1
Solutions for Chapter 11
The Chemistry Maths Book
Erich Steiner
University of Exeter
Second Edition 2008
Solutions
Chapter 11 First-order differential equations
11.1
Concepts
11.2
Solution of a differential equation
11.3
Separable equations
11.4
Separable equations in chemical kinetics
11.5
First-order linear equations
11.6
An example of linear equations in chemical kinetics
11.7
Electric circuits
© E Steiner 2008
2
Solutions for Chapter 11
Section 11.2
State the order of the differential equation and verify that the given function is a solution:
1.
dy
− 2 y = 2;
dx
y = e2 x − 1
First order.
y = e2 x − 1 →
dy
− 2 y = 2e 2 x − 2( e2 x − 1) = 2
dx
Therefore
2.
d2y
dx 2
dy
= 2e 2 x
dx
+ 4 y = 0;
y = A cos 2 x + B sin 2 x
Second order.
dy
= −2 A sin 2 x + 2 B cos 2 x
dx
d 2y
→
= −4 A cos 2 x − 4 B sin 2 x = −4 y
dx 2
y = A cos 2 x + B sin 2 x →
Therefore
3.
d3y
dx3
= 12;
d2y
+ 4y = 0
dx 2
y = 2 x3 + 3 x 2 + 4 x + 5
Third order.
y = 2 x3 + 3 x 2 + 4 x + 5 →
→
Therefore
4.
d3y
dx3
dy 3 y
+
= 3x 2 ;
dx
x
dy
= 6x2 + 6 x + 4
dx
d 2y
dx 2
= 12 x + 6
= 12
y=
x3
c
+
2 x3
First order
y=
Therefore
x3
c
dy 3 x 2 3c
+ 3 →
=
− 4
2 x
2
dx
x
dy 3 y 3x 2 3c 3 ⎛ x3
c ⎞ 3x2
3c 3 x 2
3c
+
=
− 4 + ⎜⎜
+ 3 ⎟⎟ =
− 4 +
+ 4
dx x
2
x⎝ 2 x ⎠
2
2
x
x
x
= 3x 2
© E Steiner 2008
3
Solutions for Chapter 11
Find the general solution of the differential equation:
5.
dy
= x2
dx
Integrate:
6.
∫
x 2 dx → y =
∫
dy
dx =
dx
∫
1
e −3 x dx → y = − e −3 x + c
3
d2y
dx 2
= cos 3x
Integrate twice:
d2y
∫ dx
∫
8.
1 3
x +c
3
dy
= e−3x
dx
Integrate:
7.
∫
dy
dx =
dx
d3y
dx3
2
dx =
dy
dx =
dx
∫
∫
cos 3x dx
dy 1
= sin 3 x + a
dx 3
→
1
⎡1
⎤
⎢ 3 sin 3 x + a ⎥ dx → y = − 9 cos 3x + ax + b
⎣
⎦
= 24 x
Integrate three times:
d3y
∫ dx
3
dx =
∫
24 x dx
d2y
∫ dx dx = ∫ ⎡⎣12x + a⎤⎦ dx
dy
∫ dx dx = ∫ ⎡⎣4x + ax + b⎤⎦ dx
2
2
3
© E Steiner 2008
→
→
d2y
dx 2
= 12 x 2 + a
dy
= 4 x3 + ax + b
dx
→ y = x 4 + a ′x 2 + bx + c (a ′ = a 2)
4
Solutions for Chapter 11
9.
A body moves along the x direction under the influence of the force F (t ) = cos 2π t , where t is the
time. (i) Write down the equation of motion. (ii) Find the solution that satisfies the initial
conditions x(0) = 0 and x (0) = 1.
