H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
1.
Show, by means of the substitution w x 2 y , that the differential equation x
dy dx
2 y 3 xy 0
can be reduced to the form dw dx
3w
[2]
Hence find y in terms of x, given that y
1 when x 2 . 2
[4] [2010/IJC/Prelim/I/6]
[Solution] Given: w x 2 y Differentiate w.r.t x dw dy 2 xy x 2 dx dx
x (1) x :
dy dx
2 y 3xy 0 ------- (1)
x2
dy dx
2 xy 3x 2 y 0 dw 3w 0 dx dw 3w dx
dw 3w dx
1 1 dw 1 dx 3w 1 ln w x c 3 ln w 3 x 3c w e 3 x 3c
(shown)
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
w Ae 3 x , where A is a constant x 2 y Ae 3 x
Given that y
1 when x 2 , 2
1 22 Ae 6 2 A 2e6
Thus,
x 2 y 2e6 e 3 x
y
2 6 3 x e x2
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
2.
(a) Verify that y = x is a particular solution of the differential equation dy x 2 y 2 , dx 2 xy
x, y 0
[2]
(b) Show that the substitution y = ux reduces the differential equation dy x 2 y 2 dx 2 xy
to the differential equation x
du 1 u 2 dx 2u
Hence find the general solution of the differential equation dy x 2 y 2 dx 2 xy
[3]
[4]
(c)Due to a rapid disease outbreak, the population of fish in a river, x (in thousands), is believed to obey the differential equation d2x 4ae 2t 2 dt where t is the time in days, and a > 0 is a constant. Given that the entire population of fish is wiped out by the disease eventually, show that the general solution of the differential equation is x ae 2t . [3] Explain the meaning of a, in the context of the question. Sketch the family of solution curves of the differential equation for a = 1 and 2. [2] [2010/CJC/Prelim/II/4] [Solution] dy
1, (a) Since y = x and dx LHS = 1 RHS =LHS
(b)
y ux dy du ux dx dx du x 2 u 2 x 2 ux dx 2 x(ux) ux
du dx du x dx x
du 1 u 2 dx 2u 2 1 u u 2u 1 u 2 2u 2 2u
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
x
du 1 u 2 ( shown ) dx 2u
2u
1 dx x 2u 1 1 u 2 du x dx
1 u
2
du
ln(1 u 2 ) ln x C A 1 u2 x A u2 1 x 2 y A 1 x2 x 2 2 y x Ax
d 2x 2 t (c) dt 2 4ae dx 2ae 2t C dt x ae 2t Ct D
Since entire population is wiped out by the disease eventually, as t , x 0 Hence, C = 0, D = 0. x ae 2t
a represents the initial population of the fish (in thousands). x 2 e 2 t
x e 2t
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
3.
2 A particular solution of a differential equation is given by ( x y ) 2 xy 2 Show that ( y y )
dy x dx
2 3 y. 3 [2]
A second, related, family of curves is given by the differential equation dy y2 y dx By means of the substitution y ux , show that the general solution for y, in x
terms of x, is y
x , where c is an arbitrary constant. xc
[3]
Sketch, on a single diagram, three distinct members of the second family of solution curves, stating clearly the coordinates of the points where the curves cross the axes and the equations of any asymptotes. [5] [2010/JJC/Prelim/II/2] [Solution] 2 3 y 3 2 ( x y ) 2 2 xy y 3 3 2 y 2 x2 y3 3 dy dy 2 y 2 x 2 y 2 dx dx dy ( y y 2 ) x (shown) dx ( x y ) 2 2 xy
Alternative: dy dy dy 2( x y ) 1 2 x 2 y 2 y 2 dx dx dx dy ( y y 2 ) x (shown) dx
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
dy y2 y dx use y ux dy du xu dx dx 2 du x x u ux ux dx du x u u2 x u dx 1 du 1 u 2 dx 1 u 2 du 1 dx 1 x c , where c is an arbitrary constant u x xc y x c y 1 xc xc x
Family of solution curves:
c 1, y 1 c 0, y 1 c 1, y 1
1 x 1
1 x 1
c 1 c 1 y
y 1
c0
0
x 1
x 1
x
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
4.
