OCR Further Pure Core 1 2022 June — Question 8 13 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2022
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSystems of differential equations
TypeVerify or show auxiliary equation result
DifficultyStandard +0.8 This is a Further Maths coupled differential equations question requiring elimination to derive a second-order ODE, solving with complementary function and particular integral, and analyzing long-term behavior with hyperbolic functions. While systematic, it demands multiple techniques (elimination, auxiliary equations, initial conditions, limits of hyperbolic functions) across several parts, placing it moderately above average difficulty.
Spec4.10h Coupled systems: simultaneous first order DEs

8 A biologist is studying the effect of pesticides on crops. On a certain farm pesticide is regularly applied to a particular crop which grows in soil. Over time, pesticide is transferred between the crop and the soil at a rate which depends on the amount of pesticide in both the crop and the soil. The amount of pesticide in the crop after \(t\) days is \(x\) grams. The amount of pesticide in the soil after \(t\) days is \(y\) grams. Initially, when \(t = 0\), there is no pesticide in either the crop or the soil. At first it is assumed that no pesticide is lost from the system. The biologist further assumes that pesticide is added to the crop at a constant rate of \(k\) grams per day, where \(k > 6\). After collecting some initial data, the biologist suggests that for \(t \geqslant 0\), this situation can be modelled by the following pair of first order linear differential equations. \(\frac { d x } { d t } = - 2 x + 78 y + k\) \(\frac { d y } { d t } = 2 x - 78 y\)
    1. Show that \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 80 \frac { d x } { d t } = 78 \mathrm { k }\).
    2. Determine the particular solution for \(x\) in terms of \(k\) and \(t\). If more than 250 grams of pesticide is found in the crop, then it will fail food safety standards.
    3. The crop is tested 50 days after the pesticide is first added to it. Explain why, according to this model, the crop will fail food safety standards as a result of this test. Further data collection suggests that some pesticide decays in the soil and so is lost from the system. The model is refined in light of this data. The particular solution for \(x\) for this refined model is \(\mathrm { x } = \mathrm { k } \left( 20 - \mathrm { e } ^ { - 41 \mathrm { t } } \left( 20 \cosh ( \sqrt { 1677 } \mathrm { t } ) + \frac { 819 } { \sqrt { 1677 } } \sinh ( \sqrt { 1677 } \mathrm { t } ) \right) \right.\).
  1. Given now that \(k < 12\), determine whether the crop will fail food safety standards in the long run according to this refined model. In the refined model, it is still assumed that pesticide is added to the crop at a constant rate.
  2. Suggest a reason why it might be more realistic to model the addition of pesticide as not being at a constant rate.

Question 8:
AnswerMarks Guidance
8(a) (i)
=−2 +78 ( 2x−78y )
dt2 dt
.
dx dx 
=−2 +156x−78 +2x−k
dt  dt 
d2x dx
⇒ =−80 +78k
dt2 dt
d2x dx
⇒ +80 =78k
AnswerMarks
dt2 dtM1
A1
AnswerMarks
[2]3.3
1.1dx dy
Differentiate and substitute
dt dt
AG convincingly shown after second substitution.