(i) By Newton’s second law of motion, for a body of mass m,
d 2x
F (t ) = m
dt 2
= cos 2π t
(ii) Integrating twice with respect to time t,
d 2x
∫ dt
∫
2
dt =
1
m
∫
dx
dt =
dt
∫
cos 2π t dt
→
1
dx
sin 2π t + a
=
dt 2π m
⎡ 1
⎤
1
sin 2π t + a ⎥ dt → x(t ) = − 2 cos 2π t + at + b
⎢
4π m
⎣ 2π m
⎦
By the initial conditions:
x(0) = 0 = −
1
2
4π m
+b → b =
1
4π 2 m
⎛ dx ⎞
x (0) = ⎜ ⎟ = 1 = a
⎝ dt ⎠t =0
Therefore
x(t ) =
1
4π 2 m
(1 − cos 2πt ) + t
Verify that the given function is a solution of the differential equation, and determine the particular
solution for the given initial condition:
10. x
dy
= 2 y;
dx
y = cx 2 ;
y = 24 when x = 2
y = cx 2 →
dy
dy
= 2cx → x
= 2cx 2 = 2 y
dx
dx
The initial condition is y = 24 when x = 2 .
Therefore
24 = 4c → c = 6
and
y = 6x2
© E Steiner 2008
5
Solutions for Chapter 11
11.
dy
+ 2 xy = 0;
dx
2
y = ce − x ;
y = ce − x
2
y = 2 when x = 2
2
2
dy
dy
= −2cxe− x = −2 xy →
+ 2cxe− x = 0
dx
dx
→
The initial condition is y = 2 when x = 2 .
12.
Therefore
2 = ce −4 → c = 2e4
and
y = 2e 4− x
dy
+ 2 y + 2 = 0;
dx
2
y = ce−2 x − 1;
y = ce −2 x − 1 →
y (0) = 4
dy
dy
= −2ce −2 x = −2( y + 1) →
+ 2y + 2 = 0
dx
dx
The initial condition is y = 4 when x = 0 .
Therefore
4 = c −1 → c = 5
and
y = 5e −2 x − 1
Section 11.3
Find the general solution of the differential equation:
13.
14.
dy 3x 2
: Put y dy = 3 x 2 dx .
=
dx
y
Then
∫ y dy = ∫ 3x dx
and
y 2 = 2 x3 + c
2
→
1 2
y = x3 + c′
2
dy
dy
= 4 xy 2 : Put 2 = 4 x dx
dx
y
Then
∫y ∫
and
y=
dy
2
© E Steiner 2008
=
4 x dx → −
−1
2 x2 + c
1
= 2 x2 + c
y
6
Solutions for Chapter 11
15.
dy
dy
= 3 x 2 y : Put
= 3x 2 dx .
dx
y
Then
∫ y = ∫ 3x dx
and
y = ex
16. y 2
17.
dy
2
3
+c
= ae x
→ ln y = x3 + c
3
( a = ec )
dy
= e x : Put y 2 dy = e x dx .
dx
Then
∫
and
y 3 = 3e x + c
y 2 dy =
∫
1 3
y = e x + c′
3
e x dx →
(c = 3c ′)
dy
= y ( y − 1)
dx
∫
dy
= dx →
y ( y − 1)
Now
∫ y( y −1) dy = ∫ ⎢⎣ ( y −1) − y ⎥⎦ dy = ln( y −1) − ln y = ln ⎜⎝
Therefore
⎛ y −1 ⎞
ln ⎜
⎟ = x+c →
⎝ y ⎠
1
⎡
dy
=
y ( y − 1)
∫ dx = x + c
We have
1⎤
1
y −1
= ae x
y
Then, solving for y,
y −1
1
=a → y=
y
1 − ae x
18.
dy y
dy dx
=
= : Put
dx x
y
x
Then
∫ ∫
and
y
= ea = c → y = cx
x
© E Steiner 2008
dy
=
y
dx
y
→ ln y = ln x + a → ln = a
x
x
⎛ y − 1⎞
⎟
y ⎠
7
Solutions for Chapter 11
Solve the initial value problems:
19.
dy y + 2
=
;
dx x − 3
y (0) = 1
Separation of variables and integration gives
dy
dx
=
→
y + 2 x−3
∫
dy
=
y+2
∫
dx
→ ln( y + 2) = ln( x − 3) + c
x −3
Then, putting, c = ln a ,
ln( y + 2) = ln a ( x − 3) → y + 2 = a ( x − 3)
The initial condition is y = 1 when x = 0 .
