Transcript Advance Engingeering Mathematics

Chapter 2 Differential Equations of First Order
2.1 Introduction
The general first-order equation is given by
F ( x, y , y ' )  0
(1)
where x and y are independent and dependent variables,
respectively.
2.2 The Linear Equation
a0 ( x) y'a1 ( x) y  f ( x)
(1)
y' p( x) y  q( x)
( 2)
where p(x) and q(x) are continuous over the x interval of interest.
1
2.2.1. Homogeneous case.
In the case where q(x) is zero, Eq. (2) can be reduced to
y' p( x) y  0,
(3)
Which is called the homogeneous version of (2).
dy
 y +  p(x)dx  0
( 4)
ln y    p ( x) dx  C
y ( x)  Be 
 p ( x ) dx
y ( x)  Ae 
 p ( x ) dx
( 6)
x
p ( ) d 

a
y( x)  be

(7 )
2
Example 1
( x  2) y  xy  0
'
3
2.2.2. Integrating factor method
To solve Eq. (2) through the integrating factor method by
multiplying both sides by a yet to be determined function s(x).
sy'spy  sq
(16)
The idea is to seek s(x) so that
d
(s y )  s q
dx
(18)
And the left-hand side of (16) is a derivative:
d
sy 'spy  (sy )
dx
(17)
σ(x) is called an integrating factor.
4
How to find s(x)? Writing out the right-hand side of (17) gives
sy'spy  sy's ' y
(17' )
If we choose s(x) so that
sp  s '
(19)
Eq. (17’) is satisfied identically
p ( x ) dx

s ( x)  e
(20)
Putting Eq. (20) into (18), we have
p ( x ) dx
d  p ( x ) dx

(e
y)  e
q( x)
dx
5
p ( x ) dx
p ( x ) dx


e
y  e
q( x)dx  C
y ( x)  e 
 p ( x ) dx
p ( x ) dx

( e
q ( x)dx  C )
=Ce 
 p ( x ) dx
e 
 p ( x ) dx

(21)
p ( x ) dx

e
q( x)dx
=y h  y p
x
y ( x)  e a

p ( ) d 

p ( ) d 

a
( e
q ( )d   b
x
a
(24)
Whereas (21) was the general solution to (2), we call (24)
a particular solution since it corresponds to one particular
solution curve, the solution curve through the point (a, b).
6
Example 2
xy  2 y  x3
7
Some special differential equations
1. Bernoulli’s equation
y' p( x) y  q( x) y n
(9.1)
v'(1  n) p( x)v  (1  n)q( x)
(9.2)
2. Riccati’s equation
y'  p( x) y  q( x) y  r ( x)
2
(11.1)
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3. d’Alembert-Lagrange equation
y  xf ( p)  g ( p)
y  xf ( y )  g ( y )
y  f ( y)  xf ( y) y  g ( y) y
because y   p
p  f ( p )  xf ( p ) p  g ( p ) p
p  f ( p )  [ xf ( p )  g ( p )] p
dx
f ( p )
g '( p )

x
dp p  f ( p )
p  f ( p)
(13.1)
(13.2)
(13.3)
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2.3 Applications of the Linear Equation
2.3.1. Electrical circuits
In the case of electrical circuits the relavent underlying
physics is provided by Kirchhoff’s laws
Kirchhoff’s current law: The algebraic sum of the current
approaching (or leaving) any point of a circuit is zero.
Kirchhoff’s voltage law: The algebraic sum of the voltage
drops around any loop of a circuit is zero.
The current through a given control surface is the charge
per unit time crossing that surface. Each electron carries a
negative charge of 1.6 x 10-19 coulomb, and each proton
carries an equal positive charge. Current is measured in
Amperes. By convention, a current is counted as positive in a
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given direction if it is the flow of positive charge in that direction.
An electric current flows due to a difference in the electric
potential, or voltage, measured in volts.
For a resistor, the voltage drop E(t), where t is the time
(in seconds), is proportional to the current i(t) through it:
E (t )  Ri(t )
(1)
R is called the resistance and is measured in ohms;
(1) is called Ohm’s law.
For an inductor, the voltage drop is proportional to the time
rate of change of current through it:
di (t )
E (t )  L
dt
(2)
L is called the inductance and is measured in henrys.
11
For a capacitor, the voltage drop is proportional to the
charge Q(t) on the capacitor:
1
E (t )  Q (t )
C
(3)
Where C is called the capacitance and is measured in
farads.
Physically, a capacitor is normally comprised of two plates
separated by a gap across which no current flow, and Q(t) is
the charge on one plate relative to the other. Though no current
flows across the gap, there will be a current i(t) that flows
through the circuit that links the two plates and is equal to the
time rate of change of charge on the capacitor:
dQ (t )
i (t ) 
dt
(4)
12
From Eqs. (3) and (4) it follows that the desired voltage/current
relation for a capacitor can be expressed as
1
E (t )   i (t )dt
C
(5)
According to Kirchhoff’s voltage law,
we have
(Va  Vd )  (Vb  Va )  (Vc  Vb )  (Vd  Vc )  0
(6)
gives
di 1
E (t )  Ri - L -  idt  0
dt C
(7)
d 2i
di 1
dE (t )
L 2 R  i
dt
dt C
dt
(8)
13
Example 1 RL Circuit.
If we omit the capacitor from the circuit, then (7) reduces to
the first-order linear equation
di
L  Ri  E (t )
(10)
dt
tR
 R
d  t  d  E ( )


