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THE UNIVERSITY OF MELBOURNE
DEPARTMENT of MATHEMATICS AND STATISTICS
620-151 INTRODUCTION TO BIOMEDICAL
MATHEMATICS
EXAMINATION — Semester 1, 2004
Solutions.
Page 2 of 22
1. Use strict Gauss-Jordan elimination with augmented matrices on each of the systems of linear equations below to find all the solution(s), if one or more exist, for
(x1 , x2 , x3 ).
In the case of a single solution, your final matrix should allow you to answer the
question without further algebraic manipulation: your augmented matrix must be
in Reduced Row Echelon form. If multiple solutions exist, state them all. If no
solution exists, state why such a deduction can be made from your final matrix.
(a)
3x − 15y + 21z = 54
−2x + 8y − 12z = −32
−x + 6y − 5z = −5
(b)
−3x + 15y − 48z = −48
4x − 22y + 70z = 70
2x − 4y + 14z = 16
Show the row operations used at each step. Check your answers.
(a) Write as augmented matrix


3 −15 21
54
 −2 8 −12 −32 
−1 6
−5 −5
Divide R1 by 3

18
1 −5 7
 −2 8 −12 −32 
−1 6 −5 −5

Add 2*R1 to R2. Add R1 to R3


1 −5 7 18
 0 −2 2 4 
0 1 2 13
Divide R2 by -2

1 −5 7 18
 0 1 −1 −2 
0 1
2 13

Add 5*R2 to R1. Add -1*R2 to R3


8
1 0 2
 0 1 −1 −2 
0 0 3 15
Divide R3 by 3

8
1 0 2
 0 1 −1 −2 
0 0 1
5

Page 3 of 22
Add R3 to R2. Add -2*R3 to R1


1 0 0 −2
 0 1 0 3 
0 0 1 5
Hence (x1 , x2 , x3 ) = (−2, 3, 5). Check the answer:
3 × −2 −15 × 3 +21 × 5 = −6 − 45 + 105 = 54
−2 × −2 +8 × 3 −12 × 5 = 4 + 24 − 60 = −32
−1 × −2 +6 × 3 −5 × 5
= 2 + 18 − 25
= −5
• Write as augmented matrix


−3 15 −48 −48
 4 −22 70
70 
16
2 −4 14
Divide R1 by -3

1 −5 16 16
 4 −22 70 70 
2 −4 14 16

Add -4*R1 to R2. Add -2*R1 to R3.


1 −5 16
16
 0 −2 6
6 
0 6 −18 −16
Divide R2 by -2

16
1 −5 16
 0 1 −3 −3 
0 6 −18 −16

Add 5*R2 to R1. Add -6*R2 to R3


1 0 1
1
 0 1 −3 −3 
0 0 0
2
STOP! We see that the last row is equivalent to the equation 0 = 2 which does
not make sense. Hence the system of equations is inconsistent and does not
have a solution.
Page 4 of 22
2. (a) Given the matrices


7 −8
A= 3
5
−9 4
and
0 −1 −5 5
B=
,
−7 1 −1 3
find the product AB.
(b) Let U and V be the following two matrices:




1
2
3
1 −4 −3
11
15  .
and
V = 3
U = −3 −7 −6
−4 −14 −19
2
6
5
Given that UV = V U = I, where I is the 3 × 3 multiplicative identity matrix,
solve the (x, y, z) in terms of α in the following system:
2x − 8y − 6z = −6α
−6x − 14y − 12z = 4α
4x + 12y + 10z = 2α
Do not use Gauss-Jordan elimination. Check your answer.
(a) Do very explicit matrix multiplication. A is a 3 × 2 matrix and B is a 2 × 4
matrix. Hence the product AB will be a 3 × 4 matrix:


7 × 0 + −8 × −7 7 × −1 + −8 × 1 7 × −5 + −8 × −1 7 × 5 + −8 × 3
3 × −1 + 5 × 1
3 × −5 + 5 × −1
3×5+5×3 
−9 × 0 + 4 × −7 −9 × −1 + 4 × 1 −9 × −5 + 4 × −1 −9 × 5 + 4 × 3
AB =  3 × 0 + 5 × −7
which is


56 −15 −27 11
AB = −35 2 −20 30 
−28 13
41 −33
(b) Write the system as a matrix product

  