An innovation is introduced into a community of 100 farmers at time t 0 . Let x denote the number of farmers who have adopted the innovation at time t. Assume that x is a continuous function of time. The rate at which the number of farmers in that community who adopted the innovation at a particular instant is proportional to the product of the number of farmers who have already adopted and the number of farmers who have not adopted the innovation. Initially, one farmer adopted the innovation and the rate at which the number of farmers who adopted the innovation is one farmer per unit time. dx (i) Show that [1] k(100x x 2 ) , where k is to be determined. dt (ii) Find the particular solution of x , in terms of t. [5] (iii) Sketch the graph of x versus t, for t 0 .
[2]
(iv) Using the graph in (iii) or otherwise, find the time taken for 75% of the population of farmers to adopt the innovation, leaving your answer to 2 decimal places. [1] (v) Give a reason why the model may not be suitable. [Solution] dx k ( x)(100 x) (i) dt dx At t 0 , x 1, 1 dt 1 k (1)(99) k therefore
(ii)
1 99
dx 1 (100 x x 2 ) dt 99
dx 1 (100 x x 2 ) dt 99
1 dx x(100 x)
1 100
1 dt 99
1 1 dx x 100 x
ln x
1 dt 99
[1] [2010/NYJC/Prelim/I/10]
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
ln
x 100
x , where A is a constant. 100
When t 0 , x 1 , A x
1 99
(iii) 100
x
0
t
(iv) Using GC, 5.64 years. (v) The farmers may be influenced by adoption of innovation from other sources, e.g. mass media, besides farmers. Or any other reasonable answer.
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
5.
In an experiment, Andy and Bob attempt to devise a formula to describe how the volume of water, V m3, in a tank, changes with time at t hours. (i) Andy gives his formula as dV 1 V 82 . Given that V 1 when t 0, dt 60 V t
show that V 3 8 7e 20 . Sketch this solution curve. (ii) However, Bob believes that it is more likely to be
[4]
d 2V 12t 2 2. Given that 2 dt
dV 0 when t 0 , show that the general solution for V can be expressed as dt 1 1 V (t 2 ) 2 C , where C is a constant. 2 4 Hence, or otherwise, sketch on a single diagram, two distinct members of the family of solution curves. [5]
(iii) It is also given that
when in Bob’s model. Suppose that the water V 1 t 0 in the tank does not overflow, explain, using your diagrams in parts (i) and (ii), why Andy’s model is more appropriate compared to Bob’s model. [2] [2010/DHS/Prelim/II/3]
[Solution] dV 1 8 1 V 3 8 (V 2 ) ( 2 ) dt 60 V 60 V (i) 2 V dV 1 3 V 8 dt 60 1 1 ln | V 3 8 | t C ' 3 60 t
C ''
| V 3 8 | e 20
V 3 8 Ae 20 ,
t
,
C '' 3C ' A eC ''
When t 0, V 1, A 7,
t
3
V = 8 7e 20 V
(ii)
d 2V 12t 2 12 2 dt
8 20ln( )or 2.67 7
t
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
dV 4t 3 2t C1 dt dV When t 0, 0, C1 0 dt dV 4t 3 2t dt V t 4 t 2 C , C is a constant.
V t4 t2 C 1 1 (t 2 )2 (C ) 2 4 V C
C
1 4 1 (II)C 4 1 (I)C
t
When t 0, V 1, then C2 1 . 4 (iii) Therefore given the above initial condition, Bob’s model corresponds to solution curve type (I) in part (ii). Therefore in Bob’s model, the volume of water approaches infinity in the long run (not realistic) whereas in Andy’s model, the volume of water reasonably diminishes to zero in the long run/after some time. Thus, Andy’s model is more appropriate than Bob’s model.
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
6.