20.
Therefore
3 = −3a → a = −1
and
y = 1− x
dy x 2 − 1
=
;
dx 2 y + 1
y (0) = −1
Separation of variables and integration gives
∫
(2 y + 1) dy =
∫
( x 2 − 1) dx → y 2 + y =
1 3
x −x+c
3
The initial condition is y = −1 when x = 0 . Then c = 0
and
21.
y2 + y =
dy y ( y + 1)
=
;
dx x( x − 1)
We have
∫
1 3
x −x
3
y (2) = 1
∫
dy
dx
=
y ( y + 1)
x( x − 1)
and, by the method of partial fractions,
∫
Then
⎡1
1 ⎤
⎢ −
⎥ dy =
⎣ y y + 1⎦
∫
⎡ 1
1⎤
y
⎛ x −1 ⎞
⎢ x − 1 − x ⎥ dx → ln y + 1 = ln a ⎜ x ⎟
⎝
⎠
⎣
⎦
y
⎛ x −1 ⎞
= a⎜
⎟.
y +1
⎝ x ⎠
The initial condition is y = 1 when x = 2. . Then a = 1
and
© E Steiner 2008
y
x −1
=
→ xy = xy − y + x − 1 → y = x − 1
y +1
x
8
Solutions for Chapter 11
22.
dy
= ex+ y ;
dx
y (0) = 0
Separation of variables and integration gives
e − y dy = e x dx →
∫e
−y
dy =
∫ e dx
x
→ − e− y = e x + c
The initial condition is y = 0 when x = 0 .
Then
− e − y = e x + c → − 1 = 1 + c → c = −2
Therefore
e− y = 2 − e x
and, taking the log of each side,
y = − ln(2 − e x )
Solve the initial value problems:
23.
dy x + y
=
;
dx
x
y (1) = 2
Let
u=y x
Then
dy x + y
y
=
= 1+
→ 1 + u = f (u )
dx
x
x
The general solution is given by equation (11.19),
∫ f (u) − u = ln x + c
du
The left side is
∫
Therefore
y
= ln x + c → y = x (ln x + c)
x
du
=
f (u ) − u
∫
du = u =
y
x
The initial condition, y = 2 when x = 1 , gives c = 2 . The particular solution of the differential
equation is therefore
y = x (ln x + 2)
© E Steiner 2008
9
Solutions for Chapter 11
24. 2 xy
dy
= −( x 2 + y 2 );
dx
y (1) = 0
dy
dy
( x2 + y2 )
1⎛ x y⎞
= −( x 2 + y 2 ) →
=−
=− ⎜ + ⎟
dx
dx
2 xy
2⎝ y x ⎠
We have
2 xy
Let
u=y x
Then
dy
1⎛1
1⎛
1⎞
⎞
= − ⎜ + u ⎟ = f (u ); f (u ) − u = − ⎜ 3u + ⎟
dx
2⎝u
2⎝
u⎠
⎠
By equation (11.19), the general solution is
∫ f (u) − u = ln x + c
du
The left side is
∫
∫
∫
Therefore
1 ⎛ 3y 2 ⎞
− ln ⎜ 2 + 1⎟ = ln x + c
⎟
3 ⎜⎝ x
⎠
du
du
udu
1
= −2
=2
= − ln(3u 2 + 1)
2
f (u ) − u
3u + 1 u
3
3u + 1
→ ln(3 y 2 + x 2 ) = − ln ax
( ln a = 3c)
and the general solution of the differential equation is
ln(3 y 2 + x 2 ) = ln
1
ax
→ 3 y 2 + x2 =
⎤
1
1⎡ 1
→ y2 = ⎢ − x2 ⎥
ax
3 ⎣ ax
⎦
The initial condition, y = 0 when x = 1 , gives a = 1 . The particular solution of the homogeneous
differential equation is therefore
⎤
1 ⎡1
y 2 = ⎢ − x2 ⎥
3⎣x
⎦
25. xy 3
dy
= x4 + y4 ;
dx
y (2) = 0
dy
dy x3 y 1
y
= x4 + y 4 →
= 3 + = 3 + u = f (u ) where u =
dx
dx y
x u
x
We have
xy 3
Then
∫ f (u) − u = ∫ u du = 4u
du
3
1
4
and the general solution of the differential equation is
y4
4x4
= ln x + c
The initial condition, y = 0 when x = 2 , gives c = − ln 2 .