Alternative method for first 4 marks:
AnswerMarks
M1Integrating wrt and then using integrating
factor
𝑥𝑥̇ +80𝑥𝑥 = 78𝑘𝑘𝑘𝑘 +𝑐𝑐
∫80d𝑡𝑡 ∫80d𝑡𝑡
AnswerMarks Guidance
e (𝑥𝑥̇ +80𝑥𝑥) = (78𝑘𝑘𝑘𝑘 + 𝑐𝑐)eM1 𝑥𝑥̈ +80𝑥𝑥̇ = 78𝑘𝑘 𝑘𝑘
Integrating wrt𝑥𝑥̈ +80𝑥𝑥̇ = 78𝑘𝑘 𝑘𝑘
d 80𝑡𝑡 80𝑡𝑡
(e 𝑥𝑥) = (78𝑘𝑘𝑘𝑘+𝑐𝑐)e
AnswerMarks
d𝑘𝑘Integrating wrt
80𝑡𝑡 80𝑡𝑡
e 𝑥𝑥 = �(78𝑘𝑘𝑘𝑘 +𝑐𝑐)e d𝑘𝑘+𝑑𝑑
AnswerMarks Guidance
80𝑡𝑡M1 Integrating by p𝑘𝑘arts (78kt + c may be separated)
80𝑡𝑡 (78𝑘𝑘𝑘𝑘+𝑐𝑐)e 78𝑘𝑘 80𝑡𝑡
e 𝑥𝑥 = −� e d𝑘𝑘
80 80
+𝑑𝑑
AnswerMarks Guidance
80𝑡𝑡A1 GS of form
−80𝑡𝑡 39
𝑥𝑥 = 𝐴𝐴+𝐵𝐵e +40𝑘𝑘𝑘𝑘
80𝑡𝑡 (78𝑘𝑘𝑘𝑘+𝑐𝑐)e 39𝑘𝑘 80𝑡𝑡
e 𝑥𝑥 = − e +𝑑𝑑
80 3200
39 𝑐𝑐 39𝑘𝑘 −80𝑡𝑡
𝑥𝑥 = 𝑘𝑘𝑘𝑘+ − +𝑑𝑑e
40 80 3200
−80𝑡𝑡 39
= 𝐴𝐴+𝐵𝐵e + 𝑘𝑘𝑘𝑘
40
M1
A1
GS of form
Alternative method for this M mark:
AnswerMarks
M1Finding GS for and using to find another
equation for and
𝑦𝑦 𝑥𝑥 = 0,𝑦𝑦 = 0
78𝑦𝑦 = 𝑥𝑥̇ +2𝑥𝑥 −𝑘𝑘
−80𝑡𝑡 1 78
When
= −78𝐵𝐵e − 𝑘𝑘 +2𝐴𝐴+ 𝑘𝑘𝑘𝑘
AnswerMarks Guidance
40 40Finding GS for and using to find another
𝑘𝑘 = 0,𝑥𝑥 = 0,𝑦𝑦 = 0M1 equation for and
1
AnswerMarks Guidance
0 = 2𝐴𝐴−78𝐵𝐵 − 𝑘𝑘[7] 𝐴𝐴 𝐵𝐵
(iii)40
156000k
When t =50, x>
3200
( because e−4000 <1 so 1−e−4000 >0 )
i.e. x>48.75k >292.5 for k >6
AnswerMarks Guidance
i.e. x>250 so fails safety food standardsB1
[1]3.4 Use of t = 50 must be seen to give a term in k
Must reference k > 6.
Condone the idea of ignoring exponential term as negligible
Sight of 292.5 >250 earns the mark
AnswerMarks
(b)1677=40.95...<41
( ( ) ( ))
⇒e−41t acosh 1677t +bsinh 1677t
contains only negative exponentials when
expanded
so as t →∞, the exponential/hyperbolic parts
→ 0 so x → 20k < 240 < 250 since k < 12 so
yes, food safety standards are met in the long
AnswerMarks
run.M1
A1
AnswerMarks
[2]3.4
3.2aTurning function into exponentials. Could see eg e–0.0488t and
e–81.95t. Condone error(s) in coefficients.
Allow argument such as e-41t dominates
Argument must be complete and correct but could be based on
sufficiently large values of t rather than formal limiting
process.
SC B1 argument that 20k = 240 < 250 if 2nd term is assumed
to tend to 0
AnswerMarks
(c)Pesticide is likely to be added periodically,
eg, during the day; or depending on the
weather/time of year; or only when it’s
needed; or the amount of pesticide added
changes as the amount of crop
changes/grows or if it is subject to pest
AnswerMarks Guidance
attackB1
[1]3.5a The idea that regularly is not the same as continuously (and
may not even mean “at a constant average rate”).
Other sensible answers possible, but must be in context.