2 −8 −6
x
−6α
−6 −14 −12 y  =  4α 
4
12
10
z
2α
We see that the matrix on the left-hand side is simply 2U. Hence we can
simplify the expression a bit:
 
 
x
−6
2U y  = α  4 
z
2
Since V U = I we multiply both sides of the equation on the left by V :
 
 
x
−6



2V U y = αV 4 
z
2
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This gives:
or
 
 
 
x
x
−6
2I y  = 2 y  = αV  4 
z
z
2
 
 
x
−3
y  = αV  2 
z
1
Multiplying out the right-hand side gives:
 


x
4
y  = α  28 
z
−35
Hence (x, y, z) = (4α, 28α, −35α). Checking the answer to be sure:
2 × 4α − 8 × 28α − 6 × −35α = α(8 − 224 + 210) = 6α
−6 × 4α − 14 × 28α − 12 × −35α = α(−24 − 392 + 420) = 4α
4 × 4α + 12 × 28α + 10 × −35α = α(16 − 336 − 350) = 2α
Page 6 of 22
3. (a) Two variables, x and y, are related by the equation:
x3 y 2
− loge (x + y) = 1
8
When x = 2 and y = −1, x is increasing at a rate of 4 units per minute. What
is the rate of change of y at x = 2 and y = −1?
(b) Find the slope of the curve
2
(x2 + y)exy − 3e−4 = 0
at (x, y) = (−1, 2).
(a) Since we need to find dy
and we are given dx
we differentiate both sides of the
dt
dt
equation with respect to t:
d
d
d x3 y 2
− log(x + y) =
1
dt
8
dt
dt
x3 d 2 y 2 d 3
1 d
y +
x −
(x + y) = 0
8 dt
8 dt
x + y dt
1
dx dy
x3 dy y 2 2 dx
2y
+ 3x
−
+
= 0
8
dt
8
dt
x + y dt
dt
dx dy
x3 y dy 3x2 y 2 dx
1
+
−
+
= 0
4 dt
8 dt
x + y dt
dt
We don’t need to isolate dy
yet. Substitute x = 2, y = −1 and
dt
above:
dy
−8 dy 12
1
4+
= 0
+ 4−
4 dt
8
1
dt
dy
dy
−2 + 6 − 4 −
= 0
dt
dt
Now isolate
dy
:
dx
dx
dt
= 4 into the
dy
= 2/3
dt
dy
, so we differentiate both sides of the equation with respect
(b) We need to find dx
to x.
d 2
d
2
(3e−4 ) = 0
(x + y)exy −
dx
dx
d
2 d
2
(x2 + y) (exy ) + exy
(x2 + y) + 0 = 0
dx
dx dy
2
xy 2
2
xy 2 d
2x +
(xy ) + e
= 0
(x + y)e
dx
dx
d 2
dy
2
2
(x + y) x y + y + 2x +
= 0
dx
dx
dy
dy
2
2
+ y + 2x +
= 0
(x + y) 2xy
dx
dx
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dy
yet. Substitute in x = −1 and y = 2 into the above and
Do not isolate dx
dy
then isolate dx :
dy
dy
+4 −2+
= 0
(1 + 2) −4
dx
dx
dy
dy
−12
+ 12 − 2 +
= 0
dx
dx
Hence
dy
dx
= 10/11 when (x, y) = (−1, 2).
Page 8 of 22
4. Find the antiderivative of the following functions:
29 − x
+ 5x − 14
3 cos(x) − 3 sin(x)
(b) f (x) =
cos(x) + sin(x)
(c) f (x) = x loge xπ
(a) f (x) =
x2
(a) This looks like a partial fractions question. Factorise as much as possible:
x2
29 − x
29 − x
=
+ 5 − 14
(x − 2)(x + 7)
Convert it to partial fraction form:
A
B
A(x + 7) + B(x − 2)
29 − x
=
+
=
(x − 2)(x + 7)
x−2 x+7
(x − 2)(x + 7)
Look at the numerators: 29 − x = A(x + 7) + B(x − 2). When x = −7 this
gives 36 = −9B and so B = −4. When x = 2 this gives 27 = 9A and so
A = 3. Hence
3
4
29 − x
=
−
(x − 2)(x + 7)
x−2 x+7
Check that we have the correct form:
3
4
3(x + 7) − 4(x − 2)
3x + 21 − 4x + 8
29 − x
=
−
=
=
.
(x − 2)(x + 7)
x−2 x+7
(x − 2)(x + 7)
x2 + 5x − 14
Now integrate:
Z
The integral
R
Z
dx
x+a
x2
29 − x
dx =
x2 + 5 − 14
Z
3dx
−
x−2
Z
4dx
x+7
= loge |x + a| and so
29 − x
dx = 3 loge |x − 2| − 4 loge |x + 7| + C
+ 5 − 14
where C is a constant. Simplifying a bit further gives:
Z
(x − 2)3 29 − x
+C
dx = loge x2 + 5 − 14
(x + 7)4 (b) We see that the numerator is (almost) the derivative of the denominator.
Hence we substitute u = cos(x) + sin(x), and then du
= cos(x) − sin(x). Thus
dx
Z
Z
3 cos(x) − 3 sin(x)
3 du
dx =
dx
cos(x) + sin(x)
u dx
Z
3
du
=
u
= 3 loge |u| + C
= 3 loge | cos(x) + sin(x)| + C
where C is some constant.
Page 9 of 22
(c) First simplify:
f (x) = x loge xπ = πx loge x
This doesn’t look like a partial fraction or a simple substitution, so it is probably an integration by parts.
Z
Z
du
dv
u dx = uv − v dx
dx
dx
Try with u = loge x and
dv
dx
= x. This means that
du
= 1/x
dx
and v = x2 /2.
Substitute this into the integrate-by-parts formula:
Z 2
Z
x 1
x2
loge x −
dx
x loge x =
2
2 x
Z
x2
x
loge x −
dx
=
2
2
x2
x2
=
loge x −
+C
2
4
Hence
Z
x loge xπ =
where C is some constant.
πx2
(2 loge x − 1) + C
4
Page 10 of 22
5. Use the simplex method to solve the following standard maximum problem.
Maximise
h = 8x1 + 5x2 +4x3
subject to
3x2 + 6x3 ≤ 12
2x1 + x2 + 4x3 ≤ 6
5x1 + 2x2 + 4x3 ≤ 16
with x1 ≥ 0, x2 ≥ 0 and x3 ≥ 0.
At each step show clearly the row operation(s) that you perform and clearly circle
the pivot element. Inspect your final tableau and state the maximum possible value
of h and all the values of (x1 , x2 , x3 ) for which this maximum occurs. Check your
answer.
First add slack variables, s1 , s2 , s3 :
3x2 + 6x3 +s1 = 12
2x1 + x2 + 4x3 +s2 = 6
5x1 + 2x2 + 4x3 +s3 = 16
with the constraint s1 , s2 , s3 ≥ 0.
Form the simplex tableau:


