CAIE FP1 2012 November — Question 12 OR

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionNovember
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeAsymptotic behavior for large values
DifficultyChallenging +1.2 This is a standard second-order linear differential equation with constant coefficients and a particular integral requiring the method of undetermined coefficients. While it involves multiple steps (complementary function, particular integral, applying initial conditions, and asymptotic analysis), each step follows routine procedures taught in Further Pure 1. The asymptotic behavior analysis is straightforward since the complementary function contains exponentially decaying terms due to complex roots with negative real part. The question is slightly above average difficulty due to its length and the final asymptotic part, but requires no novel insight beyond standard FP1 techniques.
Spec4.10e Second order non-homogeneous: complementary + particular integral4.10f Simple harmonic motion: x'' = -omega^2 x

Obtain the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 13 x = 75 \cos 2 t$$ Given that \(x = 5\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\) when \(t = 0\), find \(x\) in terms of \(t\). Show that, for large positive values of \(t\) and for any initial conditions, $$x \approx 5 \cos ( 2 t - \phi ) ,$$ where the constant \(\phi\) is such that \(\tan \phi = \frac { 4 } { 3 }\).

Question 12 (EITHER):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\) and \(\mathbf{B}\mathbf{e}=\mu\mathbf{e}\); \(\mathbf{AB}\mathbf{e} = \mathbf{A}\mu\mathbf{e} = \mu\mathbf{A}\mathbf{e} = \mu\lambda\mathbf{e} = \lambda\mu\mathbf{e}\)M1A1 Use of eigenvalue results
\(\begin{pmatrix}3&2&2\\-2&-2&-2\\1&2&2\end{pmatrix}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix} \Rightarrow \lambda=0\)B1 Missing eigenvalue of \(\mathbf{A}\)
\(\begin{pmatrix}3&2&2\\-2&-2&-2\\1&2&2\end{pmatrix}\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}1\\0\\-1\end{pmatrix} \Rightarrow \lambda=1\)B1 Part marks: 2
\(\lambda=2 \Rightarrow \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2&2\\-2&-4&-2\end{vmatrix}=\begin{pmatrix}4\\-2\\0\end{pmatrix}\sim\begin{pmatrix}2\\-1\\0\end{pmatrix}\)M1A1 Finds missing eigenvector of \(\mathbf{A}\) — Part marks: 2
\(\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix} \Rightarrow \mu=0\)B1
\(\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}-3\\0\\3\end{pmatrix} \Rightarrow \mu=-3\)B1
\(\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}2\\-1\\0\end{pmatrix}=\begin{pmatrix}-4\\2\\0\end{pmatrix} \Rightarrow \mu=-2\)B1
\(\mathbf{C}\) has eigenvalues: \(0\times0=0\), \(1\times(-3)=-3\), \(2\times(-2)=-4\)B2,1,0 1 mark for one correct, 2 for all three
\(\mathbf{P}=\begin{pmatrix}0&1&2\\1&0&-1\\-1&-1&0\end{pmatrix}\), \(\mathbf{D}=\begin{pmatrix}0&0&0\\0&9&0\\0&0&16\end{pmatrix}\)B1 M1A1 Part marks: 8
Question 12 (OR):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(m^2+6m+13=0 \Rightarrow m=-3\pm2i\); \(x=e^{-3t}(A\cos 2t+B\sin 2t)\)M1A1 Finds complementary function
\(x=p\cos 2t+q\sin 2t\); \(\dot{x}=-2p\sin 2t+2q\cos 2t\); \(\ddot{x}=-4p\cos 2t-4q\sin 2t\)M1 Finds particular integral
\((9p+12q)\cos 2t+(9q-12p)\sin 2t=75\cos 2t \Rightarrow p=3,\; q=4\)M1A1A1
\(x=e^{-3t}(A\cos 2t+B\sin 2t)+3\cos 2t+4\sin 2t\)A1B1 General solution — Part marks: 7
\(x=5\) when \(t=0 \Rightarrow 5=A+3 \Rightarrow A=2\)
\(\dot{x}=-3e^{-3t}(A\cos 2t+B\sin 2t)+e^{-3t}(-2A\sin 2t+2B\cos 2t)-6\sin 2t+8\cos 2t\)M1A1 Uses initial conditions
\(\dot{x}=0\) when \(t=0 \Rightarrow 0=-6+8+2B \Rightarrow B=-1\)A1 Part marks: 4
\(x=e^{-3t}(2\cos 2t-\sin 2t)+3\cos 2t+4\sin 2t\)
As \(t\to\infty\), \(e^{-3t}\to 0\); \(\therefore x\approx 3\cos 2t+4\sin 2t\)M1 Obtains limit
\(\therefore x\approx 5\!\left(\frac{3}{5}\cos 2t+\frac{4}{5}\sin 2t\right)=5\cos\!\left(2t-\tan^{-1}\!\tfrac{4}{3}\right)\)M1A1 AG — Part marks: 3
# Question 12 (EITHER):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{A}\mathbf{e}=\lambda\mathbf{e}$ and $\mathbf{B}\mathbf{e}=\mu\mathbf{e}$; $\mathbf{AB}\mathbf{e} = \mathbf{A}\mu\mathbf{e} = \mu\mathbf{A}\mathbf{e} = \mu\lambda\mathbf{e} = \lambda\mu\mathbf{e}$ | M1A1 | Use of eigenvalue results |
| $\begin{pmatrix}3&2&2\\-2&-2&-2\\1&2&2\end{pmatrix}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix} \Rightarrow \lambda=0$ | B1 | Missing eigenvalue of $\mathbf{A}$ |
| $\begin{pmatrix}3&2&2\\-2&-2&-2\\1&2&2\end{pmatrix}\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}1\\0\\-1\end{pmatrix} \Rightarrow \lambda=1$ | B1 | **Part marks: 2** |
| $\lambda=2 \Rightarrow \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2&2\\-2&-4&-2\end{vmatrix}=\begin{pmatrix}4\\-2\\0\end{pmatrix}\sim\begin{pmatrix}2\\-1\\0\end{pmatrix}$ | M1A1 | Finds missing eigenvector of $\mathbf{A}$ — **Part marks: 2** |
| $\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix} \Rightarrow \mu=0$ | B1 | |
| $\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}-3\\0\\3\end{pmatrix} \Rightarrow \mu=-3$ | B1 | |
| $\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}2\\-1\\0\end{pmatrix}=\begin{pmatrix}-4\\2\\0\end{pmatrix} \Rightarrow \mu=-2$ | B1 | |
| $\mathbf{C}$ has eigenvalues: $0\times0=0$, $1\times(-3)=-3$, $2\times(-2)=-4$ | B2,1,0 | 1 mark for one correct, 2 for all three |
| $\mathbf{P}=\begin{pmatrix}0&1&2\\1&0&-1\\-1&-1&0\end{pmatrix}$, $\mathbf{D}=\begin{pmatrix}0&0&0\\0&9&0\\0&0&16\end{pmatrix}$ | B1 M1A1 | **Part marks: 8** |