Vertices of ABC satisfy z3 – 1 (= 0)
(Complex number represented by) M =
1
−2
Vertices of LMN satisfy
3
8𝑧𝑧 +1 (= 0)
(z3 – 1)( (=0)
3
8𝑧𝑧 +1)
AnswerMarks
6 3B1
B1
M1
M1
A1
AnswerMarks
[5]1.1
3.1a
3.1a
2.1
AnswerMarks
2.2aOr (= 0) Or all three values
2 4
3𝜋𝜋i 3𝜋𝜋i
stated
(𝑧𝑧−1)�𝑧𝑧−e ��𝑧𝑧−e �
Or one of
1 1
Attempt at poly 1 nom𝜋𝜋iial rel 1 ati3n 𝜋𝜋 g i to LM 1 N. − 3𝜋𝜋i
𝑀𝑀 = 2e ,𝐿𝐿 = 2e ,𝑁𝑁 = 2e
suffices for this mark.
1 1
Attem 1 pt 𝜋𝜋 p i roduct 1 of 3th 𝜋𝜋 e i ir two c 1 ub − i3c 𝜋𝜋 f i actors.
�𝑧𝑧−2e ��𝑧𝑧−2e ��𝑧𝑧−2e �
A0 without justification of 8z3 + 1 = 0. Must see = 0.
8𝑧𝑧 −7𝑧𝑧 −1 = 0
AnswerMarks
Alternative methodB1
B1Finding A, B, C
Finding L, M, N
 1 3   1 3 
A ( 1 ) , B− + i, C− − i
   
 2 2   2 2 
1 3  1 3   1
L   + i  ,N   − i  , M − 
4 4  4 4   2
AnswerMarks Guidance
Quadratic satisfying B, C z2 +z+1=0M1 Combining to give a 6th degree polynomial
Multiplying out some terms to give quadratics or cubics
Must see = 0
Quadratic satisfying L,N 4z2 −2z+1=0
Quadratic satisfying A, M 2z2 −z−1=0
Eqn satisfying all 6 points
M1
( )( )( )
z2 +z+1 4z2 −2z+1 2z2 −z−1 =0
⇒8z6 −7z3−1=0
A1
B1
B1
Finding A, B, C
Finding L, M, N
Combining to give a 6th degree polynomial
Multiplying out some terms to give quadratics or cubics
Must see = 0
Alternative methods
By rotational symmetry Because
2 3
𝜋𝜋i
3
�e � =
Calculation of vertices of L, M and N. Use of
1
1 + ω + ω2 = 0 (and ω3 = 1) to simplify eg (z
– ½(ω + 1))(z – ½(ω2 + 1)) ))(z – ½(ω + ω2))
= (z + ½ω2))(z + ½ω) ))(z + ½) etc or to find
sum/product of roots.
AnswerMarks Guidance
8(a) (ii)
𝑦𝑦̈ = −2(𝑦𝑦̇ +78𝑦𝑦)+156𝑦𝑦+2𝑘𝑘−78𝑦𝑦̇
𝑦𝑦̈ +80𝑦𝑦̇ = 2𝑘𝑘
2
𝜆𝜆 +80𝜆𝜆 = 0
Complementary function is
𝜆𝜆 = 0,−80
Trial function is y −80𝑡𝑡
𝑦𝑦 = 𝐴𝐴+𝐵𝐵e
= 𝑎𝑎𝑘𝑘
So 80
𝑦𝑦̇ = 𝑎𝑎,𝑦𝑦̈ = 0
1
General solution is
AnswerMarks
𝑎𝑎 = 2𝑘𝑘 ⇔𝑎𝑎 = 40𝑘𝑘M1
M1
AnswerMarks
M13.1a
3.3
AnswerMarks
2.2a𝑦𝑦 𝑥𝑥.
Auxiliary equation for their DE
Correct trial function for their CF
Note the GS for in the main mark scheme achieves A1 but here
A1 is not awarded until GS for found.
