# IB DP Maths Topic 9.5 First-order differential equations HL Paper 3

## Question

(a)     Show that the solution of the differential equation

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

given that $$y = \frac{\pi }{4}{\text{ when }}x = \pi {\text{, is }}y = \arctan (1 + \sin x){\text{.}}$$

(b)     Determine the value of the constant a for which the following limit exists

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\arctan (1 + \sin x) – a}}{{{{\left( {x – \frac{\pi }{2}} \right)}^2}}}$

and evaluate that limit.

## Markscheme

(a)     this separable equation has general solution

$$\int {{{\sec }^2}y{\text{d}}y = \int {\cos x{\text{d}}x} }$$     (M1)(A1)

$$\tan y = \sin x + c$$     A1

the condition gives

$$\tan \frac{\pi }{4} = \sin \pi + c \Rightarrow c = 1$$     M1

the solution is $$\tan y = 1 + \sin x$$     A1

$$y = \arctan (1 + \sin x)$$     AG

[5 marks]

(b)     the limit cannot exist unless $$a = \arctan \left( {1 + \sin \frac{\pi }{2}} \right) = \arctan 2$$     R1A1

in that case the limit can be evaluated using l’Hopital’s rule (twice) limit is

$$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\left( {\arctan (1 + \sin x)} \right)}^\prime }}}{{2\left( {x – \frac{\pi }{2}} \right)}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{y’}}{{2\left( {x – \frac{\pi }{2}} \right)}}$$     M1A1

where y is the solution of the differential equation

the numerator has zero limit (from the factor $$\cos x$$ in the differential equation)     R1

so required limit is

$$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{y”}}{2}$$     M1A1

finally,

$$y” = – \sin x{\cos ^2}y – 2\cos x\cos y\sin y \times y'(x)$$     M1A1

since $$\cos y\left( {\frac{\pi }{2}} \right) = \frac{1}{{\sqrt 5 }}$$     A1

$$y” = – \frac{1}{5}{\text{ at }}x = \frac{\pi }{2}$$     A1

the required limit is $$– \frac{1}{{10}}$$     A1

[12 marks]

Total [17 marks]

## Examiners report

Many candidates successfully obtained the displayed solution of the differential equation in part(a). Few complete solutions to part(b) were seen which used the result in part(a). The problem can, however, be solved by direct differentiation although this is algebraically more complicated. Some successful solutions using this method were seen.

## Question

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

(a)     Use Euler’s method with a step length of 0.25 to find an estimate for the value of y when x = 2 .

(b)     (i)     Solve the differential equation giving your answer in the form $$y = f(x)$$ .

(ii)     Find the value of y when x = 2 .

## Markscheme

(a)     Using an increment of 0.25 in the x-values     A1

Note: The A1 marks are awarded for final column.

$$\Rightarrow y(2) \approx – 0.304$$     A1

[7 marks]

(b)     (i)     let y = vx     M1

$$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}$$     (A1)

$$\Rightarrow v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{{v^2}{x^2} + {x^2}}}{{2{x^2}}}$$     (M1)

$$\Rightarrow x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{1 – 2v + {v^2}}}{2}$$     (A1)

$$\Rightarrow x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{{{(1 – v)}^2}}}{2}$$     A1

$$\Rightarrow \int {\frac{2}{{{{(1 – v)}^2}}}{\text{d}}v = \int {\frac{1}{x}{\text{d}}x} }$$     M1

$$\Rightarrow 2{(1 – v)^{ – 1}} = \ln x + c$$     A1A1

$$\Rightarrow \frac{2}{{1 – \frac{y}{x}}} = \ln x + c$$

when $$x = 1,{\text{ }}y = – 1 \Rightarrow c = 1$$     M1A1

$$\Rightarrow \frac{{2x}}{{x – y}} = \ln x + 1$$

$$\Rightarrow y = x – \frac{{2x}}{{1 + \ln x}}{\text{ }}\left( { = \frac{{x\ln x – x}}{{1 + \ln x}}} \right)$$     M1A1

(ii)     when $$x = 2,{\text{ }}y = – 0.362\,\,\,\,\,\left( {{\text{accept 2}} – \frac{4}{{1 + \ln 2}}} \right)$$     A1

[13 marks]

Total [20 marks]

## Examiners report

Part (a) was well done by many candidates, but a number were penalised for not using a sufficient number of significant figures. Part (b) was started by the majority of candidates, but only the better candidates were able to reach the end. Many were unable to complete the question correctly because they did not know what to do with the substitution y = vx and because of arithmetic errors and algebraic errors.