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# Question 12 (OR):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $m^2+6m+13=0 \Rightarrow m=-3\pm2i$; $x=e^{-3t}(A\cos 2t+B\sin 2t)$ | M1A1 | Finds complementary function |
| $x=p\cos 2t+q\sin 2t$; $\dot{x}=-2p\sin 2t+2q\cos 2t$; $\ddot{x}=-4p\cos 2t-4q\sin 2t$ | M1 | Finds particular integral |
| $(9p+12q)\cos 2t+(9q-12p)\sin 2t=75\cos 2t \Rightarrow p=3,\; q=4$ | M1A1A1 | |
| $x=e^{-3t}(A\cos 2t+B\sin 2t)+3\cos 2t+4\sin 2t$ | A1B1 | General solution — **Part marks: 7** |
| $x=5$ when $t=0 \Rightarrow 5=A+3 \Rightarrow A=2$ | | |
| $\dot{x}=-3e^{-3t}(A\cos 2t+B\sin 2t)+e^{-3t}(-2A\sin 2t+2B\cos 2t)-6\sin 2t+8\cos 2t$ | M1A1 | Uses initial conditions |
| $\dot{x}=0$ when $t=0 \Rightarrow 0=-6+8+2B \Rightarrow B=-1$ | A1 | **Part marks: 4** |
| $x=e^{-3t}(2\cos 2t-\sin 2t)+3\cos 2t+4\sin 2t$ | | |
| As $t\to\infty$, $e^{-3t}\to 0$; $\therefore x\approx 3\cos 2t+4\sin 2t$ | M1 | Obtains limit |
| $\therefore x\approx 5\!\left(\frac{3}{5}\cos 2t+\frac{4}{5}\sin 2t\right)=5\cos\!\left(2t-\tan^{-1}\!\tfrac{4}{3}\right)$ | M1A1 | AG — **Part marks: 3** |
Obtain the general solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 13 x = 75 \cos 2 t$$

Given that $x = 5$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0$ when $t = 0$, find $x$ in terms of $t$.

Show that, for large positive values of $t$ and for any initial conditions,

$$x \approx 5 \cos ( 2 t - \phi ) ,$$

where the constant $\phi$ is such that $\tan \phi = \frac { 4 } { 3 }$.

\hfill \mbox{\textit{CAIE FP1 2012 Q12 OR}}