𝑥𝑥
𝑥𝑥
−80𝑡𝑡 1
Alternative method
𝑦𝑦 = 𝐴𝐴+𝐵𝐵e +40𝑘𝑘𝑘𝑘
AnswerMarks Guidance
M1M1 Integrating wrt and then using integrating factor
𝑦𝑦̇ +80𝑦𝑦 = 2𝑘𝑘𝑘𝑘+𝑐𝑐
∫80d𝑡𝑡 ∫80d𝑡𝑡
AnswerMarks Guidance
e (𝑦𝑦̇ +80𝑦𝑦)= (2𝑘𝑘𝑘𝑘+ 𝑐𝑐)eM1 𝑦𝑦̈ +78𝑦𝑦̇ = 2𝑘𝑘 𝑘𝑘
Integrating wrt
d 80𝑡𝑡 80𝑡𝑡
(e 𝑦𝑦)= (2𝑘𝑘𝑘𝑘+𝑐𝑐)e
d𝑘𝑘
80𝑡𝑡 80𝑡𝑡
e 𝑦𝑦 = �(2𝑘𝑘𝑘𝑘+𝑐𝑐)e d𝑘𝑘+𝑑𝑑
AnswerMarks Guidance
80𝑡𝑡M1 Integrating by p𝑘𝑘arts
80𝑡𝑡 (2𝑘𝑘𝑘𝑘+𝑐𝑐)e 2 80𝑡𝑡
e 𝑦𝑦 = −� 𝑘𝑘e d𝑘𝑘+𝑑𝑑
80 80
80𝑡𝑡
80𝑡𝑡 (2𝑘𝑘𝑘𝑘+𝑐𝑐)e 1 80𝑡𝑡
e 𝑦𝑦 = − 𝑘𝑘e +𝑑𝑑
80 3200
1 1 1 −80𝑡𝑡
𝑦𝑦 = 𝑘𝑘𝑘𝑘+ 𝑐𝑐− 𝑘𝑘+𝑑𝑑e
40 80 3200
−80𝑡𝑡 1
= 𝐴𝐴+𝐵𝐵e + 𝑘𝑘𝑘𝑘
M1
𝑦𝑦̈ +78𝑦𝑦̇ = 2𝑘𝑘 𝑘𝑘
Integrating wrt
−80𝑡𝑡 1
𝑦𝑦̇ = −80𝐵𝐵e + 𝑘𝑘
40
1
𝑥𝑥 = (𝑦𝑦̇ +78𝑦𝑦)
2
−80𝑡𝑡 39 1
When
= 39𝐴𝐴−𝐵𝐵e + 𝑘𝑘𝑘𝑘+ 𝑘𝑘
40 80
𝑘𝑘 = 0,𝑥𝑥 = 0,𝑦𝑦 = 0, 𝑦𝑦̇ = 0+0 = 0
0 = 𝐴𝐴+𝐵𝐵
1
0 = 39𝐴𝐴−𝐵𝐵+ 𝑘𝑘
80
1
0 = −80𝐵𝐵+ 𝑘𝑘
40
1 1
Particular solution for is
𝐵𝐵 = 𝑘𝑘 ,𝐴𝐴 = − 𝑘𝑘
3200 3200
𝑥𝑥
39 1 −80𝑡𝑡 39 1
𝑥𝑥 = − 𝑘𝑘− 𝑘𝑘e + 𝑘𝑘𝑘𝑘+ 𝑘𝑘
3200 3200 40 80
1 −80𝑡𝑡
AnswerMarks
= 𝑘𝑘(1−e +3120𝑘𝑘)A1
M1
M1
A1
AnswerMarks
[7]1.1
3.1a
1.1
AnswerMarks
3.3GS of form
−80𝑡𝑡 39
𝑥𝑥 =𝐶𝐶 +𝐷𝐷e +40𝑘𝑘𝑘𝑘
Using two of to find simultaneous equations for
and (M1 for each equation).
