IB DP Maths AA: Topic: AHL 5.18: First order differential equations: IB style Questions HL Paper 2

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]

A.a.

\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{{(x + 2)}^{n + 1}}}}{{{3^{n + 1}}(n + 1)}} \times \frac{{{3^n}n}}{{{{(x + 2)}^n}}}} \right]\)     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]

A.b.

\((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\)

\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} = \frac{{(u + 3{v^3})}}{{2v}} = \frac{u}{{2v}} + \frac{{3{v^2}}}{2}\)     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

a.

make the increments smaller or any specific correct instruction – for example change increment from \(0.2\) to \(0.1\)     A1

b.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} – y = 2x – 1\)

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

c.

(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

d.

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]

d.(i)     Find the general solution of the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{x}{y} + \frac{y}{x}\), where \(xy \ne 0\).

(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]

a.

(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)     M16/5/FURMA/HP2/ENG/TZ0/04.b.ii/M     A1A1

Note:     A1 for general shape including two branches and symmetry;

A1 for values of the intercepts.

(rectangular) hyperbola     A1

[5 marks]

b.

(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]

c.

(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)     M16/5/FURMA/HP2/ENG/TZ0/04.d.iii/M     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]

d.

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]

a.i.

\(\int {{{\sec }^3}x{\text{d}}x = \int {\sec x \times {{\sec }^2}x{\text{d}}x} } \)     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]

a.ii.

\({\text{int factor}} = {{\text{e}}^{\int {\tan x{\text{d}}x} }}\)     (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]

b.i.

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]

b.ii.

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]

b.iii.

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]

a.i.

   A2

[2 marks]

a.ii.

   A2

[2 marks]

a.iii.

the slope is the same everywhere     A1

[1 mark]

b.i.

all points that have the same \(x\) coordinate have the same slope    A1

[1 mark]

b.ii.

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]

c.

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]

d.
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