# IB DP Maths Topic 9.5 Geometric interpretation using slope fields, including identification of isoclines HL Paper 3

## Question

Consider the differential equation $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = {x^2} + {y^2}$$ where y =1 when x = 0 .

Use Euler’s method with step length 0.1 to find an approximate value of y when x = 0.4.

[7]
a.

Write down, giving a reason, whether your approximate value for y is greater than or less than the actual value of y .

[1]
b.

## Markscheme

use of $$y \to y + h\frac{{{\text{d}}y}}{{{\text{d}}x}}$$     (M1)

approximate value of y = 1.57     A1

Note: Accept values in the tables correct to 3 significant figures.

[7 marks]

a.

the approximate value is less than the actual value because it is assumed that $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$ remains constant throughout each interval whereas it is actually an increasing function     R1

[1 mark]

b.

## Examiners report

Most candidates were familiar with Euler’s method. The most common way of losing marks was either to round intermediate answers to insufficient accuracy or simply to make an arithmetic error. Many candidates were given an accuracy penalty for not rounding their answer to three significant figures. Few candidates were able to answer (b) correctly with most believing incorrectly that the step length was a relevant factor.

a.

Most candidates were familiar with Euler’s method. The most common way of losing marks was either to round intermediate answers to insufficient accuracy or simply to make an arithmetic error. Many candidates were given an accuracy penalty for not rounding their answer to three significant figures. Few candidates were able to answer (b) correctly with most believing incorrectly that the step length was a relevant factor.

b.

## Question

A particle moves in a straight line with velocity v metres per second. At any time t seconds, $$0 \leqslant t < \frac{{3\pi }}{4}$$, the velocity is given by the differential equation $$\frac{{{\text{d}}v}}{{{\text{d}}t}} + {v^2} + 1 = 0$$  .

It is also given that v = 1 when t = 0 .

Find an expression for v in terms of t .

[7]
a.

Sketch the graph of v against t , clearly showing the coordinates of any intercepts, and the equations of any asymptotes.

[3]
b.

(i)     Write down the time T at which the velocity is zero.

(ii)     Find the distance travelled in the interval [0, T] .

[3]
c.

Find an expression for s , the displacement, in terms of t , given that s = 0 when t = 0 .

[5]
d.

Hence, or otherwise, show that $$s = \frac{1}{2}\ln \frac{2}{{1 + {v^2}}}$$.

[4]
e.

## Markscheme

$$\frac{{{\text{d}}v}}{{{\text{d}}t}} = – {v^2} – 1$$

attempt to separate the variables     M1

$$\int {\frac{1}{{1 + {v^2}}}{\text{d}}v = \int { – 1{\text{d}}t} }$$     A1

$$\arctan v = – t + k$$     A1A1

Note: Do not penalize the lack of constant at this stage.

when t = 0, v = 1     M1

$$\Rightarrow k = \arctan 1 = \left( {\frac{\pi }{4}} \right) = (45^\circ )$$     A1

$$\Rightarrow v = \tan \left( {\frac{\pi }{4} – t} \right)$$     A1

[7 marks]

a.

A1A1A1

Note: Award A1 for general shape,

A1 for asymptote,

A1 for correct t and v intercept.

Note: Do not penalise if a larger domain is used.

[3 marks]

b.

(i)     $$T = \frac{\pi }{4}$$     A1

(ii)     area under curve $$= \int_0^{\frac{\pi }{4}} {\tan \left( {\frac{\pi }{4} – t} \right){\text{d}}t}$$     (M1)

$$= 0.347\left( { = \frac{1}{2}\ln 2} \right)$$     A1

[3 marks]

c.

$$v = \tan \left( {\frac{\pi }{4} – t} \right)$$

$$s = \int {\tan \left( {\frac{\pi }{4} – t} \right){\text{d}}t}$$     M1

$$\int {\frac{{\sin \left( {\frac{\pi }{4} – t} \right)}}{{\cos \left( {\frac{\pi }{4} – t} \right)}}} {\text{ d}}t$$     (M1)

$$= \ln \cos \left( {\frac{\pi }{4} – t} \right) + k$$     A1

when $$t = 0,{\text{ }}s = 0$$

$$k = – \ln \cos \frac{\pi }{4}$$     A1

$$s = \ln \cos \left( {\frac{\pi }{4} – t} \right) – \ln \cos \frac{\pi }{4}\left( { = \ln \left[ {\sqrt 2 \cos \left( {\frac{\pi }{4} – t} \right)} \right]} \right)$$     A1

[5 marks]

d.