𝑥𝑥 = 𝑦𝑦 = 𝑦𝑦̇ = 0 𝐴𝐴
𝐵𝐵
AnswerMarks Guidance
8(a) (i)
D.E. is +80y =78k where y =
dt dt
dy
I.F. e80t ⇒e80t +80e80ty =78ke80t
dt
d ( )
⇒ e80ty =78ke80t
dt
39k 39k
⇒e80ty = e80t + A⇒ y = + Ae−80t
40 40
dx 39k 39kt A
⇒ = + Ae−80t ⇒ x= − e−80t +B
AnswerMarks
dt 40 40 80M1
A1
M1
A1
PMT
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Question 8:
8 | (a) | (i) | d2x dx
=−2 +78 ( 2x−78y )
dt2 dt
.
dx dx 
=−2 +156x−78 +2x−k
dt  dt 
d2x dx
⇒ =−80 +78k
dt2 dt
d2x dx
⇒ +80 =78k
dt2 dt | M1
A1
[2] | 3.3
1.1 | dx dy
Differentiate and substitute
dt dt
AG convincingly shown after second substitution.
Alternative method for first 4 marks:
M1 | Integrating wrt and then using integrating
factor
𝑥𝑥̇ +80𝑥𝑥 = 78𝑘𝑘𝑘𝑘 +𝑐𝑐
∫80d𝑡𝑡 ∫80d𝑡𝑡
e (𝑥𝑥̇ +80𝑥𝑥) = (78𝑘𝑘𝑘𝑘 + 𝑐𝑐)e | M1 | 𝑥𝑥̈ +80𝑥𝑥̇ = 78𝑘𝑘 𝑘𝑘
Integrating wrt | 𝑥𝑥̈ +80𝑥𝑥̇ = 78𝑘𝑘 𝑘𝑘
d 80𝑡𝑡 80𝑡𝑡
(e 𝑥𝑥) = (78𝑘𝑘𝑘𝑘+𝑐𝑐)e
d𝑘𝑘 | Integrating wrt
80𝑡𝑡 80𝑡𝑡
e 𝑥𝑥 = �(78𝑘𝑘𝑘𝑘 +𝑐𝑐)e d𝑘𝑘+𝑑𝑑
80𝑡𝑡 | M1 | Integrating by p𝑘𝑘arts (78kt + c may be separated)
80𝑡𝑡 (78𝑘𝑘𝑘𝑘+𝑐𝑐)e 78𝑘𝑘 80𝑡𝑡
e 𝑥𝑥 = −� e d𝑘𝑘
80 80
+𝑑𝑑
80𝑡𝑡 | A1 | GS of form
−80𝑡𝑡 39
𝑥𝑥 = 𝐴𝐴+𝐵𝐵e +40𝑘𝑘𝑘𝑘
80𝑡𝑡 (78𝑘𝑘𝑘𝑘+𝑐𝑐)e 39𝑘𝑘 80𝑡𝑡
e 𝑥𝑥 = − e +𝑑𝑑
80 3200
39 𝑐𝑐 39𝑘𝑘 −80𝑡𝑡
𝑥𝑥 = 𝑘𝑘𝑘𝑘+ − +𝑑𝑑e
40 80 3200
−80𝑡𝑡 39
= 𝐴𝐴+𝐵𝐵e + 𝑘𝑘𝑘𝑘
40
M1
A1
GS of form
Alternative method for this M mark:
M1 | Finding GS for and using to find another
equation for and
𝑦𝑦 𝑥𝑥 = 0,𝑦𝑦 = 0
78𝑦𝑦 = 𝑥𝑥̇ +2𝑥𝑥 −𝑘𝑘
−80𝑡𝑡 1 78
When
= −78𝐵𝐵e − 𝑘𝑘 +2𝐴𝐴+ 𝑘𝑘𝑘𝑘
40 40 | Finding GS for and using to find another
𝑘𝑘 = 0,𝑥𝑥 = 0,𝑦𝑦 = 0 | M1 | equation for and
1
0 = 2𝐴𝐴−78𝐵𝐵 − 𝑘𝑘 | [7] | 𝐴𝐴 𝐵𝐵
(iii) | 40
156000k
When t =50, x>
3200
( because e−4000 <1 so 1−e−4000 >0 )
i.e. x>48.75k >292.5 for k >6
i.e. x>250 so fails safety food standards | B1
[1] | 3.4 | Use of t = 50 must be seen to give a term in k
Must reference k > 6.