x1 x2 x3 s1 s2 s3 P RHS
 0
3
6 1 0 0 0
12 



 2
1
4
0
1
0
0
6


 5
2
4 0 0 1 0
16 
−8 −5 −4 0 0 0 1
0
−8 is the most negative element of the bottom row, so the first column is the pivot
column. Work out the ratios:


x1 x2 x3 s1 s2 s3 P RHS Ratio
 0
3
6 1 0 0 0
12
◦ 


 (2) 1
4 0 1 0 0
6
3 


16

 5
2
4 0 0 1 0
16
5
−8 −5 −4 0 0 0 1
0
Hence the second row is the pivot row and “2” is
by 2:

x1 x2 x3 s1 s2 s3
 0
3
6 1
0
0

 1 1/2 2 0 1/2 0

 5
2
4 0
0
1
−8 −5 −4 0
0
1
Add -5*R2 to R3. Add 8*R2 to R4.

x1
x2
x3
 0
3
6

 1 1/2
2

 0 −1/2 −6
0
−1 12
the pivot element. Divide Row 2
P
0
0
0
0
s1
s2
s3 P
1
0
0 0
0 1/2 0 0
0 −5/2 1 0
0
4
0 1
RHS
12
3
16






RHS
12
3
1
24






Page 11 of 22
We have only one choice of pivot column
ratios:

x1
x2
x3 s1
s2
 0
(3)
6
1
0

 1 1/2
2 0 1/2

 0 −1/2 −6 0 −5/2
0
−1 12 0
4
Hence the first row is the pivot row

x1
x2
x3
 0
1
2

 1 1/2
2

 0 −1/2 −6
0
−1 12
— the second column. Work out the

s3 P RHS Ratio
0 0
12
4 

0 0
3
6 

1 0
1
◦ 
0 1
24
and “3” is the pivot element. Divide R1 by 3:

s1
s2
s3 P RHS
1/3
0
0 0
4 

0
1/2 0 0
3 

0 −5/2 1 0
1 
0
4
0 1
24
Add -1/2*R1 to R2. Add 1/2*R1 to R3. Add 1*R1 to R4.