## Question

Given that $$\frac{{{\text{d}}y}}{{{\text{d}}x}} – 2{y^2} = {{\text{e}}^x}$$ and y = 1 when x = 0, use Euler’s method with a step length of 0.1 to find an approximation for the value of y when x = 0.4. Give all intermediate values with maximum possible accuracy.

## Markscheme

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^x} + 2{y^2}$$     (A1)

required approximation = 3.85     A1

[8 marks]

## Examiners report

Most candidates seemed familiar with Euler’s method. The most common way of losing marks was either to round intermediate answers to insufficient accuracy despite the advice in the question or simply to make an arithmetic error. Many candidates were given an accuracy penalty for not rounding their answer to three significant figures.

## 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

The real and imaginary parts of a complex number $$x + {\text{i}}y$$ are related by the differential equation $$(x + y)\frac{{{\text{d}}y}}{{{\text{d}}x}} + (x – y) = 0$$.

By solving the differential equation, given that $$y = \sqrt 3$$ when x =1, show that the relationship between the modulus r and the argument $$\theta$$ of the complex number is $$r = 2{{\text{e}}^{\frac{\pi }{3} – \theta }}$$.

## Markscheme

$$(x + y)\frac{{{\text{d}}y}}{{{\text{d}}x}} + (x – y) = 0$$

$$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{y – x}}{{x + y}}$$

let $$y = vx$$     M1

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}$$     A1

$$v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{vx – x}}{{x + vx}}$$     (A1)

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

$$\int {\frac{{v + 1}}{{1 + {v^2}}}{\text{d}}v = – \int {\frac{1}{x}{\text{d}}x} }$$     M1

$$\int {\frac{v}{{1 + {v^2}}}{\text{d}}v + \int {\frac{1}{{1 + {v^2}}}{\text{d}}v = – \int {\frac{1}{x}{\text{d}}x} } }$$     M1

$$\Rightarrow \frac{1}{2}\ln \left| {1 + {v^2}} \right| + \arctan v = – \ln \left| x \right| + k$$     A1A1

Notes: Award A1 for $$\frac{1}{2}\ln \left| {1 + {v^2}} \right|$$, A1 for the other two terms.

Do not penalize missing k or missing modulus signs at this stage.

$$\Rightarrow \frac{1}{2}\ln \left| {1 + \frac{{{y^2}}}{{{x^2}}}} \right| + \arctan \frac{y}{x} = – \ln \left| x \right| + k$$     M1

$$\Rightarrow \frac{1}{2}\ln 4 + \arctan \sqrt 3 = – \ln 1 + k$$     (M1)

$$\Rightarrow k = \ln 2 + \frac{\pi }{3}$$     A1

$$\Rightarrow \frac{1}{2}\ln \left| {1 + \frac{{{y^2}}}{{{x^2}}}} \right| + \arctan \frac{y}{x} = – \ln \left| x \right| + \ln 2 + \frac{\pi }{3}$$

attempt to combine logarithms     M1

$$\Rightarrow \frac{1}{2}\ln \left| {\frac{{{y^2} + {x^2}}}{{{x^2}}}} \right| + \frac{1}{2}\ln \left| {{x^2}} \right| = \ln 2 + \frac{\pi }{3} – \arctan \frac{y}{x}$$

$$\Rightarrow \frac{1}{2}\ln \left| {{y^2} + {x^2}} \right| = \ln 2 + \frac{\pi }{3} – \arctan \frac{y}{x}$$     (A1)

$$\Rightarrow \sqrt {{y^2} + {x^2}} = {{\text{e}}^{\ln 2 + \frac{\pi }{3}\arctan \frac{y}{x}}}$$     (A1)

$$\Rightarrow \sqrt {{y^2} + {x^2}} = {{\text{e}}^{\ln 2}} \times {{\text{e}}^{\frac{\pi }{3} – \arctan \frac{y}{x}}}$$     A1

$$\Rightarrow r = 2{{\text{e}}^{\frac{\pi }{3} – \theta }}$$     AG

[15 marks]

## Examiners report

Most candidates realised that this was a homogeneous differential equation and that the substitution $$y = vx$$ was the way forward. Many of these candidates reached as far as separating the variables correctly but integrating $$\frac{{v + 1}}{{{v^2} + 1}}$$ proved to be too difficult for many candidates – most failed to realise that the expression had to be split into two separate integrals. Some candidates successfully evaluated the arbitrary constant but the combination of logs and the subsequent algebra necessary to obtain the final result proved to be beyond the majority of candidates.