Condone the idea of ignoring exponential term as negligible
Sight of 292.5 >250 earns the mark
(b) | 1677=40.95...<41
( ( ) ( ))
⇒e−41t acosh 1677t +bsinh 1677t
contains only negative exponentials when
expanded
so as t →∞, the exponential/hyperbolic parts
→ 0 so x → 20k < 240 < 250 since k < 12 so
yes, food safety standards are met in the long
run. | M1
A1
[2] | 3.4
3.2a | Turning function into exponentials. Could see eg e–0.0488t and
e–81.95t. Condone error(s) in coefficients.
Allow argument such as e-41t dominates
Argument must be complete and correct but could be based on
sufficiently large values of t rather than formal limiting
process.
SC B1 argument that 20k = 240 < 250 if 2nd term is assumed
to tend to 0
(c) | Pesticide is likely to be added periodically,
eg, during the day; or depending on the
weather/time of year; or only when it’s
needed; or the amount of pesticide added
changes as the amount of crop
changes/grows or if it is subject to pest
attack | B1
[1] | 3.5a | The idea that regularly is not the same as continuously (and
may not even mean “at a constant average rate”).
Other sensible answers possible, but must be in context.
Vertices of ABC satisfy z3 – 1 (= 0)
(Complex number represented by) M =
1
−2
Vertices of LMN satisfy
3
8𝑧𝑧 +1 (= 0)
(z3 – 1)( (=0)
3
8𝑧𝑧 +1)
6 3 | B1
B1
M1
M1
A1
[5] | 1.1
3.1a
3.1a
2.1
2.2a | Or (= 0) Or all three values
2 4
3𝜋𝜋i 3𝜋𝜋i
stated
(𝑧𝑧−1)�𝑧𝑧−e ��𝑧𝑧−e �
Or one of
1 1
Attempt at poly 1 nom𝜋𝜋iial rel 1 ati3n 𝜋𝜋 g i to LM 1 N. − 3𝜋𝜋i
𝑀𝑀 = 2e ,𝐿𝐿 = 2e ,𝑁𝑁 = 2e
suffices for this mark.
1 1
Attem 1 pt 𝜋𝜋 p i roduct 1 of 3th 𝜋𝜋 e i ir two c 1 ub − i3c 𝜋𝜋 f i actors.
�𝑧𝑧−2e ��𝑧𝑧−2e ��𝑧𝑧−2e �
A0 without justification of 8z3 + 1 = 0. Must see = 0.
8𝑧𝑧 −7𝑧𝑧 −1 = 0
Alternative method | B1
B1 | Finding A, B, C
Finding L, M, N
 1 3   1 3 
A ( 1 ) , B− + i, C− − i
   
 2 2   2 2 
1 3  1 3   1
L   + i  ,N   − i  , M − 
4 4  4 4   2
Quadratic satisfying B, C z2 +z+1=0 | M1 | Combining to give a 6th degree polynomial
Multiplying out some terms to give quadratics or cubics
Must see = 0
Quadratic satisfying L,N 4z2 −2z+1=0
Quadratic satisfying A, M 2z2 −z−1=0
Eqn satisfying all 6 points
M1
( )( )( )
z2 +z+1 4z2 −2z+1 2z2 −z−1 =0
⇒8z6 −7z3−1=0
A1
B1
B1
Finding A, B, C
Finding L, M, N
Combining to give a 6th degree polynomial
Multiplying out some terms to give quadratics or cubics
Must see = 0
Alternative methods
By rotational symmetry Because
2 3
𝜋𝜋i
3
�e � =
Calculation of vertices of L, M and N. Use of
1
1 + ω + ω2 = 0 (and ω3 = 1) to simplify eg (z
– ½(ω + 1))(z – ½(ω2 + 1)) ))(z – ½(ω + ω2))
= (z + ½ω2))(z + ½ω) ))(z + ½) etc or to find
sum/product of roots.