METHOD 1

$$\frac{\pi }{4} – t = \arctan v$$     M1

$$t = \frac{\pi }{4} – \arctan v$$

$$s = \ln \left[ {\sqrt 2 \cos \left( {\frac{\pi }{4} – \frac{\pi }{4} + \arctan v} \right)} \right]$$

$$s = \ln \left[ {\sqrt 2 \cos (\arctan v)} \right]$$     M1A1

$$s = \ln \left[ {\sqrt 2 \cos \left( {\arccos \frac{1}{{\sqrt {1 + {v^2}} }}} \right)} \right]$$     A1

$$= \ln \frac{{\sqrt 2 }}{{\sqrt {1 + {v^2}} }}$$

$$= \frac{1}{2}\ln \frac{2}{{1 + {v^2}}}$$     AG

METHOD 2

$$s = \ln \cos \left( {\frac{\pi }{4} – t} \right) – \ln \cos \frac{\pi }{4}$$

$$= – \ln \sec \left( {\frac{\pi }{4} – t} \right) – \ln \cos \frac{\pi }{4}$$     M1

$$= – \ln \sqrt {1 + {{\tan }^2}\left( {\frac{\pi }{4} – t} \right)} – \ln \cos \frac{\pi }{4}$$     M1

$$= – \ln \sqrt {1 + {v^2}} – \ln \cos \frac{\pi }{4}$$     A1

$$= \ln \frac{1}{{\sqrt {1 + {v^2}} }} + \ln \sqrt 2$$     A1

$$= \frac{1}{2}\ln \frac{2}{{1 + {v^2}}}$$     AG

METHOD 3

$$v\frac{{dv}}{{ds}} = – {v^2} – 1$$     M1

$$\int {\frac{v}{{{v^2} + 1}}dv = – \int {1ds} }$$     M1

$$\frac{1}{2}\ln ({v^2} + 1) = – s + k$$     A1

when $$s = 0\,,{\text{ }}t = 0 \Rightarrow v = 1$$

$$\Rightarrow k = \frac{1}{2}\ln 2$$     A1

$$\Rightarrow s = \frac{1}{2}\ln \frac{2}{{1 + {v^2}}}$$     AG

[4 marks]

e.

## Examiners report

This proved to be the most challenging question in section B with only a very small number of candidates producing fully correct answers. Many candidates did not realise that part (a) was a differential equation that needed to be solved using a method of separating the variables. Without this, further progress with the question was difficult. For those who did succeed in part (a), parts (b) and (c) were relatively well done. For the minority of candidates who attempted parts (d) and (e) only the best recognised the correct methods.

a.

This proved to be the most challenging question in section B with only a very small number of candidates producing fully correct answers. Many candidates did not realise that part (a) was a differential equation that needed to be solved using a method of separating the variables. Without this, further progress with the question was difficult. For those who did succeed in part (a), parts (b) and (c) were relatively well done. For the minority of candidates who attempted parts (d) and (e) only the best recognised the correct methods.

b.

This proved to be the most challenging question in section B with only a very small number of candidates producing fully correct answers. Many candidates did not realise that part (a) was a differential equation that needed to be solved using a method of separating the variables. Without this, further progress with the question was difficult. For those who did succeed in part (a), parts (b) and (c) were relatively well done. For the minority of candidates who attempted parts (d) and (e) only the best recognised the correct methods.

c.

This proved to be the most challenging question in section B with only a very small number of candidates producing fully correct answers. Many candidates did not realise that part (a) was a differential equation that needed to be solved using a method of separating the variables. Without this, further progress with the question was difficult. For those who did succeed in part (a), parts (b) and (c) were relatively well done. For the minority of candidates who attempted parts (d) and (e) only the best recognised the correct methods.

d.

This proved to be the most challenging question in section B with only a very small number of candidates producing fully correct answers. Many candidates did not realise that part (a) was a differential equation that needed to be solved using a method of separating the variables. Without this, further progress with the question was difficult. For those who did succeed in part (a), parts (b) and (c) were relatively well done. For the minority of candidates who attempted parts (d) and (e) only the best recognised the correct methods.

e.

## Question

Consider the differential equation $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{y^2}}}{{1 + x}}$$, where x > −1 and y = 1 when x = 0 .

