AQA FP3 2016 June — Question 7 10 marks

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2016
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeParticular solution with initial conditions
DifficultyChallenging +1.2 This is a standard second-order linear differential equation with constant coefficients requiring both complementary function (CF) and particular integral (PI). While it involves multiple terms (exponential and trigonometric) and applying two initial conditions, the techniques are entirely routine for FP3: the auxiliary equation gives CF = A cos 2x + B sin 2x, and the PI terms follow standard forms. The 10 marks reflect length rather than conceptual difficulty. This is moderately above average due to being Further Maths content with some algebraic manipulation, but remains a textbook exercise.
Spec4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral

7 Find the solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = 10 \mathrm { e } ^ { 4 x } + 8 \sin 2 x + 4 \cos 2 x$$ given that \(y = 2.5\) when \(x = 0\) and \(y = \frac { \pi } { 4 }\) when \(x = \frac { \pi } { 4 }\).
[0pt] [10 marks]

Question 7:
AnswerMarks Guidance
AnswerMark Guidance
Aux eqn \(m^2+4=0\)M1 PI by correct values of \(m\) seen/used
\((y_{CF}=)\; A\sin 2x + B\cos 2x\)A1 \(A\sin 2x + B\cos 2x\) OE
Try \((y_{PI}=)\; ae^{4x}\)M1
\(+bx\sin 2x + cx\cos 2x\)M1
\((y''_{PI}=)\; 16ae^{4x}+(4b-4cx)\cos 2x-(4c+4bx)\sin 2x\)A1 Correct \((y''_{PI}=)\)
Substitution into \(y''+4y\), dep on 2nd and 3rd M and at least one set of differentiations in form \(ke^{4x}+(p+qx)\cos 2x+(r+sx)\sin 2x\)m1 Non-zero constants \(k,p,q,r,s\)
\(20a=10 \Rightarrow a=0.5\); \(-4c=8 \Rightarrow c=-2\); \(4b=4 \Rightarrow b=1\)
\((y_{PI}=)\; 0.5e^{4x}\)B1 \(0.5e^{4x}\) term in PI
\(+x\sin 2x - 2x\cos 2x\)B1 \(+x\sin 2x - 2x\cos 2x\) term in PI with correct \(x\sin 2x\) and \(x\cos 2x\) terms in m1 line
\(y=2.5, x=0\): \(B+0.5=2.5 \Rightarrow B=2\)
\(y=\frac{\pi}{4}, x=\frac{\pi}{4}\): \(A+\frac{e^\pi}{2}+\frac{\pi}{4}=\frac{\pi}{4}\); \(A=-\frac{e^\pi}{2}\)A1F Correct ft value of either \(A\) or \(B\); coeffs of \(\sin 2x\) and \(\cos 2x\) respectively; m1 must have been scored
\(y=2(1-x)\cos 2x + x\sin 2x - \frac{e^\pi}{2}\sin 2x + \frac{e^{4x}}{2}\)A1 ACF. Must be correct eqn and ALL previous 9 marks scored
# Question 7:
| Answer | Mark | Guidance |
|--------|------|----------|
| Aux eqn $m^2+4=0$ | M1 | PI by correct values of $m$ seen/used |
| $(y_{CF}=)\; A\sin 2x + B\cos 2x$ | A1 | $A\sin 2x + B\cos 2x$ OE |
| Try $(y_{PI}=)\; ae^{4x}$ | M1 | |
| $+bx\sin 2x + cx\cos 2x$ | M1 | |
| $(y''_{PI}=)\; 16ae^{4x}+(4b-4cx)\cos 2x-(4c+4bx)\sin 2x$ | A1 | Correct $(y''_{PI}=)$ |
| Substitution into $y''+4y$, dep on 2nd and 3rd M and at least one set of differentiations in form $ke^{4x}+(p+qx)\cos 2x+(r+sx)\sin 2x$ | m1 | Non-zero constants $k,p,q,r,s$ |
| $20a=10 \Rightarrow a=0.5$; $-4c=8 \Rightarrow c=-2$; $4b=4 \Rightarrow b=1$ | | |
| $(y_{PI}=)\; 0.5e^{4x}$ | B1 | $0.5e^{4x}$ term in PI |
| $+x\sin 2x - 2x\cos 2x$ | B1 | $+x\sin 2x - 2x\cos 2x$ term in PI with correct $x\sin 2x$ and $x\cos 2x$ terms in m1 line |
| $y=2.5, x=0$: $B+0.5=2.5 \Rightarrow B=2$ | | |
| $y=\frac{\pi}{4}, x=\frac{\pi}{4}$: $A+\frac{e^\pi}{2}+\frac{\pi}{4}=\frac{\pi}{4}$; $A=-\frac{e^\pi}{2}$ | A1F | Correct ft value of either $A$ or $B$; coeffs of $\sin 2x$ and $\cos 2x$ respectively; m1 must have been scored |
| $y=2(1-x)\cos 2x + x\sin 2x - \frac{e^\pi}{2}\sin 2x + \frac{e^{4x}}{2}$ | A1 | ACF. Must be correct eqn and ALL previous 9 marks scored |

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7 Find the solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = 10 \mathrm { e } ^ { 4 x } + 8 \sin 2 x + 4 \cos 2 x$$

given that $y = 2.5$ when $x = 0$ and $y = \frac { \pi } { 4 }$ when $x = \frac { \pi } { 4 }$.\\[0pt]
[10 marks]

\hfill \mbox{\textit{AQA FP3 2016 Q7 [10]}}