8 | (a) | (ii) | 𝑦𝑦̈ = 2𝑥𝑥̇ −78𝑦𝑦̇ = 2(−2𝑥𝑥+78𝑦𝑦+𝑘𝑘)−78𝑦𝑦̇
𝑦𝑦̈ = −2(𝑦𝑦̇ +78𝑦𝑦)+156𝑦𝑦+2𝑘𝑘−78𝑦𝑦̇
𝑦𝑦̈ +80𝑦𝑦̇ = 2𝑘𝑘
2
𝜆𝜆 +80𝜆𝜆 = 0
Complementary function is
𝜆𝜆 = 0,−80
Trial function is y −80𝑡𝑡
𝑦𝑦 = 𝐴𝐴+𝐵𝐵e
= 𝑎𝑎𝑘𝑘
So 80
𝑦𝑦̇ = 𝑎𝑎,𝑦𝑦̈ = 0
1
General solution is
𝑎𝑎 = 2𝑘𝑘 ⇔𝑎𝑎 = 40𝑘𝑘 | M1
M1
M1 | 3.1a
3.3
2.2a | 𝑦𝑦 𝑥𝑥.
Auxiliary equation for their DE
Correct trial function for their CF
Note the GS for in the main mark scheme achieves A1 but here
A1 is not awarded until GS for found.
𝑥𝑥
𝑥𝑥
−80𝑡𝑡 1
Alternative method
𝑦𝑦 = 𝐴𝐴+𝐵𝐵e +40𝑘𝑘𝑘𝑘
M1 | M1 | Integrating wrt and then using integrating factor | Integrating wrt and then using integrating factor
𝑦𝑦̇ +80𝑦𝑦 = 2𝑘𝑘𝑘𝑘+𝑐𝑐
∫80d𝑡𝑡 ∫80d𝑡𝑡
e (𝑦𝑦̇ +80𝑦𝑦)= (2𝑘𝑘𝑘𝑘+ 𝑐𝑐)e | M1 | 𝑦𝑦̈ +78𝑦𝑦̇ = 2𝑘𝑘 𝑘𝑘
Integrating wrt
d 80𝑡𝑡 80𝑡𝑡
(e 𝑦𝑦)= (2𝑘𝑘𝑘𝑘+𝑐𝑐)e
d𝑘𝑘
80𝑡𝑡 80𝑡𝑡
e 𝑦𝑦 = �(2𝑘𝑘𝑘𝑘+𝑐𝑐)e d𝑘𝑘+𝑑𝑑
80𝑡𝑡 | M1 | Integrating by p𝑘𝑘arts
80𝑡𝑡 (2𝑘𝑘𝑘𝑘+𝑐𝑐)e 2 80𝑡𝑡
e 𝑦𝑦 = −� 𝑘𝑘e d𝑘𝑘+𝑑𝑑
80 80
80𝑡𝑡
80𝑡𝑡 (2𝑘𝑘𝑘𝑘+𝑐𝑐)e 1 80𝑡𝑡
e 𝑦𝑦 = − 𝑘𝑘e +𝑑𝑑
80 3200
1 1 1 −80𝑡𝑡
𝑦𝑦 = 𝑘𝑘𝑘𝑘+ 𝑐𝑐− 𝑘𝑘+𝑑𝑑e
40 80 3200
−80𝑡𝑡 1
= 𝐴𝐴+𝐵𝐵e + 𝑘𝑘𝑘𝑘
M1
𝑦𝑦̈ +78𝑦𝑦̇ = 2𝑘𝑘 𝑘𝑘
Integrating wrt
−80𝑡𝑡 1
𝑦𝑦̇ = −80𝐵𝐵e + 𝑘𝑘
40
1
𝑥𝑥 = (𝑦𝑦̇ +78𝑦𝑦)
2
−80𝑡𝑡 39 1
When
= 39𝐴𝐴−𝐵𝐵e + 𝑘𝑘𝑘𝑘+ 𝑘𝑘
40 80
𝑘𝑘 = 0,𝑥𝑥 = 0,𝑦𝑦 = 0, 𝑦𝑦̇ = 0+0 = 0
0 = 𝐴𝐴+𝐵𝐵
1
0 = 39𝐴𝐴−𝐵𝐵+ 𝑘𝑘
80
1
0 = −80𝐵𝐵+ 𝑘𝑘
40
1 1
Particular solution for is
𝐵𝐵 = 𝑘𝑘 ,𝐴𝐴 = − 𝑘𝑘
3200 3200
𝑥𝑥
39 1 −80𝑡𝑡 39 1
𝑥𝑥 = − 𝑘𝑘− 𝑘𝑘e + 𝑘𝑘𝑘𝑘+ 𝑘𝑘
3200 3200 40 80
1 −80𝑡𝑡
= 𝑘𝑘(1−e +3120𝑘𝑘) | A1
M1
M1
A1
[7] | 1.1
3.1a
1.1
3.3 | GS of form
−80𝑡𝑡 39
𝑥𝑥 =𝐶𝐶 +𝐷𝐷e +40𝑘𝑘𝑘𝑘
Using two of to find simultaneous equations for
and (M1 for each equation).
𝑥𝑥 = 𝑦𝑦 = 𝑦𝑦̇ = 0 𝐴𝐴
𝐵𝐵
8 | (a) | (i) | dy dx
D.E. is +80y =78k where y =
dt dt
dy
I.F. e80t ⇒e80t +80e80ty =78ke80t
dt
d ( )
⇒ e80ty =78ke80t
dt
39k 39k
⇒e80ty = e80t + A⇒ y = + Ae−80t
40 40
dx 39k 39kt A
⇒ = + Ae−80t ⇒ x= − e−80t +B
dt 40 40 80 | M1
A1
M1
A1
PMT
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8 A biologist is studying the effect of pesticides on crops. On a certain farm pesticide is regularly applied to a particular crop which grows in soil. Over time, pesticide is transferred between the crop and the soil at a rate which depends on the amount of pesticide in both the crop and the soil. The amount of pesticide in the crop after $t$ days is $x$ grams. The amount of pesticide in the soil after $t$ days is $y$ grams. Initially, when $t = 0$, there is no pesticide in either the crop or the soil.