## 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

A differential equation is given by $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x}$$ , where x > 0 and y > 0.

Solve this differential equation by separating the variables, giving your answer in the form y = f (x) .

[3]
a.

Solve the same differential equation by using the standard homogeneous substitution y = vx .

[4]
b.

Solve the same differential equation by the use of an integrating factor.

[5]
c.

If y = 20 when x = 2 , find y when x = 5 .

[1]
d.

## Markscheme

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} \Rightarrow \int {\frac{1}{y}{\text{d}}y = \int {\frac{1}{x}{\text{d}}x} }$$     M1

$$\Rightarrow \ln y = \ln x + c$$     A1

$$\Rightarrow \ln y = \ln x + \ln k = \ln kx$$

$$\Rightarrow y = kx$$     A1

[3 marks]

a.

$$y = vx \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}$$     (A1)

so $$v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = v$$     M1

$$\Rightarrow x\frac{{{\text{d}}v}}{{{\text{d}}x}} = 0 \Rightarrow \frac{{{\text{d}}v}}{{{\text{d}}x}} = 0\,\,\,\,\,({\text{as }}x \ne 0)$$     R1

$$\Rightarrow v = k$$

$$\Rightarrow \frac{y}{x} = k\,\,\,\,\,( \Rightarrow y = kx)$$     A1

[4 marks]

b.

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

$${\text{IF}} = {{\text{e}}^{\int {\frac{{ – 1}}{x}{\text{d}}x} }} = {{\text{e}}^{ – \ln x}} = \frac{1}{x}$$     M1A1

$${x^{ – 1}}\frac{{{\text{d}}y}}{{{\text{d}}x}} – {x^{ – 2}}y = 0$$

$$\Rightarrow \frac{{{\text{d}}[{x^{ – 1}}y]}}{{{\text{d}}x}} = 0$$     (M1)

$$\Rightarrow {x^{ – 1}}y = k\,\,\,\,\,( \Rightarrow y = kx)$$     A1

[5 marks]

c.

$$20 = 2k \Rightarrow k = 10{\text{ so }}y(5) = 10 \times 5 = 50$$     A1

[1 mark]

d.

## Examiners report

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from $$\ln y = \ln x + c$$ to $$y = x + c$$

a.

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from $$\ln y = \ln x + c$$ to $$y = x + c$$

b.

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from $$\ln y = \ln x + c$$ to $$y = x + c$$

c.

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from $$\ln y = \ln x + c$$ to $$y = x + c$$

d.

## Question

Let the differential equation $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \sqrt {x + y} ,{\text{ }}(x + y \geqslant 0)$$ satisfying the initial conditions y = 1 when x = 1. Also let y = c when x = 2 .

Use Euler’s method to find an approximation for the value of c , using a step length of h = 0.1 . Give your answer to four decimal places.

[6]
a.

You are told that if Euler’s method is used with = 0.05 then $$c \simeq 2.7921$$ , if it is used with = 0.01 then $$c \simeq 2.8099$$ and if it is used with = 0.005 then $$c \simeq 2.8121$$.

Plot on graph paper, with h on the horizontal axis and the approximation for c on the vertical axis, the four points (one of which you have calculated and three of which have been given). Use a scale of 1 cm = 0.01 on both axes. Take the horizontal axis from 0 to 0.12 and the vertical axis from 2.76 to 2.82.

[3]
b.

Draw, by eye, the straight line that best fits these four points, using a ruler.

[1]
c.

Use your graph to give the best possible estimate for c , giving your answer to three decimal places.

[2]
d.

## Markscheme

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

$${x_n} = 1 + 0.1n,{\text{ }}{y_{n + 1}} = {y_n} + 0.1\sqrt {{x_n} + {y_n}}$$     (M1)(M1)(A1)

Note: If they have not written down formulae but have $${x_1} = 1.1$$ and $${y_1} = 1.14142…$$ award M1M1A1.

gives by GDC $${x_{10}} = 2,{\text{ }}{y_{10}} = 2.770114792…$$     (M1)(A1)

so $$a \simeq 2.7701{\text{ (4dp)}}$$     A1     N6

Note: Do not penalize over-accuracy.