0L
0 L
i(t )  e
d  i0 
 0 e
L


1 t R ( t ) / L
 Rt / L
i (t )  i0e
  e
E ( )d
L 0

(11)
Case 1: If E(T) is a constant, then Eq.(11) gives
E0
i (t )  i0 e

(1  e  Rt / L )
R
E0
E0  Rt / L
i (t ) 
 (i0  )e
R
R
 Rt / L
or
(12)
(13)
14
Case 2: If E(T)=Eosinwt and i0 =o, then Eq.(11) gives
E0w L
R
 Rt / L
i(t )  2
(e

sin wt  cos wt )
2
R  (w L)
wL
(14)
Rt
di

L  Ri  E (t )
i p   (t )e L
dt
R
di
d E

0
L
L  Ri  E 0 sin wt
 (t )   e
sin wtdt
L
dt
Rt
E0 L
E0
di R

sin wt de L
 i
sin wt

L R
dt L
L
Rt
Rt
2 2
E
w
t
w
L
di R
0
L
L
find ih  let
 i  0  ( sin wt e  cos wt e )  2 
R
R
R
dt L
Rt
Rt
RT
E
w
L
R
0

L
L

(
sin
w
t
e

cos
w
t
e
)
L
2
2 2
 ih  Ce
R  w L wL
tR
tR
  d
d E

0
0 L
0 L
ip  e
e
sin wtdt

15
L
t
0
Example 2 Radioactive decay
The disintegration of a given nucleus, within the mass, is
independent of the past or future disintegrations of the other
nuclei, for then the number of nuclei disintegrating, per unit time,
will be proportional to the total number of nuclei present:
dN
 kN
dt
(21)
k s known as the disintegration constant, or decay rate.
Let’s multiply both sides of Eq. (21) by the atomic mass, in
which Eq. (21) becomes the simple first-order linear equation
dm
(22)
  km
dt
m(t) is the total mass. Solve Eq.(22), we have
m(t )  m0ekt
(23)
16
m0
 m0 e  kT
2
m(t )  m0 2t T
(24)
Thus, if t =T, 2T, 3T, …, then m(t) = m0, m0/2, m0 /4,
and so on.
2.3.3. Population dynamics
According to the simplest model, the rate of change dN/dt
is proportional to the population N:
dN
N
dt
(25)
κ is the net birth/death rate.
17
Solving Eq. (25), we have
t
N (t )  N0e
(26)
We expect that κ will not really be a constant but will vary
with N. In particular, we expect it to decrease as N increases.
As a simple model of such behavior, let κ = a – bN.
Then Eq. (25) is to be replaced by the following Eq.
dN
 (a  bN ) N
dt
(27)
The latter is known as the logistic equation, or the Verhulst
equation.
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2.3.4. Mixing problems
Considering a mixing tank with an flow of Q(t) gallons per
minute and an equal outflow, where t is the time.The inflow
is at a constant concentration c1 of a particular solute, and
the tank is constantly stirred. So that the concentration c(t)
within the tank is uniform. Let v is a constant. Find the
instantaneous mass of solute x(t) in the tank.
Q(t): Inflow with flow rate (gal/min)
C(t): Uniform concentration within the tank (ib/gal)
C1(t):Constant concentration of solute at inlet (lb/gal)
X(t): Instantaneous mass of solute (ib)
V: Volume of the tank
x
dx
dc
c   x  cV 
V
V
dt
dt
Rate of increase of mass of solute within V
=Rate in – Rate out
dx
dc
dc Q
Q
 Q(t )c1 (t )  Q(t )c(t )  V
 Qc  Qc1 
 c  c1
dt
dt
dt V
V
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2.4 Separable Equations
2.4.1. Separable equations (page 46-48)
Direct
If f(x,y) can be expressed as a function of x times a function
of y, that is
y  X ( x)Y ( y)
(3)
then we say that the differential equation is separable.
1
 Y ( y) ydx   X  x  dx
dy
(5)
 Y ( y)   X  x  dx
Example 1
y   y
2
20
Example 2 Solve the initial-value problem
4x
y 
1  2e y
y(0)=1
Observe that if we use the definite integrals.
 1  2e dy  
y
1
y
x
0
4 xdx
21
Example 3 Solve the equation listed in the following
y( y  2)
y' 
x( y  1)
(17)