Newton’s Law of Cooling states that the rate at which the temperature of a body falls is proportional to the amount by which its temperature exceeds that of its surroundings. At time t minutes after cooling commences, the temperature of the body is o C . Assuming that the room temperature remains constant at 30 o C and the body has an initial temperature of 90 o C , show that 30 60e kt , where k is an arbitrary constant. [5] Given that it takes 8 minutes for the temperature of the body to drop from 90 o C to 55 o C , determine how much more time is needed for the body to cool to 35 o C , leaving your answer to one decimal place. [3] [2010/MJC/Prelim/I/6] [Solution] d k 30 , k 0 dt 1 30 d k dt ln 30 kt C
30 e kt C
when t 0, 90 90 30 A A 60 30 60e kt (shown)
when t 8, 55 55 30 60e 8 k 5 e 8 k 12 1 5 k ln 8 12
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
when 35, 1 5 ln t 8 12
35 30 60e 1 5 ln t 8 12
1 12 t 22.707 e
additional time needed 22.707 8 14.7 min (1d.p.)
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
7.
The rate at which a substance evaporates is proportional to the volume of the substance which has not yet evaporated. The initial volume of the substance is A m 3 and the volume which has evaporated at time t minutes is x m 3 . Given that it
12t takes (2ln2) minutes for half of its initial volume to evaporate, show that x A 1 e . Find the additional time needed for three quarters of the substance to evaporate, giving your answer in exact form. [6] [2010/YJC/Prelim/I/11] [Solution] dx ( A x) dt
dx k ( A x) , k > 0 dt
1 dx k dt A x ln(A x) = kt + C
where C is a constant
ln(A – x) = – kt – C A – x = e -kt(B) where B = e −C is a constant x = A – B e-kt
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
When t = 0, x = 0 B = A When t = 2 ln 2, x =
1 2
A
1 A A Ae k ( 2 ln 2 ) 2
e k ( 2 ln 2 )
1 2 1 2
k (2 ln 2) ln( ) ln 2 k
1 2
x A(1 e
When x =
1 t 2
) (Ans)
3 A, 4 1
t 3 A A(1 e 2 ) 4
e
1 t 2
1 4
1 1 t ln( ) 2 ln 2 2 4
t = 4 ln 2
Additional time required = 4 ln2 – 2 ln2 = (2 ln 2) minutes (Ans)
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
8.
An economist is studying how the annual economic growth of 2 countries varies with time. The annual economic growth of a country is measured in percentage and is denoted by G and the time in years after 1980 is denoted by t. Both G and t are taken to be continuous variables. (i) Country A is a developing country and the economist found that G and t are can be modeled by the differential equation
dG G 1 . Given that, when dt 2
t 0 , G 0 , find G in terms of t.
[4] (ii) Comment on the suitability of the above differential equation model to forecast the future economic growth of Country A. [1] (iii) Country B is a developed country and the economist found that G and t can dG G 1 . dt 2 Given that Country B has been experiencing decreasing economic growth during the period of study, sketch a member of the family of solution curves of the differential equation model for Country B. Hence, comment on the economic growth of Country B in the long term. [2] [2010/ACJC/Prelim/I/6] be modeled by the differential equation
[Solution] dG G 1 dt 2 1 1 G 1 dG 2 dt ln G 1 0.5t C G 1 e0.5t C G 1 Ae0.5t , where A eC When t 0 , G 0 , A 1 G 1 e0.5t Examples of possible comments: The model is not suitable because … The economist is assuming that that there are no fluctuations in the economic growth in the future. The economist is assuming that the country will enjoy perpetual economic growth in the long term. The economist is assuming Country A is always experiencing positive and increasing economic growth in the future.
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
Factors affecting economic growth remains unchanged. G t
0
1
G1Be0.5t,B1
In the long term, Country B is expected to be still in recession with an economic growth decreasing towards -1%.
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
9.