[6 marks]

a.

points drawn on graph above     A1A1A1

Note: Award A1 for scales, A1 for 2 points correctly plotted, A1 for other 2 points correctly plotted (second and third A1 dependent on the first being correct).

[3 marks]

b.

suitable line of best fit placed on graph     A1

[1 mark]

c.

letting $${\text{h}} \to {\text{0}}$$ we approach the y intercept on the graph so     (R1)

$$c \simeq 2.814{\text{ (3dp)}}$$     A1

Note: Accept 2.815.

[2 marks]

d.

## Examiners report

Part (a) was done well. We would recommend that candidates write down the equation they are using, in this case, $${y_{n + 1}} = {y_n} + 0.1\sqrt {{x_n} + {y_n}}$$ , to ensure they get all the method marks. Beyond this the answer is all that is needed (or if a student wishes to show working, simply each of the values of $${{x_n}}$$ and $${{y_n}}$$) . Many candidates wasted a lot of time by writing out values of each part of the function, perhaps indicating they did not how to do it more quickly using their calculators.

Part (b) Surprisingly when drawing the graph a lot of candidates had (0.01, 2.8099) closer to 2.80 than 2.81

Most realised that the best possible estimate was given by the y-intercept of the line they had drawn.

a.

Part (a) was done well. We would recommend that candidates write down the equation they are using, in this case, $${y_{n + 1}} = {y_n} + 0.1\sqrt {{x_n} + {y_n}}$$ , to ensure they get all the method marks. Beyond this the answer is all that is needed (or if a student wishes to show working, simply each of the values of $${{x_n}}$$ and $${{y_n}}$$) . Many candidates wasted a lot of time by writing out values of each part of the function, perhaps indicating they did not how to do it more quickly using their calculators.

Part (b) Surprisingly when drawing the graph a lot of candidates had (0.01, 2.8099) closer to 2.80 than 2.81

Most realised that the best possible estimate was given by the y-intercept of the line they had drawn.

b.

Part (a) was done well. We would recommend that candidates write down the equation they are using, in this case, $${y_{n + 1}} = {y_n} + 0.1\sqrt {{x_n} + {y_n}}$$ , to ensure they get all the method marks. Beyond this the answer is all that is needed (or if a student wishes to show working, simply each of the values of $${{x_n}}$$ and $${{y_n}}$$) . Many candidates wasted a lot of time by writing out values of each part of the function, perhaps indicating they did not how to do it more quickly using their calculators.

Part (b) Surprisingly when drawing the graph a lot of candidates had (0.01, 2.8099) closer to 2.80 than 2.81

Most realised that the best possible estimate was given by the y-intercept of the line they had drawn.

c.

Part (a) was done well. We would recommend that candidates write down the equation they are using, in this case, $${y_{n + 1}} = {y_n} + 0.1\sqrt {{x_n} + {y_n}}$$ , to ensure they get all the method marks. Beyond this the answer is all that is needed (or if a student wishes to show working, simply each of the values of $${{x_n}}$$ and $${{y_n}}$$) . Many candidates wasted a lot of time by writing out values of each part of the function, perhaps indicating they did not how to do it more quickly using their calculators.

Part (b) Surprisingly when drawing the graph a lot of candidates had (0.01, 2.8099) closer to 2.80 than 2.81

Most realised that the best possible estimate was given by the y-intercept of the line they had drawn.

d.

## Question

Consider the differential equation $$\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\tan x = {\cos ^2}x$$, given that y = 2 when x = 0.

Use Euler’s method with a step length of 0.1 to find an approximation to the value of y when x = 0.3.

[5]
a.

(i)     Show that the integrating factor for solving the differential equation is $$\sec x$$.

(ii)     Hence solve the differential equation, giving your answer in the form $$y = f(x)$$.

[10]
b.

## Markscheme

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

Note: Award A1 for $$y(0.1)$$ and A1 for $$y(0.2)$$

$$y(0.3) = 2.23$$     A2

[5 marks]

a.