Therefore
© E Steiner 2008
y 4 = 4 x 4 ln
x
2
10
Solutions for Chapter 11
26. Show that a differential equation of the form
dy
= f (ax + by + c)
dx
is reduced to separable form by means of the substitution u = ax + by .
Let
u = ax + by
Then
du
dy
= a+b
= a + f (u + c)
dx
dx
Therefore, by separation of variables,
du
= dx →
a + f (u + c)
∫ a + f (u + c) = x + c
du
Use the method of Exercise 26 to find the general solution:
27.
dy
= 2x + y + 3
dx
Let
u = 2x + y
Then
du
dy
= 2+
= u +5
dx
dx
Then, by separation of variables,
∫ u + 5 = ∫ dx
du
Therefore
28.
→ ln (u + 5) = x + c → u + 5 = e x + c = ae x
( a = ec )
y = ae x − 2 x − 5
dy
x− y
=
dx x − y + 2
Let
u = x− y
Then
du
dy
x− y
u
2
= 1−
= 1−
= 1−
=
dx
dx
x− y + 2
u+2 u+2
Then, by separation of variables,
1
2
and
© E Steiner 2008
∫
(u + 2) du =
∫
dx →
⎞
1 ⎛ u2
⎜ + 2u ⎟⎟ = x + c
⎜
2⎝ 2
⎠
⎞
1 ⎛ ( x − y)2
+ 2( x − y ) ⎟⎟ = x + c → ( x − y )2 − 4 y = a
⎜⎜
2⎝
2
⎠
( a = 4c )
11
Solutions for Chapter 11
Section 11.4
29. Find the interval τ1 n in which the amount of reactant in a first-order decay process is reduced by
factor n.
As for the half-life (n = 2) ,
x(t + τ1 n ) =
1
x(t )
n
and, for first-order exponential decay,
x(t ) = ae− kt
x(t + τ1 n ) = ae
=e
Then
1
− kτ
= e 1n
n
and
τ1 n =
− k (t +τ1 n )
− kτ1 n
= ae− kt × e
− kτ1 n
x(t )
→ ln
1
= −kττ1 n
n
→ ln n = kττ1 n
ln n
k
30. Solve the initial value problem for the nth-order kinetic process A → products
dx
= −kx n
dt
x(0) = a
(n > 1)
By separation of variables and integration,
−
∫ x = k ∫ dt
dx
n
→
1
(n − 1) x n −1
= kt + c
The initial condition, x = a when t = 0 , gives c =
Therefore
© E Steiner 2008
1
x
n −1
−
1
a
n −1
= (n − 1)kt
1
(n − 1)a n −1
12
Solutions for Chapter 11
31. The reversible reaction A U B, first-order in both directions, has rate equation
dx
= k1 (a − x) − k−1 x
dt
Find x(t ) for initial condition x(0) = 0 .
dx
= k1 (a − x) − k−1 x = k1a − (k1 + k−1 ) x
dt
We have
Then, by separation of variables and integration,
∫
dx
=
k1a − (k1 + k−1 ) x
∫
dt →
−1
ln ⎡ k1a − (k1 + k−1 ) x ⎤⎦ = t + c
k1 + k−1 ⎣
The initial condition, x = 0 when t = 0 gives c =
Therefore
⎡
⎤
k1a
k1a
= e( k1 + k−1 )t
ln ⎢
⎥ = (k1 + k−1 )t →
k1a − (k1 + k−1 ) x
⎣⎢ k1a − (k1 + k−1 ) x ⎦⎥
x=
and
−1
ln ⎡ k1a ⎤ .
