) = 1,{\text{ }}f'(0) = 0\) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 + \frac{{{\text{d}}y}}{{{\text{d}}x}} \Rightarrow {f^2}(0) = 2\) A1
\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} \Rightarrow {f^3}(0) = 2\) A1
hence \(y = 1 + {x^2} + \frac{{{x^3}}}{3} + \ldots \) A1
Note: Accuracy marks are independent of each other.
METHOD 2
using Maclaurin series for \({{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots \) M1
\(y = – 1 – 2x + 2\left( {1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots } \right)\) M1A1
\(y = 1 + {x^2} + \frac{{{x^3}}}{3} + \ldots \) A1
(ii) when \(x = 1,{\text{ }}y = 1 + 1 + \frac{1}{3} = \frac{7}{3} = 2.33\) A1
Question
Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{x}{y}\), where \(y \ne 0\).
a.Find the general solution of the differential equation, expressing your answer in the form \(f(x,{\text{ }}y) = c\), where \(c\) is a constant.[3]
b.(i) Hence find the particular solution passing through the points \((1,{\rm{ \pm }}\sqrt 2 )\).
(ii) Sketch the graph of your solution and name the type of curve represented.[5]
c.(i) Write down the particular solution passing through the points \((1,{\text{ }} \pm 1)\).
(ii) Give a geometrical interpretation of this solution in relation to part (b).[3]
(ii) Find the particular solution passing through the point \((1,{\text{ }}\sqrt 2 )\).
(iii) Sketch the particular solution.
(iv) The graph of the solution only contains points with \(\left| x \right| > a\).
Find the exact value of \(a,{\text{ }}a > 0\).[12]
▶️Answer/Explanation
Markscheme
attempt to separate the variables M1
\(\int {y\frac{{{\text{d}}y}}{{{\text{d}}x}}{\text{d}}x = \int {x{\text{d}}x} } \) A1
Note: Accept \(\int {y{\text{d}}y = \int {x{\text{d}}x} } \).
obtain \(\frac{1}{2}{y^2} = \frac{1}{2}{x^2} + {\text{ constant }}( \Rightarrow {y^2} – {x^2} = c)\) A1
[3 marks]
(i) substitute the coordinates for both points M1
\({( \pm \sqrt 2 )^2} – {1^2} = 1\)
obtain \({y^2} – {x^2} = 1\) or equivalent A1
(ii) 
A1A1
Note: A1 for general shape including two branches and symmetry;
A1 for values of the intercepts.
(rectangular) hyperbola A1
[5 marks]
(i) \({y^2} – {x^2} = 0\) A1
(ii) the two straight lines \(y = \pm x\) A1
form the asymptotes to the hyperbola found above, or equivalent A1
[3 marks]
(i) the equation is homogeneous, so attempt to substitute \(y = vx\) M1
as a first step write \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x\frac{{{\text{d}}v}}{{{\text{d}}x}} + v\) (A1)
then \(x\frac{{{\text{d}}v}}{{{\text{d}}x}} + v = \frac{1}{v} + v\) A1
attempt to solve the resulting separable equation M1
\(\int {v{\text{d}}v = \int {\frac{1}{x}{\text{d}}x} } \) A1
obtain \(\frac{1}{2}{v^2} = \ln \left| x \right| + {\text{ constant}} \Rightarrow {y^2} = 2{x^2}\ln \left| x \right| + c{x^2}\) A1
(ii) substituting the coordinates (M1)
obtain \(c = 2 \Rightarrow {y^2} = 2{x^2}\ln \left| x \right| + 2{x^2}\) A1
(iii) 
A1
(iv) since \({y^2} > 0\) and \({x^2} \ne 0\) R1
\(\ln \left| x \right| > – 1 \Rightarrow \left| x \right| > {{\text{e}}^{ – 1}}\) A1
\(a = {{\text{e}}^{ – 1}}\) A1
Note: The R1 may be awarded for a correct reason leading to subsequent correct work.
[12 marks]
Question
Consider the differential equation
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\tan x = 2{\sec ^2}x,{\text{ }}0 \leqslant x < \frac{\pi }{2}\), given that \(y = 1\) when \(x = 0\).