(i)     $${\text{IF}} = {{\text{e}}^{\left( {\int {\tan x{\text{d}}x} } \right)}}$$     (M1)

$${\text{IF}} = {{\text{e}}^{\left( {\int {\frac{{\sin x}}{{\cos x}}{\text{d}}x} } \right)}}$$     (M1)

Note: Only one of the two (M1) above may be implied.

$$= {{\text{e}}^{( – \ln \cos x)}}{\text{ (or }}{{\text{e}}^{(\ln \sec x)}})$$     A1

$$= \sec x$$     AG

(ii)     multiplying by the IF     (M1)

$$\sec x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\sec x\tan x = \cos x$$     (A1)

$$\frac{{\text{d}}}{{{\text{d}}x}}(y\sec x) = \cos x$$     (A1)

$$y\sec x = \sin x + c$$     A1A1

putting $$x = 0,{\text{ }}y = 2 \Rightarrow c = 2$$

$$y = \cos x(\sin x + 2)$$     A1

[10 marks]

b.

## Examiners report

Most candidates knew Euler’s method and were able to apply it to the differential equation to answer part (a). Some candidates who knew Euler’s method completed one iteration too many to arrive at an incorrect answer. Surprisingly few candidates were able to efficiently use their GDCs to answer this question and this led to many final answers that were incorrect due to rounding errors.

a.

Most candidates were able to correctly derive the Integration Factor in part (b) but some lost marks due to not showing all the steps that would be expected in a “show that” question. The differential equation was solved correctly by a significant number of candidates but there were errors when candidates multiplied by $$\sec x$$ before the inclusion of the arbitrary constant.

b.

## Question

Consider the differential equation $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{x + \sqrt {xy} }}$$, for $$x,{\text{ }}y > 0$$.

(a)     Use Euler’s method starting at the point $$(x,{\text{ }}y) = (1,{\text{ }}2)$$, with interval $$h = 0.2$$, to find an approximate value of y when $$x = 1.6$$.

(b)     Use the substitution $$y = vx$$ to show that $$x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{v}{{1 + \sqrt v }} – v$$.

(c)     (i)     Hence find the solution of the differential equation in the form $$f(x,{\text{ }}y) = 0$$, given that $$y = 2$$ when $$x = 1$$.

(ii)     Find the value of $$y$$ when $$x = 1.6$$.

## Markscheme

(a)     let $$f(x,{\text{ }}y) = \frac{y}{{x + \sqrt {xy} }}$$

$$y(1.2) = y(1) + 0.2f(1,{\text{ }}2){\text{ }}( = 2 + 0.1656 \ldots )$$     (M2)(A1)

$$= 2.1656 \ldots$$     A1

$$y(1.4) = 2.1656 \ldots + 0.2f(1.2,{\text{ }}2.1256 \ldots ){\text{ }}( = 2.1656 \ldots + 0.1540 \ldots )$$     (M1)

Note: M1 is for attempt to apply formula using point $$\left( {1.2,{\text{ }}y(1.2)} \right)$$.

$$= 2.3197 \ldots$$     A1

$$y(1.6) = 2.3197 \ldots + 0.2f(1.4,{\text{ }}2.3197 \ldots ){\text{ }}( = 2.3297 \ldots + 0.1448 \ldots )$$

$$= 2.46$$   (3sf)     A1     N3

[7 marks]

(b)     $$y = vx \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}$$     (M1)

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{x + \sqrt {xy} }} \Rightarrow v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{vx}}{{x + \sqrt {v{x^2}} }}$$     M1

$$\Rightarrow v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{vx}}{{x + x\sqrt v }}{\text{ (as }}x > 0)$$     A1

$$\Rightarrow x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{v}{{1 + \sqrt v }} – v$$     AG

[3 marks]

(c)     (i)     $$x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{v}{{1 + \sqrt v }} – v$$

$$x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{ – v\sqrt v }}{{1 + \sqrt v }} \Rightarrow \frac{{1 + \sqrt v }}{{ – v\sqrt v }}{\text{d}}v = \frac{1}{x}{\text{d}}x$$     M1

$$\int {\frac{{1 + \sqrt v }}{{ – v\sqrt v }}{\text{d}}v = \int {\frac{1}{x}{\text{d}}x} }$$     (M1)

$$\frac{2}{{\sqrt v }} – \ln v = \ln x + C$$     A1A1

Note: Do not penalize absence of $$+ C$$ at this stage; ignore use of absolute values on $$v$$and $$x$$ (which are positive anyway).

