Edexcel F2 2021 January — Question 6 12 marks

Exam BoardEdexcel
ModuleF2 (Further Pure Mathematics 2)
Year2021
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeParticular solution with initial conditions
DifficultyChallenging +1.2 This is a standard Further Maths second-order differential equation with constant coefficients requiring both complementary function (complex roots) and particular integral (trigonometric forcing). While it involves multiple techniques, the method is entirely routine for F2 students with no novel problem-solving required. The initial conditions add computational work but no conceptual difficulty.
Spec4.10e Second order non-homogeneous: complementary + particular integral4.10f Simple harmonic motion: x'' = -omega^2 x
6. (a) Determine the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 6 \cos x$$ (b) Find the particular solution for which \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) at \(x = 0\)
Question 6:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(m^2+2m+5=0 \Rightarrow m=-1\pm 2i\)M1
CF: \(y = e^{-x}(A\cos 2x + B\sin 2x)\) OR \(y=e^{-x}(Pe^{2ix}+Qe^{-2ix})\)A1
PI: \(y = a\cos x + b\sin x\)B1
\(-a\cos x - b\sin x - 2a\sin x + 2b\cos x + 5a\cos x + 5b\sin x = 6\cos x\)M1
\(-b-2a+5b=0 \quad -a+2b+5a=6\)M1
\(a=\frac{6}{5}, \quad b=\frac{3}{5}\)A1
GS: \(y = \text{their CF} + \frac{6}{5}\cos x + \frac{3}{5}\sin x\)A1ft (7)
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(x=0, y=0: \quad 0 = A+\frac{6}{5} \Rightarrow A=-\frac{6}{5}\)M1
\(y' = -e^{-x}(A\cos 2x+B\sin 2x)+e^{-x}(-2A\sin 2x+2B\cos 2x) -\frac{6}{5}\sin x+\frac{3}{5}\cos x\)M1A1ft
\(x=0, \frac{dy}{dx}=0: \quad 0 = +\frac{6}{5}+2B+\frac{3}{5} \Rightarrow B=-\frac{9}{10}\)dM1
PS: \(y = e^{-x}\left(-\frac{6}{5}\cos 2x - \frac{9}{10}\sin 2x\right)+\frac{6}{5}\cos x+\frac{3}{5}\sin x\)A1 (5)
ALT (using complex exponentials):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(x=0, y=0: \quad 0=P+Q+\frac{6}{5}\)M1
\(\frac{dy}{dx} = e^{-x}(2iPe^{2ix}-2iQe^{-2ix})-e^{-x}(Pe^{2ix}+Qe^{-2ix})-\frac{6}{5}\sin x+\frac{3}{5}\cos x\)M1A1ft
\(0=2iP-2iQ+\frac{9}{5}\)
\(P+Q=-\frac{6}{5}, \quad P-Q=\frac{9}{10}i\)
\(P=\frac{1}{2}\left(-\frac{6}{5}+\frac{9}{10}i\right), \quad Q=\frac{1}{2}\left(-\frac{6}{5}-\frac{9}{10}i\right)\)dM1
PS: \(y=\frac{1}{2}e^{-x}\left(-\frac{6}{5}+\frac{9}{10}i\right)e^{2ix}+\frac{1}{2}e^{-x}\left(-\frac{6}{5}-\frac{9}{10}i\right)e^{-2ix}+\frac{6}{5}\cos x+\frac{3}{5}\sin x\)A1 (5) 
# Question 6:

## Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $m^2+2m+5=0 \Rightarrow m=-1\pm 2i$ | M1 | |
| CF: $y = e^{-x}(A\cos 2x + B\sin 2x)$ OR $y=e^{-x}(Pe^{2ix}+Qe^{-2ix})$ | A1 | |
| PI: $y = a\cos x + b\sin x$ | B1 | |
| $-a\cos x - b\sin x - 2a\sin x + 2b\cos x + 5a\cos x + 5b\sin x = 6\cos x$ | M1 | |
| $-b-2a+5b=0 \quad -a+2b+5a=6$ | M1 | |
| $a=\frac{6}{5}, \quad b=\frac{3}{5}$ | A1 | |
| GS: $y = \text{their CF} + \frac{6}{5}\cos x + \frac{3}{5}\sin x$ | A1ft (7) | |

## Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $x=0, y=0: \quad 0 = A+\frac{6}{5} \Rightarrow A=-\frac{6}{5}$ | M1 | |
| $y' = -e^{-x}(A\cos 2x+B\sin 2x)+e^{-x}(-2A\sin 2x+2B\cos 2x) -\frac{6}{5}\sin x+\frac{3}{5}\cos x$ | M1A1ft | |
| $x=0, \frac{dy}{dx}=0: \quad 0 = +\frac{6}{5}+2B+\frac{3}{5} \Rightarrow B=-\frac{9}{10}$ | dM1 | |
| PS: $y = e^{-x}\left(-\frac{6}{5}\cos 2x - \frac{9}{10}\sin 2x\right)+\frac{6}{5}\cos x+\frac{3}{5}\sin x$ | A1 (5) | |

**ALT** (using complex exponentials):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $x=0, y=0: \quad 0=P+Q+\frac{6}{5}$ | M1 | |
| $\frac{dy}{dx} = e^{-x}(2iPe^{2ix}-2iQe^{-2ix})-e^{-x}(Pe^{2ix}+Qe^{-2ix})-\frac{6}{5}\sin x+\frac{3}{5}\cos x$ | M1A1ft | |
| $0=2iP-2iQ+\frac{9}{5}$ | | |
| $P+Q=-\frac{6}{5}, \quad P-Q=\frac{9}{10}i$ | | |
| $P=\frac{1}{2}\left(-\frac{6}{5}+\frac{9}{10}i\right), \quad Q=\frac{1}{2}\left(-\frac{6}{5}-\frac{9}{10}i\right)$ | dM1 | |
| PS: $y=\frac{1}{2}e^{-x}\left(-\frac{6}{5}+\frac{9}{10}i\right)e^{2ix}+\frac{1}{2}e^{-x}\left(-\frac{6}{5}-\frac{9}{10}i\right)e^{-2ix}+\frac{6}{5}\cos x+\frac{3}{5}\sin x$ | A1 (5) | |
6. (a) Determine the general solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 6 \cos x$$

(b) Find the particular solution for which $y = 0$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ at $x = 0$\\

\hfill \mbox{\textit{Edexcel F2 2021 Q6 [12]}}
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