Use Euler’s method, with a step length of 0.1, to find an approximate value of when x = 0.5.

[7]
a.

(i)     Show that $$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{{2{y^3} – {y^2}}}{{{{(1 + x)}^2}}}$$.

(ii)     Hence find the Maclaurin series for y, up to and including the term in $${x^2}$$ .

[8]
b.

(i)     Solve the differential equation.

(ii)     Find the value of a for which $$y \to \infty$$ as $$x \to a$$.

[6]
c.

## Markscheme

attempt the first step of

$${y_{n + 1}} = {y_n} + (0.1)f({x_n},\,{y_n})$$ with $${y_0} = 1,{\text{ }}{x_0} = 0$$     (M1)

$${y_1} = 1.1$$     A1

$${y_2} = 1.1 + (0.1)\frac{{{{1.1}^2}}}{{1.1}} = 1.21$$     (M1)A1

$${y_3} = 1.332(0)$$     (A1)

$${y_4} = 1.4685$$     (A1)

$${y_5} = 1.62$$     A1

[7 marks]

a.

(i)     recognition of both quotient rule and implicit differentiation     M1

$$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{{(1 + x)2y\frac{{{\text{d}}y}}{{{\text{d}}x}} – {y^2} \times 1}}{{{{(1 + x)}^2}}}$$     A1A1

Note: Award A1 for first term in numerator, A1 for everything else correct.

$$= \frac{{(1 + x)2y\frac{{{y^2}}}{{1 + x}} – {y^2} \times 1}}{{{{(1 + x)}^2}}}$$     M1A1

$$= \frac{{2{y^3} – {y^2}}}{{{{(1 + x)}^2}}}$$     AG

(ii)     attempt to use $$y = y(0) + x\frac{{{\text{d}}y}}{{{\text{d}}x}}(0) + \frac{{{x^2}}}{{2!}}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}(0) + …$$     (M1)

$$= 1 + x + \frac{{{x^2}}}{2}$$     A1A1

Note: Award A1 for correct evaluation of $$y(0),{\text{ }}\frac{{dy}}{{dx}}(0),{\text{ }}\frac{{{d^2}y}}{{d{x^2}}}(0)$$, A1 for correct series.

[8 marks]

b.

(i)     separating the variables $$\int {\frac{1}{{{y^2}}}{\text{d}}y = \int {\frac{1}{{1 + x}}{\text{d}}x} }$$     M1

obtain $$– \frac{1}{y} = \ln (1 + x) + (c)$$     A1

impose initial condition $$– 1 = \ln 1 + c$$     M1

obtain $$y = \frac{1}{{1 – \ln (1 + x)}}$$     A1

(ii)     $$y \to \infty$$ if $$\ln (1 + x) \to 1$$ , so a = e – 1     (M1)A1

Note: To award A1 must see either $$x \to e – 1$$ or a = e – 1 . Do not accept x = e – 1.

[6 marks]

c.

## Examiners report

Most candidates had a good knowledge of Euler’s method and were confident in applying it to the differential equation in part (a). A few candidates who knew the Euler’s method completed one iteration too many to arrive at an incorrect answer but this was rare. Nearly all candidates who applied the correct technique in part (a) correctly calculated the answer. Most candidates were able to attempt part (b) but some lost marks due to a lack of rigour by not clearly showing the implicit differentiation in the first line of working. Part (c) was reasonably well attempted by many candidates and many could solve the integrals although some did not find the arbitrary constant meaning that it was not possible to solve (ii) of the part (c).

a.

Most candidates had a good knowledge of Euler’s method and were confident in applying it to the differential equation in part (a). A few candidates who knew the Euler’s method completed one iteration too many to arrive at an incorrect answer but this was rare. Nearly all candidates who applied the correct technique in part (a) correctly calculated the answer. Most candidates were able to attempt part (b) but some lost marks due to a lack of rigour by not clearly showing the implicit differentiation in the first line of working. Part (c) was reasonably well attempted by many candidates and many could solve the integrals although some did not find the arbitrary constant meaning that it was not possible to solve (ii) of the part (c).

b.

Most candidates had a good knowledge of Euler’s method and were confident in applying it to the differential equation in part (a). A few candidates who knew the Euler’s method completed one iteration too many to arrive at an incorrect answer but this was rare. Nearly all candidates who applied the correct technique in part (a) correctly calculated the answer. Most candidates were able to attempt part (b) but some lost marks due to a lack of rigour by not clearly showing the implicit differentiation in the first line of working. Part (c) was reasonably well attempted by many candidates and many could solve the integrals although some did not find the arbitrary constant meaning that it was not possible to solve (ii) of the part (c).

c.