By considering integration as the reverse of differentiation, show that for
a.i.\(0 \leqslant x < \frac{\pi }{2}\)
\[\int {\sec x{\text{d}}x = \ln (\sec x + \tan x) + C.} \][4]
a.ii.Hence, using integration by parts, show that
\[\int {{{\sec }^3}x{\text{d}}x = \frac{1}{2}\left( {\sec x\tan x + \ln (\sec x + \tan x)} \right) + C.} \][4]
b.i.Find an integrating factor and hence solve the differential equation, giving your answer in the form \(y = f(x)\).[9]
b.ii.Starting with the differential equation, show that
\[\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = 2{\sec ^2}x\tan x.\][3]
b.iii.Hence, by using your calculator to draw two appropriate graphs or otherwise, find the \(x\)-coordinate of the point of inflection on the graph of \(y = f(x)\).[4]
▶️Answer/Explanation
Markscheme
\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\ln (\sec x + \tan x)} \right) = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}}\) M1
\( = \sec x\) A1
therefore \(\int {\sec x{\text{d}}x = \ln (\sec x + \tan x) + C} \) AG
[4 marks]
\(\int {{{\sec }^3}x{\text{d}}x = \int {\sec x \times {{\sec }^2}x{\text{d}}x} } \) M1
\( = \sec x\tan x – \int {\sec x{{\tan }^2}x{\text{d}}x} \) A1A1
\( = \sec x\tan x – \int {\sec x({{\sec }^2}x – 1){\text{d}}x} \) A1
\( = \sec x\tan x – \int {{{\sec }^3}x{\text{d}}x + \int {\sec x{\text{d}}x} } \)
\( = \sec x\tan x – \int {{{\sec }^3}x{\text{d}}x + \ln (\sec x + \tan x)} \) A1
\(2\int {{{\sec }^3}x{\text{d}}x = \left( {\sec x\tan x + \ln (\sec x + \tan x)} \right)} \) A1
therefore
\(\int {{{\sec }^3}x{\text{d}}x = \frac{1}{2}\left( {\sec x\tan x + \ln (\sec x + \tan x)} \right) + C} \) AG
[4 marks]
\({\text{int factor}} = {{\text{e}}^{\int {\tan x{\text{d}}x} }}\) (M1)
\( = {{\text{e}}^{\ln \sec x}}\) (A1)
\( = \sec x\) A1
the differential equation can be written as
\(\frac{{\text{d}}}{{{\text{d}}x}}(y\sec x) = 2{\sec ^3}x\) M1A1
integrating,
\(y\sec x = \sec x\tan x + \ln (\sec x + \tan x) + C\) A1
putting \(x = 0,{\text{ }}y = 1,\) M1
\(C = 1\) A1
the solution is \(y = \cos x\left( {\sec x\tan x + \ln (\sec x + \tan x) + 1} \right)\) A1
[??? marks]
differentiating the differential equation,
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + \frac{{{\text{d}}y}}{{{\text{d}}x}}\tan x + y{\sec ^2}x = 4{\sec ^2}x\tan x\) A1A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + (2{\sec ^2}x – y\tan x)\tan x + y{\sec ^2}x = 4{\sec ^2}x\tan x\) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = 2{\sec ^2}x\tan x\) AG
[??? marks]
at a point of inflection, \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 0\) so \(y = 2{\sec ^2}x\tan x\) (M1)
therefore the point of inflection can be found as the point of intersection of the graphs of \(y = \cos x\left( {\sec x\tan x + \ln (\sec x + \tan x) + 1} \right)\)
and \(y = 2{\sec ^2}x\tan x\) (M1)
drawing these graphs on the calculator, \(x = 0.605\) A2
[??? marks]
Question
Draw slope fields for the following cases for \( – 2 \leqslant x \leqslant 2,\,\, – 2 \leqslant y \leqslant 2\)
Explain what isoclines tell you about the slope field in the following case:
a.i.\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2\).[2]
a.ii.\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + 1\).[2]
a.iii.\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x – 1\).[2]
b.i.\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \) constant.[1]
b.ii.\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( x \right)\).[1]
The slope field for the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + y\) for \( – 4 \leqslant x \leqslant 4,\,\, – 4 \leqslant y \leqslant 4\) is shown in the following diagram.
c.Explain why the slope field indicates that the only linear solution is \(y = – x – 1\).[2]
d.Given that all the isoclines from a slope field of a differential equation are straight lines through the origin, find two examples of the differential equation.[4]
▶️Answer/Explanation
Markscheme
A2
[2 marks]
A2
[2 marks]
A2
[2 marks]
the slope is the same everywhere A1
[1 mark]
all points that have the same \(x\) coordinate have the same slope A1
[1 mark]
this is where a straight line appears on the slope field A1
There is no other straight line, all the other solutions are curves A1
[2 marks]
given \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( {x,\,y} \right)\), the isoclines are \(f\left( {x,\,y} \right) = k\) (M1)
here the isoclines are \(y = kx\) (or \(x = ky\)) (A1)
any two differential equations of the correct form, for example
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ky}}{x},\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{kx}}{y},\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = {\text{sin}}\left( {\frac{y}{x}} \right),\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = {\text{sin}}\left( {\frac{x}{y}} \right)\) A1A1
[4 marks]