x1 x2 x3
s1
s2
s3 P RHS
 0 1 2
1/3
0
0 0
4

 1 0 1 −1/6 1/2 0 0
1

 0 0 −5 1/6 −5/2 1 0
3
0 0 14 1/3
4
0 1
28






Here we must stop because there are no negative elements in the bottom row. We
can now read off the solution.
P = 28 − 14x3 − s1 /3 − 4s2
Hence the maximum P is 28 and x3 = s1 = s2 = 0. Reading off the other variables
we see that x1 = 1 and x2 = 4 and s3 = 3. ie the maximum P is 28 occurring at
(x1 , x2 , x3 ) = (1, 4, 0) and (s1 , s2 , s3 ) = (0, 0, 3).
Checking our answer:
P = 8 × 1 + 4 × 5 = 28
and the inequalities:
3 × 4 + 6 × 0 = 12
2×1+1×4+4×0 = 6
5 × 1 + 2 × 4 + 4 × 0 = 13
This also agrees with the slack variables.
Page 12 of 22
6. (a) Find the Taylor Polynomial of degree 2 for the function
1
about the point x = 0.
f (x) = loge √
x2 + e
Express all coefficients as vulgar fractions, NOT decimals.
(b) (i) Consider the initial value problem
dy
= xy 2 + 2 loge (y)
dx
with y = 1 when x = 1.
Find the Taylor Polynomial of degree 3 for y(x) about x = 1.
Express all coefficients as vulgar fractions, NOT decimals.
(ii) Estimate the value of y at x = 1.1 to 3 decimal places using the Taylor
polynomial above.
(a) First simplify the function:
1
f (x) = − loge (x2 + e)
2
In order to find the Taylor polynomial of degree 2 we need to find f 0 and f 00 :
1 1
d 2
x
(x + e) = − 2
2
2 x + e dx
x +e
d
dx
2
2
(x + e) dx − x dx (x + e)
f 00 (x) = −
(x2 + e)2
x2 + e − 2x2
x2 − e
= −
=
(x2 + e)2
(x2 − e)2
f 0 (x) = −
Hence at x = 0 we have
f (0) = −1/2
f 0 (0) = 0
f 00 (0) = −1/e
Substitute this into the Taylor polynomial formula:
1
P (x + a) = f (a) + f 0 (a)(x − a) + f 00 (a)(x − a)2
2
1 2
P (a) = −1/2 + 0 − a
2e
And so our degree 2 Taylor polynomial approximation to the function f (x)
about x = 0 is
1
−1/2 − x2
2e
Page 13 of 22
(b-i) In order to compute the Taylor polynomial of degree 3 about x = 1 we need
to know y(1), y 0(1), y 00(1) and y 000 (1). From the DE we have
dy
= xy 2 + 2 loge y
dx
and so setting x = 1 and y = 1 we get
y 0(1) = 1 + 2 log 1 = 1
Differentiating both sides of the DE by x gives:
y 00 (x) = 2xy(x)y 0(x) + y(x)2 +
2y 0(x)
y(x)
Substituting x = 1, y(1) = 1 and y 0 (1) = 1 gives
y 00(1) = 2 + 1 + 2 = 5
Differentiate again:
y 00 y − (y 0)2
y2
y 00y − (y 0 )2
= 4yy 0 + 2xyy 00 + 2x(y 0)2 + 2
y2
y 000 = 2yy 0 + 2x(y 0)2 + 2xyy 00 + 2yy 0 + 2
Substitute x = 1, y(1) = 1, y 0(1) = 1, y 00(1) = 5 into this:
y 000 (1) = 4 + 10 + 2 + 2
5−1
= 24
1
We can now form the Taylor polynomial about x = 1:
1
1
P (x) = y(1) + y 0(1)(x − 1) + y 00(1)(x − 1)2 + y 000 (1)(x − 1)3
2
6
5
2
3
= 1 + (x − 1) + (x − 1) + 4(x − 1)
2
(b-ii) Using the above Taylor polynomial we can put x = 1.1 and get an approximation of y(1.1):
5
× 0.01 + 4 × 0.001
2
= 1 + 0.1 + 0.025 + 0.004 = 1.129
y(1.1) = 1 + 0.1 +
Page 14 of 22
7. Solve graphically the following non-standard linear programming problem: Draw a
graph with feasible region clearly marked and with all its corner points calculated.
Write down all basic feasible solutions and write down the values of (x, y) for which
P takes its maximum value.