$$2\sqrt {\frac{x}{y}} – \ln \frac{y}{x} = \ln x + C$$ as $$y = vx \Rightarrow v = \frac{y}{x}$$     M1

$$y = 2$$ when $$x = 1 \Rightarrow \sqrt 2 – \ln 2 = 0 + C$$     M1

$$2\sqrt {\frac{x}{y}} – \ln \frac{y}{x} = \ln x + \sqrt 2 – \ln 2$$

$$2\sqrt {\frac{x}{y}} – \ln \frac{y}{x} – \ln x – \sqrt 2 + \ln 2 = 0$$   $$\left( {2\sqrt {\frac{x}{y}} – \ln y – \sqrt 2 + \ln 2 = 0} \right)$$     A1

(ii)     $$2\sqrt {\frac{{1.6}}{y}} – \ln \frac{y}{{1.6}} – \ln 1.6 – \sqrt 2 + \ln 2 = 0$$     (M1)

$$y = 2.45$$     A1

[9 marks]

## Examiners report

Part (a) was well answered by most candidates. In a few cases calculation errors and early rounding errors prevented candidates from achieving full marks, but most candidates scored at least a few marks here. In part (b) some candidates failed to convincingly show the given result. Part (c) proved to be a hard question for many candidates and a significant number of candidates had difficulty manipulating the algebraic expression, and either had the incorrect expression to integrate, or incorrectly integrated the correct expression. Many candidates reached as far as separating the variables correctly but integrating proved to be too difficult for many of them although most realised that the expression on v had to be split into two separate integrals. Most candidates made good attempts to evaluate the arbitrary constant and arrived at a correct or almost correct expression (sign errors were a common error) which allowed follow through for part b (ii). In some cases however the expression obtained was too simple or was omitted and it was not possible to grant follow through marks.

## Question

Consider the functions $$f(x) = {(\ln x)^2},{\text{ }}x > 1$$ and $$g(x) = \ln \left( {f(x)} \right),{\text{ }}x > 1$$.

(i)     Find $$f'(x)$$.

(ii)     Find $$g'(x)$$.

(iii)     Hence, show that $$g(x)$$ is increasing on $$\left] {1,{\text{ }}\infty } \right[$$.

[5]
a.

Consider the differential equation

$(\ln x)\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{2}{x}y = \frac{{2x – 1}}{{(\ln x)}},{\text{ }}x > 1.$

(i)     Find the general solution of the differential equation in the form $$y = h(x)$$.

(ii)     Show that the particular solution passing through the point with coordinates $$\left( {{\text{e, }}{{\text{e}}^2}} \right)$$ is given by $$y = \frac{{{x^2} – x + {\text{e}}}}{{{{(\ln x)}^2}}}$$.

(iii)     Sketch the graph of your solution for $$x > 1$$, clearly indicating any asymptotes and any maximum or minimum points.

[12]
b.

## Markscheme

(i)     attempt at chain rule     (M1)

$$f'(x) = \frac{{2\ln x}}{x}$$     A1

(ii)     attempt at chain rule     (M1)

$$g'(x) = \frac{2}{{x\ln x}}$$     A1

(iii)     $$g'(x)$$ is positive on $$\left] {1,{\text{ }}\infty } \right[$$     A1

so $$g(x)$$ is increasing on $$\left] {1,{\text{ }}\infty } \right[$$     AG

[5 marks]

a.

(i)     rearrange in standard form:

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{2}{{x\ln x}}y = \frac{{2x – 1}}{{{{(\ln x)}^2}}},{\text{ }}x > 1$$     (A1)

integrating factor:

$${{\text{e}}^{\int {\frac{2}{{x\ln x}}{\text{d}}x} }}$$     (M1)

$$= {{\text{e}}^{\ln \left( {{{(\ln x)}^2}} \right)}}$$

$$= {(\ln x)^2}$$     (A1)

multiply by integrating factor     (M1)