## Question

Find the general solution of the differential equation $$t\frac{{{\text{d}}y}}{{{\text{d}}t}} = \cos t – 2y$$ , for t > 0 .

## Markscheme

recognise equation as first order linear and attempt to find the IF     M1

$${\text{IF}} = {{\text{e}}^{\int {\frac{2}{t}{\text{d}}t} }} = {t^2}$$     A1

solution $$y{t^2} = \int {t\cos t{\text{d}}t}$$     M1A1

using integration by parts with the correct choice of u and v     (M1)

$$\int {t\cos t{\text{d}}t = t\sin t + \cos t( + C)}$$     A1

obtain $$y = \frac{{\sin t}}{t} + \frac{{\cos t + C}}{{{t^2}}}$$     A1

[7 marks]

## Examiners report

Perhaps a small number of candidates were put off by the unusual choice of variables but in most instances it seemed that candidates who recognised the need for an integration factor could make a good attempt at this problem. Candidates who were not able to simplify the integrating factor from $${e^{2\ln t}}$$ to $${t^2}$$ rarely gained full marks. A significant number of candidates did not gain the final mark due to a lack of an arbitrary constant or not dividing the constant by the integration factor.

## Question

Consider the differential equation $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = f(x,{\text{ }}y)$$ where $$f(x,{\text{ }}y) = y – 2x$$.

Sketch, on one diagram, the four isoclines corresponding to $$f(x,{\text{ }}y) = k$$ where $$k$$ takes the values $$-1$$, $$-0.5$$, $$0$$ and $$1$$. Indicate clearly where each isocline crosses the $$y$$ axis.

[2]
a.

A curve, $$C$$, passes through the point $$(0,1)$$ and satisfies the differential equation above.

Sketch $$C$$ on your diagram.

[3]
b.

A curve, $$C$$, passes through the point $$(0,1)$$ and satisfies the differential equation above.

State a particular relationship between the isocline $$f(x,{\text{ }}y) = – 0.5$$ and the curve $$C$$, at their point of intersection.

[1]
c.

A curve, $$C$$, passes through the point $$(0,1)$$ and satisfies the differential equation above.

Use Euler’s method with a step interval of $$0.1$$ to find an approximate value for $$y$$ on $$C$$, when $$x = 0{\text{.}}5$$.

[4]
d.

## Markscheme

A1 for 4 parallel straight lines with a positive gradient     A1

A1 for correct $$y$$ intercepts     A1

[2 marks]

a.

A1 for passing through $$(0,1)$$ with positive gradient less than $$2$$

A1 for stationary point on $$y = 2x$$

A1 for negative gradient on both of the other $$2$$ isoclines     A1A1A1

[3 marks]

b.

The isocline is perpendicular to $$C$$     R1

[1 mark]

c.

$${y_{n + 1}} = {y_n} + 0.1({y_n} – 2{x_n})\;\;\;( = 1.1{y_n} – 0.2{x_n})$$     (M1)(A1)

Note:     Also award M1A1 if no formula seen but $${y_2}$$ is correct.

$${y_0} = 1,{\text{ }}{y_1} = 1.1,{\text{ }}{y_2} = 1.19,{\text{ }}{y_3} = 1.269,{\text{ }}{y_4} = 1.3359$$     (M1)

$${y_5} = 1.39{\text{ to 3sf}}$$     A1

Note:     M1 is for repeated use of their formula, with steps of $$0.1$$.

Note:     Accept $$1.39$$ or $$1.4$$ only.

[4 marks]

Total [10 marks]

d.

## Examiners report

Some candidates ignored the instruction to prove from first principles and instead used standard differentiation. Some candidates also only found a derivative from one side.

a.

Parts (b) and (c) were attempted by very few candidates. Few recognized that the gradient of the curve had to equal the value of $$k$$ on the isocline.

b.

Parts (b) and (c) were attempted by very few candidates. Few recognized that the gradient of the curve had to equal the value of $$k$$ on the isocline.

c.

Those candidates who knew the method managed to score well on this part. On most calculators a short program can be written in the exam to make Euler’s method very quick. Quite a few candidates were losing time by calculating and writing out many intermediate values, rather than just the $$x$$ and$$y$$ values.

d.