Maximise
P = 39x + 13y
subject to
21x + 7y ≤ 63
12x − 6y ≥ −24
42x − 14y ≤ 42
with x ≥ 0 and y ≥ 0. Check your answer.
Work out the intersection to create plot:
• Intersection of x = 0 and y = 0 is (0, 0).
• Intersection of y = 0 and the constraints gives (3, 0), (−2, 0) and (1, 0).
• Intersection of x = 0 and the constraints gives (0, 9), (0, 4) and (0, −3).
Use these to plot the graph:
y
(0,9)
12x−6y=−24
(1,6)
42x−14y=42
(0,4)
FR
(2,3)
21x+7y=63
(0,0)
(−2,0)
(1,0)
(0,−3)
(3,0)
x
Page 15 of 22
Work out the remaining corner points by finding the intersections of the lines:
• Intersection of 21x + 7y = 63 and 12x − 6y = −24.
−12 6 24
21 7 63
∼
2 −1 −4
3 1
9
∼
1 0 1
3 1 9
∼
6 0 12
3 −1 3
∼
2 −1 −4
1 0
7
5 0 5
3 1 9
∼
1 0 2
3 −1 3
1 0 1
0 1 6
So the intersection is at (x, y) = (1, 6).
• Intersection of 21x + 7y = 63 and 42x − 14y = 42.
21 7 63
42 −14 42
∼
3 1 9
3 −1 3
∼
∼
1 0
2
0 −1 −3
So the intersection is at (x, y) = (2, 3).
• Intersection of 12x − 6y = −24 and 42x − 14y = 42.
12 −6 −24
42 −14 42
∼
2 −1 −4
3 −1 3
∼
∼
0 −1 −18
1 0
7
So the intersection is at (7, 18).
Of these intersection points, only the following are basic feasible solutions:
• (0, 0) — P = 0
• (1, 0) — P = 39
• (2, 3) — P = 2 × 39 + 3 × 13 = 78 + 39 = 117
• (1, 6) — P = 39 + 6 × 13 = 39 + 78 = 117
• (0, 4) — P = 4 × 13 = 52
Hence the maximum value of P is 117 and occurs at two different basic feasible
solutions (2, 3) and (1, 6), and so also occurs at all the points on the line segment
between then.
The equation for this line is y = mx + c. The gradient is (6 − 3)/(1 − 2) = −3.
Hence y = c − 3x. It goes through the point (1, 6), so 6 = c − 3 and c = 9. The
equation is y + 3x = 9.
Thus the maximum is
P = 117
at all points (x, 9 − 3x) where 1 ≤ x ≤ 2
Check the answer.
• P = 39x + 13y = 39x + 13(9 − 3x) = 13 × 9 = 117.
• 21x + 7y = 21x + 7(9 − 3x) = 63 ≤ 63.
• 12x − 6y = 12x − 6(9 − 3x) = −54 + 30x ≥ −24 provided x ≥ 1.
• 42x − 14y = 42x − 14(9 − 3x) = 84x − 126 ≤ 42 provided x ≤ 2.
Page 16 of 22
8. (a) Find the general solution y(x) of the differential equation
x cos(πx)
dy
=
dx
y
(b) Solve the initial value problem, for t ≥ 1,
2t
√
dx
3
+ 3x = 12 t3 et
dt
and
x(1) = 2e + 2
Show all intermediate steps in your solution of the differential equation.
(a) This DE is separable:
1
dy
= × x cos(πx)
dx
y
Hence
Z
x cos πxdx =
Z
dy
y dx =
dx
Z
ydy
Solve the RHS first.
Z
ydy =
y2
+C
2
where C is some constant.
Now solve the LHS. This is an integration by parts question.
dv
Put u = x and dx
= cos πx. Hence du
= 1 and v = π1 sin πx. Put this into the
dx
integration by parts formula:
Z
Z
dv
du
u dx = uv − v dx
dx
dx
Z
Z
sin πx
x sin πx
−
dx
x cos πxdx =
π
π
x sin πx cos πx
=
+
+D
π
π2
where D is some constant.
Therefore we have
2x sin πx 2 cos πx
+
+B
y2 =
π
π2
where B = 2(C + D) is some constant. Solve for y:
r
2x sin πx 2 cos πx
y=±
+
+B
π
π2
(b) We solve the DE using an integrating factor. First put it in standard form:
√ 3
3
dx
+ x = 6 t et
dt
2t
Page 17 of 22
Use the integrating factor formula:
Z
3
3
