OCR FP3 2013 January — Question 6 11 marks

Exam BoardOCR
ModuleFP3 (Further Pure Mathematics 3)
Year2013
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeResonance cases requiring modified PI
DifficultyChallenging +1.8 This is a resonance problem in second-order DEs requiring the modified particular integral form (multiplying by x) when the forcing frequency matches the natural frequency. While the technique is standard for FP3, it requires recognizing resonance, correctly applying the modified PI method, and analyzing long-term behavior—going beyond routine calculation to require conceptual understanding of why resonance causes unbounded growth versus bounded oscillation in the non-resonant case.
Spec4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral

6 The differential equation \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = \sin k x\) is to be solved, where \(k\) is a constant.
  1. In the case \(k = 2\), by using a particular integral of the form \(a x \cos 2 x + b x \sin 2 x\), find the general solution.
  2. Describe briefly the behaviour of \(y\) when \(x \rightarrow \infty\).
  3. In the case \(k \neq 2\), explain whether \(y\) would exhibit the same behaviour as in part (ii) when \(x \rightarrow \infty\).

Question 6:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
PI: \(y=ax\cos 2x+bx\sin 2x\)
\(\frac{dy}{dx}=a\cos 2x-2ax\sin 2x+b\sin 2x+2bx\cos 2x\)B1 For correct \(\frac{dy}{dx}\) or better
\(\frac{d^2y}{dx^2}=-4a\sin 2x-4ax\cos 2x+4b\cos 2x-4bx\sin 2x\)
substituting into DE and comparing coefficients: \(-4a=1, 4b=0\)M1, M1 Differentiate twice and substitute; compare coefficients
\(\Rightarrow a=-\frac{1}{4}, b=0\)A1
AE: \(\lambda^2+4=0\), \(\lambda=\pm 2i\)M1 For correct auxiliary equation and attempt to solve
CF: \(A\cos 2x+B\sin 2x\)A1 oe form
GS: \(y=\left(A-\frac{1}{4}x\right)\cos 2x+B\sin 2x\)A1ft Must be real and contain 2 unknowns
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
oscillationsB1 oe (accept sketch) dep consistent with 6(i)
unboundedB1 oe; if zero, sc1 for recognition that \(x\cos 2x\) term becomes dominant
Part (iii)
AnswerMarks Guidance
AnswerMarks Guidance
If \(k\neq 2\) then PI \(y=\alpha\cos kx+\beta\sin kx\)B1
So bounded oscillationsB1 oe (accept sketch)
# Question 6:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| PI: $y=ax\cos 2x+bx\sin 2x$ | |  |
| $\frac{dy}{dx}=a\cos 2x-2ax\sin 2x+b\sin 2x+2bx\cos 2x$ | B1 | For correct $\frac{dy}{dx}$ or better |
| $\frac{d^2y}{dx^2}=-4a\sin 2x-4ax\cos 2x+4b\cos 2x-4bx\sin 2x$ | | |
| substituting into DE and comparing coefficients: $-4a=1, 4b=0$ | M1, M1 | Differentiate twice and substitute; compare coefficients |
| $\Rightarrow a=-\frac{1}{4}, b=0$ | A1 | |
| AE: $\lambda^2+4=0$, $\lambda=\pm 2i$ | M1 | For correct auxiliary equation and attempt to solve |
| CF: $A\cos 2x+B\sin 2x$ | A1 | oe form |
| GS: $y=\left(A-\frac{1}{4}x\right)\cos 2x+B\sin 2x$ | A1ft | Must be real and contain 2 unknowns |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| oscillations | B1 | oe (accept sketch) dep consistent with 6(i) |
| unbounded | B1 | oe; if zero, sc1 for recognition that $x\cos 2x$ term becomes dominant |

## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $k\neq 2$ then PI $y=\alpha\cos kx+\beta\sin kx$ | B1 | |
| So bounded oscillations | B1 | oe (accept sketch) |

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6 The differential equation $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = \sin k x$ is to be solved, where $k$ is a constant.\\
(i) In the case $k = 2$, by using a particular integral of the form $a x \cos 2 x + b x \sin 2 x$, find the general solution.\\
(ii) Describe briefly the behaviour of $y$ when $x \rightarrow \infty$.\\
(iii) In the case $k \neq 2$, explain whether $y$ would exhibit the same behaviour as in part (ii) when $x \rightarrow \infty$.

\hfill \mbox{\textit{OCR FP3 2013 Q6 [11]}}