(18)
y 1
dx
dy  
y ( y  2)
x
The solution can be expressed as
ln
y( y  2)
 2C
2
x
(21)
y( y - 2) 2C
so
=e  B (0  B< )
2
x
(22)
B is nonnegative. Thus,
y ( y - 2)
=  B  A (- <A< )
2
x
y 2  2 y  Ax2  0
(23)
yx   1  1  Ax
2
22
Example 4 Free Fall. Suppose that a body of mass m is
dropped, from rest, at time t=0. With its displacement x(t)
measured down-ward from the point of release, the equation of
motion is mx’’ = mg
(25a)
x ''  g , (0  t<) subjected to B.C.
x(0)  0 and x '(0)  0
g 2
x (t )  t
2
(25b&c)
(26)
Let us multiply Eq. (25a) by dx and integrate on x
x' ' dx 
dx '
dx ' dx
dx '
dx 
dt  x'
dt  x' dx '
dt
dt dt
dt
(28)
x' dx'  gdx
1 2
x'  gx  A
2
x'  2g x1 2
(27)
(29)
x(t ) 
1
( 2g t  C )2
4
(30a)
23
Example 5 Verhulst Population Model.
N (t )  (a  bN ) N ; N(0)=N0
(48)
dN
 (a  bN ) N   dt
1
1
1
1 1
1 1



a
a
a  bN N b ( N  ) N a N  a N
b
b
1
a 1
 ln N   ln N  t  C
a
b a
1
N
a
N
b
N
a
N
b
(49)
a
N
 et C
a
N
b
  Be  Ae
at
at
(51)
 e at  aC  Beat
(50)
aN0
(52)
N (t ) 
 at
(a  bN0 )e  bN0
24
Example 6
Example 7
( x2  y 2 )
yy '  xe
Sol:
yy '  xe e
x2
ye
 y2
 ye
y2
dy  xe dx
 y2
x2
dy   xe dx
x2
y '  4x /(1  2e )
y
y '  4 x /(1  2e y )
dy
4x

dx 1  2e y
y
4 xdx  (1  2e ) dy  (1)
2 x  y  2e  C
2
y
1  y 2 1 x2
 e  e C
2
2
 e e
x2
 y2
A
25
Indirect
(a) Homogeneous of degree zero
y
y'  f ( )
x
Hence
y
let  u, dy  xdu  udx
x
dy xdu  udx
y' 

 f (u )
dx
dx
xdu  udx  f (u )dx
xdu  [ f (u )  u ]dx
du
dx

[ f (u )  u ] x
26
Example 8
y
x
y'  3
x
y
xdu  3u

1
2
y
set
 u , so dy  xdu  udx
x
 dx
1
1 12
dx  u du
x
3
1
1 12
 x dx   3 u du
1 2 32
ln x  C 
u
33
2 y 32
ln x  C  ( )
9 x
27
(b) Almost-homogeneous equation
a1 x  b1 y  c1
y' 
a2 x  b2 y  c2
(a1 , b1 ,...., c2 constants)
(11.1)
can be reduced to homogeneous form by the change of
variables x=u+h, y=v+k, where h and k are suitably
chosen constants, provided that a1b2-a2b1≠0.
(a1x  b1 y  c1 )dx  (a2 x  b2 y  c2 )dy  0
Steps for changing the Almost-homogeneous Eq.
to homogeneous Eq.
a1 b1
(a)

a2 b2
28
1. Find the intersection (α ,β)
2. let u  ( x   ), v  ( y   )

 du  dx,
dv  dy
dy dv a1u  b1v
3. y' 


dx du a2u  b2v
or (a1u  b1v)du  (a2u  b2v)dv  0
v
4. let  t
u
dv  udt  tdu
It turns to be a homogeneous of degree zero.
29
Example 10
(2 x  5 y  3)dx  (2 x  4 y  6)dy  0
Sol:
2 x  5 y  3