dt = exp
loge t = t3/2
IF = exp
2t
2
Multiply both sides of the equation by the integrating factor:
dx 3 1/2
3
+ t x = 6t2 et
dt
2
t3/2
We see that the DE can now be written as
d
d 3/2
3
t x(t) = 6t2 et
(IF.x(t)) =
dx
dx
Integrate both sides of the equation:
3/2
t
Substitute u = t3 and
du
dt
3
6t2 et
= 3t2 :
3/2
t
x(t) =
Z
du
2eu dt
dt
Z
= 2 eu du = 2eu + C
x(t) =
Z
3
= 2et + C
Hence
3
x(t) = 2t−3/2 et + Ct−3/2
We need to find C. Set t = 1 and x(1) = 2e + 2:
x(1) = 2e + 2 = 2e + C
Hence C = 2 and our solution is
3
x(t) = 2t−3/2 et + 2t−3/2 .
Page 18 of 22
9. A population of p million after t years is modelled by the equation
dp
= 9p − p2 − 14
dt
(a) Find the equilibrium solutions and determine whether or not they are stable.
Make sure that you give a clear argument for your answer.
(b) Assume that the initial population is p = 3.141. Find the population after
t years have elapsed. State what happens as t becomes very large. Show
all intermediate steps in your solution of the differential equation. State any
constants to 3 decimal places.
(a) The equilibrium solutions are those that do not change with t. Hence they are
equivalent to finding
dp
= 9p − p2 − 14 = −(p − 7)(p − 2) = 0
dt
Hence the equilibrium populations are p = 7 and p = 2.
From the DE we see that when p < 2, dp
< 0. When 2 < p < 7, dp
> 0. And
dt
dt
dp
when p > 7, dt < 0.
Thus when p < 2 the DE implies that p is decreasing in time and so will
become smaller and so move away from 2. If p is between 2 and 7 then it will
increase towards 7. Finally, when p > 7 the DE implies that p is decreasing
with time and so p will move towards 7.
Therefore p = 2 is an unstable equilibrium and p = 7 is a stable equilibrium.
(b) We have to solve the DE. It is separable:
p2
1
dp
= −1
− 9p + 14 dt
Integrate both sides. The RHS is simply −t, while the LHS becomes
Z
Z
dp
dp
1
dt =
2
p − 9p + 14 dt
(p − 2)(p − 7)
Do a partial fraction decomposition:
1
A
B
=
+
(p − 2)(p − 7)
p−2 p−7
A(p − 7) + B(p − 2)
=
(p − 2)(p − 7)
When p = 7 the numerators give 1 = 5B and when p = 2 the numerators give
−5a = 1. Hence we have
1
1
1
=
−
(p − 2)(p − 7)
5(p − 7) 5(p − 2)
Page 19 of 22
Hence the LHS is
Z
Z
Z
dp
dp
dp
1
1
=
−
(p − 2)(p − 7)
5
p−7 5
p−2
1
1
loge |p − 7| + loge |p − 2|
=
5
5
1
p − 7 =
loge 5
p − 2
So putting both sides together we have:
p − 7
1
= −t + C
loge 5
p − 2
Isolate p:
p − 7
loge p − 2
p−7
p−2
p−7
p(1 − De−5t )
= −5t + 5C
where D = ±e5C
= De−5t
= Dpe−5t − 2De−5t
= 7 − 2De−5t
7 − 2De−5t
p(t) =
1 − De−5t
where D is some constant.
The initial population is p(0) = 3.141. We need to find D:
7 − 2D
1−D
3.141(1 − D) = 7 − 2D
D(2 − 3.141) = 7 − 3.141
p(0) = 3.141 =
Hence D ≈ −3.382 and so
p(t) =
7 + 6.764e−5t
.
1 + 3.382e−5t
When t → ∞, e−5t → 0 and we are left with
p(t) →
7
= 7.
1
So the equilibrium population is 7. We can check this by comparing against
our answer to part (a).
Page 20 of 22
10. Five grams of a chemical Q are formed by combining 2 grams of a chemical M and
3 grams of a chemical L. Initially there are 80 grams of M, 300 grams of L and
