IB DP Maths Topic 9.5 Solution of y′+P(x)y=Q(x), using the integrating factor HL Paper 3

 

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Question

Consider the differential equation

\[x\frac{{{\text{d}}y}}{{{\text{d}}x}} – 2y = \frac{{{x^3}}}{{{x^2} + 1}}.\]

(a)     Find an integrating factor for this differential equation.

(b)     Solve the differential equation given that \(y = 1\) when \(x = 1\) , giving your answer in the forms \(y = f(x)\) .

Answer/Explanation

Markscheme

(a)     Rewrite the equation in the form

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

Integrating factor \( = {{\text{e}}^{\int { – \frac{2}{x}{\text{d}}x} }}\)     M1

\( = {{\text{e}}^{ – 2\ln x}}\)     A1

\( = \frac{1}{{{x^2}}}\)     A1

Note: Accept \(\frac{1}{{{x^3}}}\) as applied to the original equation.

 

[5 marks]

 

(b)     Multiplying the equation,

\(\frac{1}{{{x^2}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} – \frac{2}{{{x^3}}}y = \frac{1}{{{x^2} + 1}}\)     (M1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{y}{{{x^2}}}} \right) = \frac{1}{{{x^2} + 1}}\)     (M1)(A1)

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

\( = \arctan x + C\)     A1

Substitute \(x = 1,{\text{ }}y = 1\) .     M1

\(1 = \frac{\pi }{4} + C \Rightarrow C = 1 – \frac{\pi }{4}\)     A1

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

[8 marks]

Total [13 marks]

Examiners report

The response to this question was often disappointing. Many candidates were unable to find the integrating factor successfully. 

Question

A curve that passes through the point (1, 2) is defined by the differential equation

\[\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x(1 + {x^2} – y){\text{ }}.\]

(a)     (i)     Use Euler’s method to get an approximate value of y when x = 1.3 , taking steps of 0.1. Show intermediate steps to four decimal places in a table.

(ii)     How can a more accurate answer be obtained using Euler’s method?

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

Answer/Explanation

Markscheme

(a)

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

 

Note: Award A2 for complete table.

   Award A1 for a reasonable attempt.

 

\(f(1.3) = 2.14\,\,\,\,\,{\text{(accept 2.141)}}\)     A1

 

(ii)     Decrease the step size     A1

[5 marks]

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

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

Integrating factor is \({{\text{e}}^{\int {2x{\text{d}}x} }} = {{\text{e}}^{{x^2}}}\)     M1A1

So, \({{\text{e}}^{{x^2}}}y = \int {(2x} {{\text{e}}^{{x^2}}} + 2x{{\text{e}}^{{x^2}}}{x^2}){\text{d}}x\)     A1

\( = {{\text{e}}^{{x^2}}} + {x^2}{{\text{e}}^{{x^2}}} – \int {2x{{\text{e}}^{{x^2}}}{\text{d}}x} \)     M1A1

\( = {{\text{e}}^{{x^2}}} + {x^2}{{\text{e}}^{{x^2}}} – {{\text{e}}^{{x^2}}} + k\)

\( = {x^2}{{\text{e}}^{{x^2}}} + k\)     A1

\(y = {x^2} + k{{\text{e}}^{ – {x^2}}}\)

\(x = 1,{\text{ }}y = 2 \to 2 = 1 + k{{\text{e}}^{ – 1}}\)     M1

\(k = {\text{e}}\)

\(y = {x^2} + {{\text{e}}^{1 – {x^2}}}\)     A1

[9 marks]

Total [14 marks]

Examiners report

Some incomplete tables spoiled what were often otherwise good solutions. Although the intermediate steps were asked to four decimal places the answer was not and the usual degree of IB accuracy was expected.

Some candidates surprisingly could not solve what was a fairly easy differential equation in part (b).

Question

(a)     Show that the solution of the homogeneous differential equation

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} + 1,{\text{ }}x > 0,\)

given that \(y = 0{\text{ when }}x = {\text{e, is }}y = x(\ln x – 1)\).

(b)     (i)     Determine the first three derivatives of the function \(f(x) = x(\ln x – 1)\).

  (ii)     Hence find the first three non-zero terms of the Taylor series for f(x) about x = 1.