At first it is assumed that no pesticide is lost from the system. The biologist further assumes that pesticide is added to the crop at a constant rate of $k$ grams per day, where $k > 6$.

After collecting some initial data, the biologist suggests that for $t \geqslant 0$, this situation can be modelled by the following pair of first order linear differential equations.\\
$\frac { d x } { d t } = - 2 x + 78 y + k$\\
$\frac { d y } { d t } = 2 x - 78 y$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\frac { d ^ { 2 } x } { d t ^ { 2 } } + 80 \frac { d x } { d t } = 78 \mathrm { k }$.
\item Determine the particular solution for $x$ in terms of $k$ and $t$.

If more than 250 grams of pesticide is found in the crop, then it will fail food safety standards.
\item The crop is tested 50 days after the pesticide is first added to it.

Explain why, according to this model, the crop will fail food safety standards as a result of this test.

Further data collection suggests that some pesticide decays in the soil and so is lost from the system. The model is refined in light of this data. The particular solution for $x$ for this refined model is\\
$\mathrm { x } = \mathrm { k } \left( 20 - \mathrm { e } ^ { - 41 \mathrm { t } } \left( 20 \cosh ( \sqrt { 1677 } \mathrm { t } ) + \frac { 819 } { \sqrt { 1677 } } \sinh ( \sqrt { 1677 } \mathrm { t } ) \right) \right.$.
\end{enumerate}\item Given now that $k < 12$, determine whether the crop will fail food safety standards in the long run according to this refined model.

In the refined model, it is still assumed that pesticide is added to the crop at a constant rate.
\item Suggest a reason why it might be more realistic to model the addition of pesticide as not being at a constant rate.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2022 Q8 [13]}}