## Question

The curves $$y = f(x)$$ and $$y = g(x)$$ both pass through the point $$(1,{\text{ }}0)$$ and are defined by the differential equations $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = x – {y^2}$$ and $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = y – {x^2}$$ respectively.

Show that the tangent to the curve $$y = f(x)$$ at the point $$(1,{\text{ }}0)$$ is normal to the curve $$y = g(x)$$ at the point $$(1,{\text{ }}0)$$.

[2]
a.

Find $$g(x)$$.

[6]
b.

Use Euler’s method with steps of $$0.2$$ to estimate $$f(2)$$ to $$5$$ decimal places.

[5]
c.

Explain why $$y = f(x)$$ cannot cross the isocline $$x – {y^2} = 0$$, for $$x > 1$$.

[3]
d.

(i)     Sketch the isoclines $$x – {y^2} = – 2,{\text{ }}0,{\text{ }}1$$.

(ii)     On the same set of axes, sketch the graph of $$f$$.

[4]
e.

## Markscheme

gradient of $$f$$ at $$(1,{\text{ }}0)$$ is $$1 – {0^2} = 1$$ and the gradient of $$g$$ at $$(1,{\text{ }}0)$$ is $$0 – {1^2} = – 1$$     A1

so gradient of normal is $$1$$     A1

= Gradient of the tangent of $$f$$ at $$(1,{\text{ }}0)$$     AG

[2 marks]

a.

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} – y = – {x^2}$$

integrating factor is $${{\text{e}}^{\int { – 1{\text{d}}x} }} = {{\text{e}}^{ – x}}$$     M1

$$y{{\text{e}}^{ – x}} = \int { – {x^2}{{\text{e}}^{ – x}}{\text{d}}x}$$     A1

$$= {x^2}{{\text{e}}^{ – x}} – \int {2x{{\text{e}}^{ – x}}{\text{d}}x}$$     M1

$$= {x^2}{{\text{e}}^{ – x}} + 2x{{\text{e}}^{ – x}} – \int {2{{\text{e}}^{ – x}}{\text{d}}x}$$

$$= {x^2}{{\text{e}}^{ – x}} + 2x{{\text{e}}^{ – x}} + 2{{\text{e}}^{ – x}} + c$$     A1

Note:     Condone missing $$+ c$$ at this stage.

$$\Rightarrow g(x) = {x^2} + 2x + 2 + c{{\text{e}}^x}$$

$$g(1) = 0 \Rightarrow c = – \frac{5}{{\text{e}}}$$     M1

$$\Rightarrow g(x) = {x^2} + 2x + 2 – 5{{\text{e}}^{x – 1}}$$     A1

[6 marks]

b.

use of $${y_{n + 1}} = {y_n} + hf'({x_n},{\text{ }}{y_n})$$     (M1)

$${x_0} = 1,{\text{ }}{y_0} = 0$$

$${x_1} = 1.2,{\text{ }}{y_1} = 0.2$$     A1

$${x_2} = 1.4,{\text{ }}{y_2} = 0.432$$     (M1)(A1)

$${x_3} = 1.6,{\text{ }}{y_3} = 0.67467 \ldots$$

$${x_4} = 1.8,{\text{ }}{y_4} = 0.90363 \ldots$$

$${x_5} = 2,{\text{ }}{y_5} = 1.1003255 \ldots$$

answer $$= 1.10033$$     A1     N3

Note:     Award A0 or N1 if $$1.10$$ given as answer.

[5 marks]

c.

at the point $$(1,{\text{ }}0)$$, the gradient of $$f$$ is positive so the graph of $$f$$ passes into the first quadrant for $$x > 1$$

in the first quadrant below the curve $$x – {y^2} = 0$$ the gradient of $$f$$ is positive     R1

the curve $$x – {y^2} = 0$$ has positive gradient in the first quadrant     R1

if $$f$$ were to reach $$x – {y^2} = 0$$ it would have gradient of zero, and therefore would not cross     R1

[3 marks]

d.

(i) and (ii)

A4

Note:     Award A1 for 3 correct isoclines.

Award A1 for $$f$$ not reaching $$x – {y^2} = 0$$.

Award A1 for turning point of $$f$$ on $$x – {y^2} = 0$$.

Award A1 for negative gradient to the left of the turning point.

Note:     Award A1 for correct shape and position if curve drawn without any isoclines.