Answer/Explanation

Markscheme

(a)     EITHER

use the substitution y = vx

\(\frac{{{\text{d}}y}}{{{\text{d}}x}}x + v = v + 1\)     M1A1

\(\int {{\text{d}}v = \int {\frac{{{\text{d}}x}}{x}} } \)

by integration

\(v = \frac{y}{x} = \ln x + c\)     A1

OR

the equation can be rearranged as first order linear

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} – \frac{1}{x}y = 1\)     M1

the integrating factor I is

\({{\text{e}}^{\int { – \frac{1}{x}{\text{d}}x} }} = {{\text{e}}^{ – \ln x}} = \frac{1}{x}\)     A1

multiplying by I gives

\(\frac{{{\text{d}}}}{{{\text{d}}x}}\left( {\frac{1}{x}y} \right) = \frac{1}{x}\)

\(\frac{1}{x}y = \ln x + c\)     A1

THEN

the condition gives c = –1

so the solution is \(y = x(\ln x – 1)\)     AG

[5 marks]

 

(b)     (i)     \(f'(x) = \ln x – 1 + 1 = \ln x\)     A1

\(f”(x) = \frac{1}{x}\)     A1

\(f”'(x) = – \frac{1}{{{x^2}}}\)     A1

(ii)     the Taylor series about x = 1 starts

\(f(x) \approx f(1) + f'(1)(x – 1) + f”(1)\frac{{{{(x – 1)}^2}}}{{2!}} + f”'(1)\frac{{{{(x – 1)}^3}}}{{3!}}\)     (M1)

\( =  – 1 + \frac{{{{(x – 1)}^2}}}{{2!}} – \frac{{{{(x – 1)}^3}}}{{3!}}\)     A1A1A1

[7 marks]

Total: [12 marks]

Examiners report

Part(a) was well done by many candidates. In part(b)(i), however, it was disappointing to see so many candidates unable to differentiate \(x(\ln x – 1)\) correctly. Again, too many candidates were able to quote the general form of a Taylor series expansion, but not how to apply it to the given function.

Question

The variables x and y are related by \(\frac{{{\text{d}}y}}{{{\text{d}}x}} – y\tan x = \cos x\) .

(a)     Find the Maclaurin series for y up to and including the term in \({x^2}\) given that

\(y = – \frac{\pi }{2}\) when x = 0 .

(b)     Solve the differential equation given that y = 0 when \(x = \pi \) . Give the solution in the form \(y = f(x)\) .

Answer/Explanation

Markscheme

(a)     from \(\frac{{{\text{d}}y}}{{{\text{d}}x}} – y\tan x + \cos x\) , \(f'(0) = 1\)     A1

now \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = y{\sec ^2}x + \frac{{{\text{d}}y}}{{{\text{d}}x}}\tan x – \sin x\)     M1A1A1A1

Note: Award A1 for each term on RHS.

 

\( \Rightarrow f”(0) = – \frac{\pi }{2}\)     A1

\( \Rightarrow y = – \frac{\pi }{2} + x – \frac{{\pi {x^2}}}{4}\)     A1

[7 marks]

 

(b)     recognition of integrating factor     (M1)

integrating factor is \({{\text{e}}^{\int { – \tan x{\text{d}}x} }}\)

\( = {{\text{e}}^{\ln \cos x}}\)     (A1)

\( = \cos x\)     (A1)

\( \Rightarrow y\cos x = \int {{{\cos }^2}x{\text{d}}x} \)     M1

\( \Rightarrow y\cos x = \frac{1}{2}\int {(1 + \cos 2x){\text{d}}x} \)     A1

\( \Rightarrow y\cos x = \frac{x}{2} + \frac{{\sin 2x}}{4} + k\)     A1

when \(x = \pi ,{\text{ }}y = 0 \Rightarrow k = – \frac{\pi }{2}\)     M1A1

\( \Rightarrow y\cos x = \frac{x}{2} + \frac{{\sin 2x}}{4} – \frac{\pi }{2}\)     (A1)

\( \Rightarrow y = \sec x\left( {\frac{x}{2} + \frac{{\sin 2x}}{4} – \frac{\pi }{2}} \right)\)     A1

[10 marks]

Total [17 marks]

Examiners report

Part (a) of the question was set up in an unusual way, which caused a problem for a number of candidates as they tried to do part (b) first and then find the Maclaurin series by a standard method. Few were successful as they were usually weaker candidates and made errors in finding the solution \(y = f(x)\) . The majority of candidates knew how to start part (b) and recognised the need to use an integrating factor, but a number failed because they missed out the negative sign on the integrating factor, did not realise that \({{\text{e}}^{\ln \cos x}} = \cos x\) or were unable to integrate \({{{\cos }^2}x}\) . Having said this, a number of candidates succeeded in gaining full marks on this question.

Question

Solve the differential equation

\[(x – 1)\frac{{{\text{d}}y}}{{{\text{d}}x}} + xy = (x – 1){{\text{e}}^{ – x}}\]

given that y = 1 when x = 0. Give your answer in the form \(y = f(x)\).