An underground storm canal has a fixed capacity of 6000 m3 and is able to discharge rainwater at a rate proportional to V , the volume of rainwater in the storm canal. On a particular stormy day, rainwater is flowing into the canal at a constant rate of 300 m3 per minute. The storm canal is initially empty. Let t be the time in minutes for which the rainwater had been flowing into the storm canal, (i) show that V
300(1 e kt ) , where k is a positive constant. k
[4]
A first alarm will be sounded at the control room when the volume of rainwater in the storm canal reaches 4500 m3 and a second alarm will be sounded when the storm canal is completely filled. Given that the first alarm was sounded 20 minutes after the rainwater started flowing into the storm canal. (ii) Find the time interval between the first and second alarm. (Assuming the weather condition remains unchanged). [3] (iii) Briefly discuss the validity of the model for large values of t . [1] [2010/AJC/Prelim/II/5] [Solution] dV 1 300 kV dV 1 dt (i) dt 300 kV 1 ln 300 kV t C k 300 kV e k ( t C ) 300 kV Ae kt When t 0 , V 0 300 0 Ae A 300 V 0
300(1 e kt ) (Shown). k
(ii) When t 20 , V 4500 , 4500
300(1 e 20 k ) 15k (1 e 20 k ) k
From the GC, k 0.030293 2nd alarm : when V 6000 300(1 e 0.030293t ) t 30.7 6000 0.030293 The residents will have 10.7 minutes between the 1st and 2nd alarm. (iii) t V
300 9903 m3 which is impossible as the canal has only 0.030293
a fixed volume of 6000 m3 . The model is not valid for large values of t.
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
10. In a chemical reaction a compound X is formed from a compound Y. The mass in grams of X and Y present at time t seconds after the start of the reaction are x and y respectively. The sum of the two masses is equal to 100 grams throughout the reaction. At any time t, the rate of formation of X is proportional to the mass of Y at that time. When t = 0, x = 5 and
dx 1.9 . dt
dx 0.02(100 x) . [2] dt (ii) Solve this differential equation, obtaining an expression for x in terms of t.[4] (i) Show that x satisfies the differential equation
(iii) Calculate the time taken for the mass of compound Y to decrease to half its initial value. [2] (iv) Sketch the solution curve obtained in part (ii) and state what happens to compounds X and Y as t becomes very large. [3] [2010/SAJC/Prelim/II/2] [Solution] (i) x + y = 100 dx dx (100 x) k (100 x) , where k is a constant/ dt dt 1.9 = k (100 – 5) k = 0.02 dx 0.02 (100 x) dt (ii)
dx 0.02 (100 x) dt 1 dx 0.02 dt 100 x ln 100 x 0.02t C
When t = 0, x = 5, ln(95) = C So, ln 100 x 0.02t ln(95) ln 100 x ln(95) 0.02t
since x 100
100 x 0.02t 95 100 x e 0.02t 95 x 100 95e 0.02 t (iii) When t = 0, x = 5 y = 95 Initial value of y = 95 Half of initial value = 47.5 When y = 47.5, x = 52.5 1n
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
52.5 100 95e 0.02 t t 34.7 (iv) x 100 x 100 95e0.02t
As t , x 100, y 0 . Compound Y will be transformed almost completely to compound X.
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
11. A long cylindrical metal bar is submerged into iced water. A researcher claims that the rate at which the length, l cm, of the bar is shrinking at any time t seconds is proportional to the volume of the bar at that instant, assuming that the crosssectional area of the bar remains constant during the shrinking process. Formulate and integrate a differential equation to show that l Ae kt , where A and k are constants. State the range of values of A and of k. [5] Given that the initial length of the bar is L, sketch a graph to show the relation between l and t. Comment on the suitability of the claim by the researcher. [3] It is later found that l and t are related by the equation l B Le kt , where B is some constant. Given that l = 0.5L when t = T, show that the length of the bar when t = 3T, is given by
0.5L B B L2
3
.
[2] [2010/RVHS/Prelim/II/4]
[Solution] dl kl dt 1 l dl k dt ln l kt C (note: l > 0)
l e kt C l Ae kt (shown) A 0 and k 0 (Ans)
When t = 0, l = L: A = L l Le kt l L
0
t
The model suggested by the researcher is not suitable as l 0 when t (i.e. the bar vanished).
H2 Math (9740) 2010 MSM: DIFFERENTIAL EQUATION
t T , l 0.5L 0.5L B Le kT
0.5 L B e kT L
When t = 3T, l B Le
3kT
=
0.5L B B L2
3
(shown)