no chemical Q. The rate of formation of Q is proportional to the product of the
unconverted amounts of M and L. It is observed that exactly 20 grams of Q have
been formed after 5 minutes.
(a) If x(t) is the amount of chemical Q in grams at time t minutes, write down an
in terms of x and t.
equation for dx
dt
(b) Find x(t) and hence deduce how much chemical Q is left after 20 minutes.
(c) How much chemical M is left after a very long time?
State decimal numbers to 2 decimal places.
(a) We are told that dx
= k(amount of M) × (amount of L). The amount of M is
dt
2
80 − 5 x and the amount of L is 300 − 35 x. Hence the DE is
2x
3x
6k
dx
= k(80 − )(300 − ) =
(200 − x)(500 − x)
dt
5
5
25
where k is some constant yet to be determined.
(b) We need to solve the DE. It is separable:
6k
1
dx
=
(200 − x)(500 − x) dt
25
Integrate both sides.
Z
Z
Z
dx
6k
1
dx
6kt
dt =
=
dt =
+C
(200 − x)(500 − x) dt
(200 − x)(500 − x)
25
25
Where C is some constant.
The RHS needs to be written in partial fraction form:
1
A
B
=
+
(200 − x)(500 − x)
200 − x 500 − x
A(500 − x) + B(200 − x)
=
(200 − x)(500 − x)
The numerators give 1 = A(500 − x) + B(200 − x). Setting x = 500 gives
B = −1/300. Setting x = 200 gives A = 1/300. Hence
1
1
1
1
=
−
(200 − x)(500 − x)
300 200 − x 500 − x
We can now do the integral:
Z
Z dx
1
1
1
dx
=
−
(200 − x)(500 − x)
300
200 − x 500 − x
1
(− log |200 − x| + log |500 − x|)
=
300
500 − x 1
=
log 300
200 − x Page 21 of 22
Put it all together:
500 − x 1
= 6kt + C
log 300
200 − x 25
500 − x = 72kt + 300C
log 200 − x 500 − x
= ± exp(72kt + 300C) = D exp(72kt)
200 − x
where D = ±e300C
We have to determine 2 constants, k and D. We know that at t = 0 there is
no Q, so x(0) = 0. Hence
500
=D
200
Therefore we can write
5
500 − x
= exp(72kt)
200 − x
2
The other piece of information is that at t = 5 there are 20 grams of Q, ie
x(5) = 20. Substitute this into the equation:
8
5
500 − 20
=
=
exp(360k)
200 − 20
3
2
16
= exp(360k)
15
1
And so k = 360
log(16/15) = 0.00018 or 72k = 51 log(16/15) ≈ 0.013.
So the solution is
500 − x
5
=
exp(72kt)
200 − x
2
1000 − 2x = 1000e72kt − 5xe72kt
x(5e72kt − 2) = 1000(e72kt − 1)
1000e72kt − 1000
x =
5e72kt − 2
where 72k ≈ 0.013. (this can be stated exactly, but don’t need it).
After 20 minutes:
1000e72k×20 − 1000
2e72k×20 − 5
1000e0.26 − 1000
=
5e0.26 − 2
1000 × 1.297 − 1000
=
5 × 1.297 − 2
297
= 66.22
=
4.485
x(20) =
there are 66.22 grams of Q. (rigorous answer is 65.85. . . — will accept some
rounding error).
Page 22 of 22
(c) After a long time e72kt → ∞ since k > 0. So we have to do some manipulation
to find the limit:
1000e72kt − 1000 e−72kt
× −72kt
5e72kt − 2
e
1000 − 1000e−72kt
=
5 − 2e−72kt
x =
Now as e−72kt → ∞ since −72k < 0 and we are left with
x(t) →
1000
= 200.
5
Hence there are 200g of Q left after a long time. To make this amount Q we
need 200 ∗ 2/5 = 80g of M. ie we use up all the M and so there is 0g left.
We can check that we haven’t run out of L — to make 200g of Q requires 80g
of M and 120g of L — this leaves 180g of L unused.