Answer/Explanation

Markscheme

writing the differential equation in standard form gives

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

\(\int {\frac{x}{{x – 1}}{\text{d}}x = \int {\left( {1 + \frac{1}{{x – 1}}} \right){\text{d}}x = x + \ln (x – 1)} } \)     M1A1

hence integrating factor is \({{\text{e}}^{x + \ln (x – 1)}} = (x – 1){{\text{e}}^x}\)     M1A1

hence, \((x – 1){{\text{e}}^x}\frac{{{\text{d}}y}}{{{\text{d}}x}} + x{{\text{e}}^x}y = x – 1\)     (A1)

\( \Rightarrow \frac{{{\text{d}}\left[ {(x – 1){{\text{e}}^x}y} \right]}}{{{\text{d}}x}} = x – 1\)     (A1)

\( \Rightarrow (x – 1){{\text{e}}^x}y = \int {(x – 1){\text{d}}x} \)     A1

\( \Rightarrow (x – 1){{\text{e}}^x}y = \frac{{{x^2}}}{2} – x + c\)     A1

substituting (0, 1), c = –1     (M1)A1

\( \Rightarrow (x – 1){{\text{e}}^x}y = \frac{{{x^2} – 2x – 2}}{2}\)     (A1)

hence, \(y = \frac{{{x^2} – 2x – 2}}{{2(x – 1){{\text{e}}^x}}}\) (or equivalent)     A1

[13 marks]

Examiners report

Apart from some candidates who thought the differential equation was homogenous, the others were usually able to make a good start, and found it quite straightforward. Some made errors after identifying the correct integrating factor, and so lost accuracy marks.

Question

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

Answer/Explanation

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

Use an integrating factor to show that the general solution for \(\frac{{{\text{d}}x}}{{{\text{d}}t}} – \frac{x}{t} =  – \frac{2}{t},{\text{ }}t > 0\) is \(x = 2 + ct\), where \(c\) is a constant.

The weight in kilograms of a dog, \(t\) weeks after being bought from a pet shop, can be modelled by the following function:

\[w(t) = \left\{ {\begin{array}{*{20}{c}} {2 + ct}&{0 \le t \le 5} \\ {16 – \frac{{35}}{t}}&{t > 5} \end{array}.} \right.\]

[4]
a.

Given that \(w(t)\) is continuous, find the value of \(c\).

[2]
b.

Write down

(i)     the weight of the dog when bought from the pet shop;

(ii)     an upper bound for the weight of the dog.

[2]
c.

Prove from first principles that \(w(t)\) is differentiable at \(t = 5\).

[6]
d.
Answer/Explanation

Markscheme

integrating factor \({e^{\int { – \frac{1}{2}{\text{d}}t} }} = {e^{ – \ln t}}\left( { = \frac{1}{t}} \right)\)     M1A1

\(\frac{x}{t} = \int { – \frac{2}{{{t^2}}}{\text{d}}t = \frac{2}{t} + c} \)     A1A1

Note:     Award A1 for \(\frac{x}{t}\) and A1 for \({\frac{2}{t} + c}\).

\(x = 2 + ct\)     AG

[4 marks]

a.

given continuity at \(x = 5\)

\(5c + 2 = 16 – \frac{{35}}{5} \Rightarrow c = \frac{7}{5}\)     M1A1

[2 marks]

b.

(i)     \(2\)     A1

(ii)     any value \( \ge 16\)     A1

Note:     Accept values less than \(16\) if fully justified by reference to the maximum age for a dog.

[2 marks]

c.

\(\mathop {\lim }\limits_{h \to 0 – } \left( {\frac{{\frac{7}{5}(5 + h) + 2 – \frac{7}{5}(5) – 2}}{h}} \right) = \frac{7}{5}\)     M1A1

\(\mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{16 – \frac{{35}}{{5 + h}} – 16 + \frac{{35}}{5}}}{h}} \right)\;\;\;\left( { = \mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{\frac{{ – 35}}{{5 + h}} + 7}}{h}} \right)} \right)\)     M1

\( = \)\(\mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{\frac{{ – 35 + 35 + 7h}}{{(5 + h)}}}}{h}} \right) = \mathop {\lim }\limits_{h \to 0 + } \left( {\frac{7}{{5 + h}}} \right) = \frac{7}{5}\)     M1A1

both limits equal so differentiable at \(t = 5\)     R1AG

Note:     The limits \(t \to 5\) could also be used.

For each value of \(\frac{7}{5}\) obtained by standard differentiation award     A1.

To gain the other 4 marks a rigorous explanation must be given on how you can get from the left and right hand derivatives to the derivative.

Note:     If the candidate works with \(t\) and then substitutes \(t = 5\) at the end award as follows

First M1 for using formula with \(t\) in the linear case, A1 for \(\frac{7}{5}\)

Award next 2 method marks even if \(t = 5\) not substituted, A1 for \(\frac{7}{5}\)

[6 marks]

Total [14 marks]

d.

Examiners report

This was generally well done. Some candidates did not realize \({e^{ – \ln t}}\) could be simplified to \(\frac{1}{t}\).

a.

This part was well done by the majority of candidates.

b.

This part was well done by the majority of candidates.

c.

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.

d.

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.
Answer/Explanation

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.
Answer/Explanation

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.

Examiners report

[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.
Answer/Explanation

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.

Examiners report

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