Question 6 - A-Level Maths

Edexcel F2 2021 June — Question 6 13 marks

Exam Board	Edexcel
Module	F2 (Further Pure Mathematics 2)
Year	2021
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Particular solution with initial conditions
Difficulty	Standard +0.8 This is a standard Further Maths second-order linear ODE with constant coefficients requiring both complementary function (solving auxiliary equation with distinct real roots) and particular integral (polynomial trial solution). Part (b) adds initial conditions requiring algebraic manipulation. While methodical, it's more demanding than typical A-level questions due to the multi-step process and being Further Maths content.
Spec	4.10d Second order homogeneous: auxiliary equation method 4.10e Second order non-homogeneous: complementary + particular integral

6. (a) Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 8 y = 2 x ^ { 2 } + x$$ (b) Find the particular solution of this differential equation for which $y = 1$ and $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 0 \text { when } x = 0$$

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Question 6:

Part (a):

Answer	Marks	Guidance
Working/Answer	Marks	Guidance
$m^2 - 6m + 8 = 0$, $(m-2)(m-4) = 0$, $m = 2, 4$	M1	Form aux equation and attempt to solve (any valid method). Equation need not be shown if CF is correct or complete solution $(m=2,4)$ is shown
$(\text{CF} =)\ Ae^{2x} + Be^{4x}$	A1	Correct CF, $y=...$ not needed
PI: $y = \lambda x^2 + \mu x + \nu$	B1	Correct form for PI
$y' = 2\lambda x + \mu$, $y'' = 2\lambda$; substituted into equation	M1	Their PI (minimum 2 terms) differentiated twice and substituted in the equation
$\lambda = \frac{1}{4},\ -12\lambda + 8\mu = 1,\ 2\lambda - 6\mu + 8\nu = 0$	M1	Coefficients equated
$\lambda = \frac{1}{4},\ \mu = \frac{1}{2},\ \nu = \frac{5}{16}$	A1A1	Any 2 values correct / All 3 values correct
$y = Ae^{2x} + Be^{4x} + \frac{1}{4}x^2 + \frac{1}{2}x + \frac{5}{16}$	A1ft	A complete solution, follow through their CF and PI. All 3 M marks must have been earned. Must start $y=...$

(8 marks)

Part (b):

Answer	Marks	Guidance
Working/Answer	Marks	Guidance
$1 = A + B + \frac{5}{16}$	M1	Substitute $y=1$ and $x=0$ in their complete solution from (a)
$\frac{dy}{dx} = 2Ae^{2x} + 4Be^{4x} + \frac{1}{2}x + \frac{1}{2}$; $0 = 2A + 4B + \frac{1}{2}$	M1	Differentiate and substitute $\frac{dy}{dx} = 0$, $x = 0$
$A = \frac{13}{8}$, $B = -\frac{15}{16}$	dM1A1	Solve the 2 equations to $A=...$ or $B=...$; depends on two previous M marks. Both values correct
$y = \frac{13}{8}e^{2x} - \frac{15}{16}e^{4x} + \frac{1}{4}x^2 + \frac{1}{2}x + \frac{5}{16}$	A1ft	Particular solution, follow through their general solution and $A$ and $B$. Must start $y=...$

(5 marks) [13 total]

# Question 6:

## Part (a):

| Working/Answer | Marks | Guidance |
|---|---|---|
| $m^2 - 6m + 8 = 0$, $(m-2)(m-4) = 0$, $m = 2, 4$ | M1 | Form aux equation and attempt to solve (any valid method). Equation need not be shown if CF is correct or complete solution $(m=2,4)$ is shown |
| $(\text{CF} =)\ Ae^{2x} + Be^{4x}$ | A1 | Correct CF, $y=...$ not needed |
| PI: $y = \lambda x^2 + \mu x + \nu$ | B1 | Correct form for PI |
| $y' = 2\lambda x + \mu$, $y'' = 2\lambda$; substituted into equation | M1 | Their PI (minimum 2 terms) differentiated twice and substituted in the equation |
| $\lambda = \frac{1}{4},\ -12\lambda + 8\mu = 1,\ 2\lambda - 6\mu + 8\nu = 0$ | M1 | Coefficients equated |
| $\lambda = \frac{1}{4},\ \mu = \frac{1}{2},\ \nu = \frac{5}{16}$ | A1A1 | Any 2 values correct / All 3 values correct |
| $y = Ae^{2x} + Be^{4x} + \frac{1}{4}x^2 + \frac{1}{2}x + \frac{5}{16}$ | A1ft | A complete solution, follow through their CF and PI. All 3 M marks must have been earned. Must start $y=...$ |

**(8 marks)**

## Part (b):

| Working/Answer | Marks | Guidance |
|---|---|---|
| $1 = A + B + \frac{5}{16}$ | M1 | Substitute $y=1$ and $x=0$ in their complete solution from (a) |
| $\frac{dy}{dx} = 2Ae^{2x} + 4Be^{4x} + \frac{1}{2}x + \frac{1}{2}$; $0 = 2A + 4B + \frac{1}{2}$ | M1 | Differentiate and substitute $\frac{dy}{dx} = 0$, $x = 0$ |
| $A = \frac{13}{8}$, $B = -\frac{15}{16}$ | dM1A1 | Solve the 2 equations to $A=...$ or $B=...$; depends on two previous M marks. Both values correct |
| $y = \frac{13}{8}e^{2x} - \frac{15}{16}e^{4x} + \frac{1}{4}x^2 + \frac{1}{2}x + \frac{5}{16}$ | A1ft | Particular solution, follow through their general solution and $A$ and $B$. Must start $y=...$ |

**(5 marks) [13 total]**

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6. (a) Find the general solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 8 y = 2 x ^ { 2 } + x$$

(b) Find the particular solution of this differential equation for which $y = 1$ and

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 0 \text { when } x = 0$$

\hfill \mbox{\textit{Edexcel F2 2021 Q6 [13]}}

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