k1 + k−1 ⎣ ⎦
k1a ⎡
1 − e( k1 + k−1 )t ⎤
⎦
k1 + k−1 ⎣
32. A third-order process A + 2B → products has rate equation
dx
= k (a − x) (b − 2 x) 2
dt
where a and b are the initial amounts of A and B, respectively. Show that the solution that satisfies
the initial condition x(0) = 0 is given by
kt =
1
(2a − b)
2
ln
a (b − 2 x)
2x
+
b(a − x) b(2a − b) (b − 2 x)
Separation of variables and integration gives
∫ (a − x) (b − 2x) = k ∫ dt = kt + c
dx
2
Now, by the method of partial fractions,
1
(a − x) (b − 2 x)
Therefore
2
=
∫ (a − x) (b − 2x)
dx
2
⎡ 1
2
2(2a − b) ⎤
−
+
⎢
⎥
(2a − b) ⎢⎣ (a − x) b − 2 x (b − 2 x) 2 ⎥⎦
1
2
=
=
© E Steiner 2008
1
(2a − b)
2
∫
⎡ 1
2
2(2a − b) ⎤
−
+
⎢
⎥ dx
2
⎣⎢ (a − x) b − 2 x (b − 2 x ) ⎦⎥
⎡
2a − b ⎤
⎢ − ln(a − x) + ln(b − 2 x) + b − 2 x ⎥ = kt + c
(2a − b) ⎣
⎦
1
2
13
Solutions for Chapter 11
The initial condition, x = 0 when t = 0 , gives
c=
Therefore
⎡
2a − b ⎤
1
b
1
ln +
⎢ − ln a + ln b + b ⎥ =
2
a b(2a − b)
(2a − b) ⎣
⎦ (2a − b)
1
2
⎡ a(b − 2 x) ⎤
2x
⎢ ln
⎥+
(2a − b) ⎣ b(a − x) ⎦ b(2a − b)(b − 2 x)
1
kt =
2
Section 11.5
Find the general solution:
33.
dy
+ 2y = 4
dx
The linear differential equation has the form (11.47) with p( x) = 2, r ( x) = 4 . Then
∫ p( x) dx = 2 x , and the integrating factor is
F ( x) = e ∫
p ( x ) dx
= e2 x
Then, by formula ,
e2 x y = 4
and
34.
∫e
2x
dx + c = 2e 2 x + c
y = 2 + ce −2 x
dy
− 4 xy = x
dx
We have p( x) = −4 x, r ( x) = x , and integrating factor F ( x ) = e ∫
Therefore, by formula ,
2
e −2 x y =
and
© E Steiner 2008
2
∫e
y = ce 2 x −
−2 x 2
1
4
2
1
x dx + c = − e −2 x + c
4
−4 x dx
2
= e −2 x .
14
Solutions for Chapter 11
35.
dy
+ 3 y = e−3 x
dx
We have p( x) = 3, r ( x) = e −3 x , and integrating factor F ( x ) = e ∫
3 dx
= e3 x .
Then, by formula ,
e3 x y =
3x
× e−3 x dx =
∫ dx = x + c
y = e −3 x ( x + c)
and
36.
∫e
dy 2 y
+
= 2 cos x
dx
x
We have p( x) = 2 x , r ( x) = 2 cos x . Then
factor is F ( x ) = e ∫
p ( x ) dx
x2 y = 2
∫
p ( x) dx = 2
∫
dx
= 2 ln x = ln x 2 and the integrating
x
2
= eln x = x 2 . Then, by formula ,
∫ x cos x dx + c
2
and integration by parts, as in Example 6.11, gives
x 2 y = 2 ⎡ x 2 sin x + 2 x cos x − 2sin x + c ⎤
⎣
⎦
y=
37.
