| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2019 |
| 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 Further Maths second-order differential equation requiring both complementary function (complex roots) and particular integral (for forced oscillation), then applying initial conditions. The multi-step process and Further Maths context place it moderately above average difficulty, though it follows a standard method once learned. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral4.10g Damped oscillations: model and interpret |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (a) | d2x dx |
| Answer | Marks |
|---|---|
| 2 | M1 |
| Answer | Marks |
|---|---|
| [11] | 1.1a |
| Answer | Marks |
|---|---|
| 3.4 | For obtaining the AE |
| Answer | Marks |
|---|---|
| For B | Or equivalent forms |
| Answer | Marks |
|---|---|
| (b) | x=e−t( Acos2t+Bsin2t )+2sint−cost |
| Answer | Marks |
|---|---|
| ⇒ x≈2sint−cost | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 3.4 | Consider behaviour of e-t |
| Answer | Marks |
|---|---|
| ft their equation from (a) | Accept this final line |
Question 11:
11 | (a) | d2x dx
+2 +5x=10sint
dt2 dt
A.E. n2 +2n+5=0⇒n=−1±2i
⇒ x=e−t( Acos2t+Bsin2t )
P.I. x=asint+bcost
dx
⇒ =acost−bsint
dt
d2x
⇒ =−asint−bcost
dt2
⇒−asint−bcost+2 ( acost−bsint )
+5 ( asint+bcost )=10sint
⇒−b+2a+5b=0, −a−2b+5a =10
i.e. a=−2b, 2a−b=5⇒b=−1, a =2
⇒G.S. x=e−t( Acos2t+Bsin2t )+2sint−cost
dx
When t =0 we have x=2and =0
dt
⇒ A−1=2⇒ A=3
dx
=−e−t( Acos2t+Bsin2t )
dt
+e−t(−2Asin2t+2Bcos2t )+2cost+sint
1
t =0⇒−A+2B+2=0⇒ B=
2
1
⇒ x=e−t 3cos2t+ sin2t +2sint−cost
2 | M1
A1
M1
M1
A1
A1
B1
B1
M1
A1
A1
[11] | 1.1a
1.1
3.1a
1.1
1.1
1.1
3.3
3.3
3.4
3.3
3.4 | For obtaining the AE
Solving AE and
interpreting
Correct form of PI
and differentiate twice
Dep on previous M
Substitute their PI with
two constants into DE
Values of a and b
Correct GS
For substituting into 2
eqns
For A
Dep on previous M
For diffn their GS. Their
value for A could be
substituted
For B | Or equivalent forms
(i.e. exponential form)
(b) | x=e−t( Acos2t+Bsin2t )+2sint−cost
As t →∞, e−t →0
⇒ x≈2sint−cost | M1
A1
[2] | 3.4
3.4 | Consider behaviour of e-t
Accept =
ft their equation from (a) | Accept this final line
for 2 marks
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance
programme your call may be recorded or monitored
Oxford Cambridge and RSA Examinations
is a Company Limited by Guarantee
Registered in England
Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA
Registered Company Number: 3484466
OCR is an exempt Charity
OCR (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 201911 A particle is suspended in a resistive medium from one end of a light spring. The other end of the spring is attached to a point which is made to oscillate in a vertical line.
The displacement of the particle may be modelled by the differential equation $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 10 \sin t$\\
where $x$ is the displacement of the particle below the equilibrium position at time $t$.\\
When $t = 0$ the particle is stationary and its displacement is 2 .
\begin{enumerate}[label=(\alph*)]
\item Find the particular solution of the differential equation.
\item Write down an approximate equation for the displacement when $t$ is large.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2019 Q11 [13]}}