[4 marks]

Total [20 marks]

e.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

## Question

Consider the differential equation $$x\frac{{{\text{d}}y}}{{{\text{d}}x}} – y = {x^p} + 1$$ where $$x \in \mathbb{R},\,x \ne 0$$ and $$p$$ is a positive integer, $$p > 1$$.

Solve the differential equation given that $$y = – 1$$ when $$x = 1$$. Give your answer in the form $$y = f\left( x \right)$$.

[8]
a.

Show that the $$x$$-coordinate(s) of the points on the curve $$y = f\left( x \right)$$ where $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$$ satisfy the equation $${x^{p – 1}} = \frac{1}{p}$$.

[2]
b.i.

Deduce the set of values for $$p$$ such that there are two points on the curve $$y = f\left( x \right)$$ where $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$$. Give a reason for your answer.

[2]
b.ii.

## Markscheme

METHOD 1

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} = {x^{p – 1}} + \frac{1}{x}$$    (M1)

integrating factor $$= {{\text{e}}^{\int { – \frac{1}{x}{\text{d}}x} }}$$     M1

$${\text{ = }}{{\text{e}}^{ – {\text{ln}}\,x}}$$     (A1)

= $$\frac{1}{x}$$     A1

$$\frac{1}{x}\frac{{{\text{d}}y}}{{{\text{d}}x}} – \frac{y}{{{x^2}}} = {x^{p – 2}} + \frac{1}{{{x^2}}}$$     (M1)

$$\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{y}{x}} \right) = {x^{p – 2}} + \frac{1}{{{x^2}}}$$

$$\frac{y}{x} = \frac{1}{{p – 1}}{x^{p – 1}} – \frac{1}{x} + C$$    A1

Note: Condone the absence of C.

$$y = \frac{1}{{p – 1}}{x^p} + Cx – 1$$

substituting $$x = 1$$, $$y = – 1 \Rightarrow C = – \frac{1}{{p – 1}}$$    M1

Note: Award M1 for attempting to find their value of C.

$$y = \frac{1}{{p – 1}}\left( {{x^p} – x} \right) – 1$$      A1

[8 marks]

METHOD 2

put $$y = vx$$ so that $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}$$    M1(A1)

substituting,       M1

$$x\left( {v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}} \right) – vx = {x^p} + 1$$     (A1)

$$x\frac{{{\text{d}}v}}{{{\text{d}}x}} = {x^{p – 1}} + \frac{1}{x}$$      M1

$$\frac{{{\text{d}}v}}{{{\text{d}}x}} = {x^{p – 2}} + \frac{1}{{{x^2}}}$$

$$v = \frac{1}{{p – 1}}{x^{p – 1}} – \frac{1}{x} + C$$     A1

Note: Condone the absence of C.

$$y = \frac{1}{{p – 1}}{x^p} + Cx – 1$$

substituting $$x = 1$$, $$y = – 1 \Rightarrow C = – \frac{1}{{p – 1}}$$    M1

Note: Award M1 for attempting to find their value of C.

$$y = \frac{1}{{p – 1}}\left( {{x^p} – x} \right) – 1$$      A1

[8 marks]

a.

METHOD 1

find $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$ and solve $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$$ for $$x$$

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{p – 1}}\left( {p{x^{p – 1}} – 1} \right)$$     M1

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow p{x^{p – 1}} – 1 = 0$$     A1

$$p{x^{p – 1}} = 1$$

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

$${x^{p – 1}} = \frac{1}{p}$$     AG

METHOD 2

substitute $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$$ and their $$y$$ into the differential equation and solve for $$x$$

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow – \left( {\frac{{{x^p} – x}}{{p – 1}}} \right) + 1 = {x^p} + 1$$     M1

$${x^p} – x = {x^p} – p{x^p}$$     A1

$$p{x^{p – 1}} = 1$$

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

$${x^{p – 1}} = \frac{1}{p}$$     AG

[2 marks]

b.i.

there are two solutions for $$x$$ when $$p$$ is odd (and $$p > 1$$     A1

if $$p – 1$$ is even there are two solutions (to $${x^{p – 1}} = \frac{1}{p}$$)

and if $$p – 1$$ is odd there is only one solution (to $${x^{p – 1}} = \frac{1}{p}$$)   R1

Note: Only award the R1 if both cases are considered.

[4 marks]

b.ii.

[N/A]

a.

[N/A]

b.i.

[N/A]

b.ii.