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

$$\frac{{\text{d}}}{{{\text{d}}x}}\left( {y{{(\ln x)}^2}} \right) = 2x – 1{\text{ }}\left( {{\text{or }}y{{(\ln x)}^2} = \int {2x – 1{\text{d}}x} } \right)$$     M1

attempt to integrate:     M1

$${(\ln x)^2}y = {x^2} – x + c$$

$$y = \frac{{{x^2} – x + c}}{{{{(\ln x)}^2}}}$$     A1

(ii)     attempt to use the point $$\left( {{\text{e, }}{{\text{e}}^2}} \right)$$ to determine c:     M1

eg, $${(\ln {\text{e}})^2}{{\text{e}}^2} = {{\text{e}}^2} – {\text{e}} + {\text{c}}$$ or $${{\text{e}}^2} = \frac{{{{\text{e}}^2} – {\text{e}} + {\text{c}}}}{{{{(\ln {\text{e}})}^2}}}$$ or $${{\text{e}}^2} = {{\text{e}}^2} – {\text{e}} + {\text{c}}$$

$${\text{c}} = {\text{e}}$$     A1

$$y = \frac{{{x^2} – x + {\text{e}}}}{{{{(\ln x)}^2}}}$$     AG

(iii)

graph with correct shape     A1

minimum at $$x = 3.1$$ (accept answers to a minimum of 2 s.f)     A1

asymptote shown at $$x = 1$$     A1

Note: y-coordinate of minimum not required for A1;

Equation of asymptote not required for A1 if VA appears on the sketch.

Award A0 for asymptotes if more than one asymptote are shown

[12 marks]

b.

[N/A]

a.

[N/A]

b.

## Question

Show that $$y = \frac{1}{x}\int {f(x){\text{d}}x}$$ is a solution of the differential equation

$$x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = f(x),{\text{ }}x > 0$$.

[3]
a.

Hence solve $$x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = {x^{ – \frac{1}{2}}},{\text{ }}x > 0$$, given that $$y = 2$$ when $$x = 4$$.

[5]
b.

## Markscheme

METHOD 1

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{1}{{{x^2}}}\int {f(x){\text{d}}x + \frac{1}{x}f(x)}$$     M1M1A1

$$x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = f(x),{\text{ }}x > 0$$     AG

Note:     M1 for use of product rule, M1 for use of the fundamental theorem of calculus, A1 for all correct.

METHOD 2

$$x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = f(x)$$

$$\frac{{{\text{d}}(xy)}}{{{\text{d}}x}} = f(x)$$     (M1)

$$xy = \int {f(x){\text{d}}x}$$     M1A1

$$y = \frac{1}{x}\int {f(x){\text{d}}x}$$     AG

[3 marks]

a.

$$y = \frac{1}{x}\left( {2{x^{\frac{1}{2}}} + c} \right)$$     A1A1

Note:     A1 for correct expression apart from the constant, A1 for including the constant in the correct position.

attempt to use the boundary condition     M1

$$c = 4$$     A1

$$y = \frac{1}{x}\left( {2{x^{\frac{1}{2}}} + 4} \right)$$     A1

Note:     Condone use of integrating factor.

[5 marks]

Total [8 marks]

b.

## Examiners report

This question allowed for several different approaches. The most common of these was the use of the integrating factor (even though that just took you in a circle). Other candidates substituted the solution into the differential equation and others multiplied the solution by $$x$$ and then used the product rule to obtain the differential equation. All these were acceptable.

a.

This was a straightforward question. Some candidates failed to use the hint of ‘hence’, and worked from the beginning using the integrating factor. A surprising number made basic algebra errors such as putting the $$+ c$$ term in the wrong place and so not dividing it by $$\chi$$.

b.

## 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 $$\frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {\frac{{2x}}{{1 + {x^2}}}} \right)y = {x^2}$$, given that $$y = 2$$ when $$x = 0$$.

Show that $$1 + {x^2}$$ is an integrating factor for this differential equation.

[5]
a.

Hence solve this differential equation. Give the answer in the form $$y = f(x)$$.

[6]
b.

## Markscheme

METHOD 1

attempting to find an integrating factor     (M1)

$$\int {\frac{{2x}}{{1 + {x^2}}}{\text{d}}x = \ln (1 + {x^2})}$$    (M1)A1

IF is $${{\text{e}}^{\ln (1 + {x^2})}}$$     (M1)A1

$$= 1 + {x^2}$$    AG

METHOD 2

multiply by the integrating factor

$$(1 + {x^2})\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2xy = {x^2}(1 + {x^2})$$    M1A1

left hand side is equal to the derivative of $$(1 + {x^2})y$$

A3

[5 marks]

a.