2 ⎡ 2
x sin x + 2 x cos x − 2sin x + c ⎤
⎦
x2 ⎣
dy
y
4
−
=
dx x 2 x 2
p( x) = − 1 x 2 , r ( x) = 4 x 2 ,
and the integrating factor is F ( x) = e ∫
e1 x y = 4
∫e
1x
×
1
x2
p dx
= e1 x . Then, by formula ,
dx + c
1
1
, du = − 2 dx .
x
x
Let
u=
Then
e1 x y = 4
and
e1 x y = −4e1 x + c → y = ce −1 x − 4
© E Steiner 2008
∫e
1x
×
1
x
2
dx + c = −4
∫ e du + c = −4e + c = −4e
u
u
1x
+c
15
Solutions for Chapter 11
38.
dy
+ (2 tan x) y = sin x
dx
We have
p( x) = 2 tan x, r ( x) = sin x .
Then
∫ p(x) dx = 2∫ tan x dx = −2 ln(cos x) = ln(1 cos x)
2
and the integrating factor is F ( x) = eln(1 cos
Then
and
39.
1
cos 2 x
=
1
cos 2 x
.
∫ cos x × sin x dx + c = cos x + c
1
1
2
(n ≠ −1)
p( x) = ax n , r ( x) = bx n , and
F ( x) = e ∫
40.
x)
y = cos x + c cos 2 x
dy
+ ax n y = bx n ,
dx
We have
y=
2
ax n dx
n+1 ( n +1)
y=
= eax
∫
n+1 ( n +1)
e ax
Let
t = x n +1 (n + 1) , dt = x n .
Then
eax
and
y=
n+1 ( n +1)
y=b
e ax
n+1 ( n +1)
Then
∫e
at
.
× bx n dx + c
dt + c =
b at
b n+1
e + c = eax ( n +1) + c
a
a
n+1
b
+ ce− ax ( n +1)
a
dy
y
+ a = xn
dx
x
We have p( x) = a x , r ( x) = x n , and integrating factor F ( x) = e ∫
Then
xa y =
and
y=
© E Steiner 2008
∫
x a × x n dx + c =
x n +1
+ cx − a
a + n +1
x a + n +1
+c
a + n +1
a x dx
= e a ln x = x a .
16
Solutions for Chapter 11
Section 11.6
k
k
k
3
1
2
41. The system of three consecutive first-order processes A ⎯⎯
→ B ⎯⎯
→ C ⎯⎯
→ D is modelled
by the set of equations
d (a − x)
= −k1 (a − x ),
dt
dy
= k1 (a − x) − k2 y,
dt
dz
= k2 y − k3 z
dt
where (a − x), y, and z are the amounts of A, B, and C, respectively, at time t. Given the initial
conditions x = y = z = 0 at t = 0 , find the amount of C as a function of t. Assume
k1 ≠ k2 , k1 ≠ k3 , k2 ≠ k3 .
For the first-order step A → B ,
d (a − x)
= −k1 (a − x)
dt
Separation of variables and integration gives
∫
∫
d (a − x)
= − k1 dt
(a − x)
→ ln(a − x) = −k1t + c → a − x = Ae− k1t ( A = ec )
The initial condition for this step is x = 0 when t = 0 . Therefore A = a and
a − x = ae− k1t
(equation 1)
For step B → C , making use of equation 1 above,
dy
dy
= k1 (a − x) − k2 y = ak1e − k1t − k2 y →
+ k2 y = ak1e− k1t
dt
dt
We have a linear equation with p(t ) = k2 , r (t ) = ak1e − k1t and integrating factor
F (t ) = e ∫
k2 dt
= e k2 t .