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

$$\frac{{\text{d}}}{{{\text{d}}x}}\left[ {(1 + {x^2})y} \right] = {x^2} + {x^4}$$

$$(1 + {x^2})y = \left( {\int {{x^2} + {x^4}{\text{d}}x = } } \right){\text{ }}\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5}( + c)$$    A1A1

$$y = \frac{1}{{1 + {x^2}}}\left( {\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} + c} \right)$$

$$x = 0,{\text{ }}y = 2 \Rightarrow c = 2$$    M1A1

$$y = \frac{1}{{1 + {x^2}}}\left( {\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} + 2} \right)$$    A1

[6 marks]

b.

[N/A]

a.

[N/A]

b.

## 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.

## Question

Consider the differential equation

$x\frac{{{\text{d}}y}}{{{\text{d}}x}} = y + \sqrt {{x^2} – {y^2}} ,{\text{ }}x > 0,{\text{ }}{x^2} > {y^2}.$

Show that this is a homogeneous differential equation.

[1]
a.

Find the general solution, giving your answer in the form $$y = f(x)$$ .

[7]
b.

## Markscheme

the equation can be rewritten as

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{y + \sqrt {{x^2} – {y^2}} }}{x} = \frac{y}{x} + \sqrt {1 – {{\left( {\frac{y}{x}} \right)}^2}}$$     A1

so the differential equation is homogeneous     AG

[1 mark]

a.

put y = vx so that $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}$$     M1A1

substituting,

$$v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = v + \sqrt {1 – {v^2}}$$     M1

$$\int {\frac{{{\text{d}}v}}{{\sqrt {1 – {v^2}} }} = \int {\frac{{{\text{d}}x}}{x}} }$$     M1

$$\arcsin v = \ln x + C$$     A1

$$\frac{y}{x} = \sin (\ln x + C)$$     A1

$$y = x\sin (\ln x + C)$$     A1

[7 marks]

b.

[N/A]

a.

[N/A]

b.

## Question

Consider the differential equation

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2{{\text{e}}^x} + y\tan x$$ , given that y = 1 when x = 0 .

The domain of the function y is $$\left[ {0,\frac{\pi }{2}} \right[$$.

By finding the values of successive derivatives when x = 0 , find the Maclaurin series for y as far as the term in $${x^3}$$ .

[6]
a.

(i)     Differentiate the function $${{\text{e}}^x}(\sin x + \cos x)$$ and hence show that

$\int {{{\text{e}}^x}\cos x{\text{d}}x = \frac{1}{2}{{\text{e}}^x}(\sin x + \cos x) + c} .$

(ii)     Find an integrating factor for the differential equation and hence find the solution in the form $$y = f(x)$$ .

[9]
b.

## Markscheme

we note that $$y(0) = 1$$ and $$y'(0) = 2$$     A1

$$y” = 2{{\text{e}}^x} + y’\tan x + y{\sec ^2}x$$     M1

$$y”(0) = 3$$     A1

$$y”’ = 2{{\text{e}}^x} + y”\tan x + 2y'{\sec ^2}x + 2y{\sec ^2}x\tan x$$     M1

$$y”'(0) = 6$$     A1

the maclaurin series solution is therefore

$$y = 1 + 2x + \frac{{3{x^2}}}{2} + {x^3} + \ldots$$     A1

[6 marks]

a.

(i)     $$\frac{{\text{d}}}{{{\text{d}}x}}\left( {{{\text{e}}^x}(\sin x + \cos x)} \right) = {{\text{e}}^x}(\sin x + \cos x) + {{\text{e}}^x}(\cos x – \sin x)$$     M1

$$= 2{{\text{e}}^x}\cos x$$     A1

it follows that

$$\int {{{\text{e}}^x}\cos x{\text{d}}x = \frac{1}{2}{{\text{e}}^x}(\sin x + \cos x) + c}$$     AG

(ii)     the differential equation can be written as

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} – y\tan x = 2{{\text{e}}^x}$$     M1

$${\text{IF}} = {{\text{e}}^{\int { – \tan x{\text{d}}x} }} = {{\text{e}}^{\ln \cos x}} = \cos x$$     M1A1

$$\cos x\frac{{{\text{d}}y}}{{{\text{d}}x}} – y\sin x = 2{{\text{e}}^x}\cos x$$     M1

integrating,

$$y\cos x = {{\text{e}}^x}(\sin x + \cos x) + C$$     A1

y = 1 when x = 0 gives C = 0     M1

therefore

$$y = {{\text{e}}^x}(1 + \tan x)$$     A1

[9 marks]

b.

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a.

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b.