Then
e k2 t y =
and
y=
∫e
k2 t
∫
× ak1e− k1t dt = ak1 e( k2 − k1 )t dt =
ak1 ( k2 − k1 )t
e
+c
k2 − k1
ak1 − k1t
e + ce− k2t
k2 − k1
The initial condition for this step is y = 0 when t = 0 . Therefore c = −
y=
© E Steiner 2008
ak1
⎡e − k1t − e − k2t ⎤
⎦
k2 − k1 ⎣
(equation 2)
ak1
and
k2 − k1
17
Solutions for Chapter 11
For step C → D , making use of equation 2,
ak k
dz
dz
= k 2 y − k3 z →
+ k3 z = k2 y = 1 2 ⎡⎣e − k1t − e− k2t ⎤⎦
dt
dt
k2 − k1
This is a linear equation with p(t ) = k3 , r (t ) =
Then
e k3t z =
and
z=
ak1k2
k2 − k1
∫
ak1k2 − k1t
⎡e − e − k2t ⎤ and integrating factor e k3t .
⎦
k2 − k1 ⎣
ak k ⎡ ( k3 − k1 )t e( k3 − k2 )t ⎤
⎡e( k3 − k1 )t − e( k3 − k2 )t ⎤ dt = 1 2 ⎢ e
−
⎥+c
⎣
⎦
k2 − k1 ⎢⎣ k3 − k1
k3 − k2 ⎥⎦
ak1k2 ⎡ e − k1
e − k 2t ⎤
−k t
−
⎢
⎥ + ce 3
k2 − k1 ⎣⎢ k3 − k1 k3 − k2 ⎦⎥
The initial condition for this step is z = 0 when t = 0 . Therefore c = −
and
the amount of substance C at time t is
z=
© E Steiner 2008
ak1k2 ⎡ e − k1t − e − k3t e − k2t − e− k3t ⎤
−
⎢
⎥
k2 − k1 ⎢⎣ k3 − k1
k3 − k2 ⎥⎦
ak1k2 ⎡ 1
1 ⎤
−
⎢
⎥,
k2 − k1 ⎣ k3 − k1 k3 − k2 ⎦
18
Solutions for Chapter 11
k
k
1
2
42. The first-order process A ⎯⎯
→ B is followed by the parallel first-order processes B ⎯⎯
→C
k
3
and B ⎯⎯
→ D, and the system is modelled by the equations
d (a − x)
= −k1 (a − x),
dt
dz
du
= k3 y
= k 2 y,
dt
dt
dy
= k1 (a − x) − (k2 + k3 ) y,
dt
where (a − x), y, z and u are the amounts of A, B, C, and D, respectively, at time t. Given the
initial conditions x = y = z = u = 0 at t = 0 find the amount of C as a function of time.
A
a−x
k1
B
y
k2
C
z
k3
D
u
A → B : as in Exercise 41.
d (a − x)
= −k1 (a − x) → a − x = ae− k1t
dt
B → C + D : as in Exercise 41, but with k2 replaced by k2 + k3 .
ak1
dy
⎡ e− k1t − e− ( k2 + k3 )t ⎤
= k1 (a − x) − (k2 + k3 ) y → y =
⎦
dt
k2 + k3 − k1 ⎣
B→D:
ak1k2
dz
⎡e − k1t − e−( k2 + k3 )t ⎤
= k2 y =
⎦
dt
k2 + k3 − k1 ⎣
Integration with respect to t gives
z=
ak1k2 ⎡ e − k1t e −( k2 + k3 )t ⎤
+
⎢−
⎥+c
k2 + k3 − k1 ⎣⎢ k1
k2 + k3 ⎦⎥
The initial condition, z = 0 when t = 0 , gives
c=
ak1k2 ⎡ 1
1 ⎤
⎢ −
⎥ . The amount of
k2 + k3 − k1 ⎣ k1 k2 + k3 ⎦
substance C at time t is therefore,
z=
© E Steiner 2008
ak1k2 ⎡1 − e − k1t 1 − e −( k2 + k3 )t ⎤
−
⎢
⎥
k2 + k3 − k1 ⎢⎣ k1
k2 + k3 ⎥⎦
19
Solutions for Chapter 11
Section 11.7
43. The current in an RL–circuit containing one resistor and one inductor is given by the equation
L
dI
+ RI = E
dt
Solve the equation for a periodic e.m.f. E (t ) = E0 sin ωt , with initial condition I (0) = 0 .
This is the problem discussed in Example 11.7, but with a periodic e.m.f. Thus, by equation (11.66),
the solution of the linear differential equation is
⎡
I (t ) = e− Rt /L ⎢
⎣
∫e
Rt /L
E
⎤
dt + c ⎥
L
⎦
and, inserting E = E0 sin ωt ,
⎡E
I (t ) = e− Rt /L ⎢ 0
⎣ L
∫
⎤
e Rt /L sin ωtdt + c ⎥
⎦
The integral is evaluated by parts, as described in Example 6.13. Let a = R /L and b = ω , then
(see also Exercises 49 and 50 in Chapter 6)
∫e
at
sin bt dt =
=
Solving for
∫e
∫
Then
at
1 at
b
e sin bt −
a
a
∫e
at
1 at
b
b2
e sin bt − 2 eat cos bt − 2
a
a
a
e at sin bt dt =
eat
a2 + b2
at
⎡E
I (t ) = e− Rt /L ⎢ 0
⎣ L
∫
⎡⎣ a sin bt − b cos bt ⎤⎦ .
⎤
e Rt /L sin ωtdt + c ⎥
⎦
⎡R
⎤ ⎫⎪
⎢ L sin ωt − ω cos ωt ⎥ + c ⎬
⎦ ⎭⎪
⎣
The initial condition, I = 0 when t = 0 , gives c =
© E Steiner 2008
∫e
sin bt dt ,
⎧⎪ e Rt /L E0 L
= e− Rt /L ⎨ 2
2 2
⎩⎪ R + ω L
Therefore
cos bt dt
I (t ) =
E0
E0ω L
2
R + ω 2 L2
.
⎡ω Le− Rt /L + R sin ωt − ω L cos ωt ⎤
⎦
R + ω 2 L2 ⎣
2
sin bt dt
20
Solutions for Chapter 11
44. Show that the current in an RC–circuit containing one resistor and one capacitor is given by the
equation
R
dI I dE
+ =
dt C dt
Solve the equation for (i) a constant e.m.f., E = E0 , (ii) a periodic e.m.f., E (t ) = E0 sin ωt .
By Kirchoff’s law and equations (11.62) and (11.64),
RI +
Q
=E
C
Then, because
dQ
=I
dt
we have
R
dI 1 dQ dE
dI I dE
+
=
→ R + =
dt C dt
dt
dt C dt
(i) For constant e.m.f.,
R
Therefore
(ii) We have
Then
dE
= 0 and the differential equation is separable. Thus
dt
∫ I = − RC ∫ dt
dI I
+ =0 →
dt C
I = I 0 e −t
RC
→ ln I = −
t
+c
RC
, where I 0 is the current at time t = 0 .
dE
= E0ω cos ωt .
dt
E = E0 sin ωt →
Eω
dI
I
+
= 0 cos ωt
dt RC
R
is a linear equation with p = 1 RC , r =
Then
1
dI
I (t ) = e−t
RC
⎧ E0ω
⎨
⎩ R
∫e
t RC
E0ω
cos ωt and integrating factor F (t ) = et
R
RC
.
⎫
cos ωt dt + c ⎬
⎭
The integral is evaluated by parts, as in Exercise 43 above and Exercise 50 in Chapter 6:
∫
eat cos bt dt =
eat
a 2 + b2
⎡⎣ a cos bt + b sin bt ⎤⎦
where a = 1 RC and b = ω . Therefore
∫
and
© E Steiner 2008
et
RC
I (t ) =
cos ωt dt =
RCe t
RC
q + (ω RC )2
⎡⎣ cos ωt + ω RC sin ωt ⎤⎦
ω E0C
⎡⎣cos ωt + ω RC sin ωt ⎤⎦ + ce−t
1 + (ω RC ) 2
RC