 ( ,  )  (1,1)
2
x

4
y

6

x  1  u
dx  du
; 

dy  dv
y 1  v
so (2u  5v )du  ( 2u  4v )dv  0    (1)
v
set  t , so dv  udt  tdu    ( 2)
u
(2  5t )du  (2  4t )dv  0
 (2  5t )du  (2  4t )(udt  tdu)  0
30
 (2  5t  2t  4t 2 )du  (2  4t )udt
du
2  4t
(
)dt
2
u
2  7t  4t
2  4t
a
b
ln u  C1  
dt   (

)dt
(1  4t )(2  t )
2  t 1  4t
P.S. a  4at  2b  bt  2  4t
 4a  b  4
2
 a

3
a  2b  2
4
b
3
2
1
 ln u  C1   ln 2  t  ln 1  4t
3
3
 ( x  4 y  3)( y  2 x  3) 2  C
31
a1 b1
c1
(b)
 m
a2 b2
c2
set a2 x  b2 y  z , so
a1 x  b1 y  c1
m z  c1
y'  f (
) f(
)  f ( z)
a2 x  b2 y  c2
z  c2
dz  a2 dx
dy  (
), thus
b2
dy dz  a2 dx
y' 

 f ( z)
dx
b2 dx
32
Example 11
( x  y)dx  (3x  3 y - 4)dy  0
set ( x  y)  z , so dy  dz  dx
zdx  (3z - 4)(dz - dx)  0
(3z - 4)dz  ( z - 3z  4)dx  0
Therefore,
3z  4
3
2
3
1
dx  (
)dz  ( 
) dz  ( 
) dz
2z  4
2 2z  4
2 z2
Integration
3
 x  c  z  ln z  2
2
3
x  c  ( x  y )  ln x  y  2
2
x 3
 (  y)  ln x  y  332  c
2 2
(c)
y '  f (ax  by  c)
set ax  by  c  t  adx  bdy  dt
dt  adx
dt  adx
bdy  dt  adx dy 
y' 
 f (t )
b
bdx
 dt  adx  bf (t )dx  dt  bf (t )  a  dx
Example 12
y '  tan ( x  y)
let ( x  y )  z so dy  dz  dx
dy dz  dx
2

 tan z
dx
dx
dz  (tan 2 z  1)dx
2
 dx  cos zdz
2
34
1  cos 2 z
xc  (
)dz
2
z sin 2 z
 
2
4
x  y sin(2 x  2 y )


2
4
 2 y  2 x  sin(2 x  2 y )  k
35
(c) Isobaric Eqs.
Example 13
3xy 2 y ' x2  2 y3  0
let y  x m v
 x2  2
 3
 y  3m
 xy 2 y '  1  2m  m  1  3m

2
2  3m  m 
3
2
3

1
3
2
3
2
set y  x v, y '  x v  x v '
3
36
4
3 2

1
3
2
3
2
3 x( x v )( x v  x v ')  x 2  2 x 2 v 3  0
3
2 3
3 2
2
2 3
2 x v  3 x v v ' x  2 x v  0
 3 xv 2 v ' 1  0
dx
 3v dv 
x
 v 3  ln x  c
2
e
 v3
 kx

2
3
(v  yx )
e

y3
x2
 kx
37
2.5 Exact equations and integrating factors
2 f
 f
 f
 ( ) ( )
xy x y
y x
2.5.1 Exact differential equations
Considering a function F(x,y)=C, the derivative of the
function is dF(x,y)=o
dF ( x, y)  M ( x, y)dx  N ( x, y)dy  0
Considering
F
F
dF ( x, y ) 
dx 
dy  0

x

y
and
M
 F
 ( )
y y x
N  F
 ( )
x x y
38
Therefore, if
M
 F
 F
N
 ( ) ( )
y y x
x y
x
(10)
Then Mdx+Ndy is an exact differential equation, and
according to the definitions
F ( x, y )
F ( x, y )
M ( x, y ) 
and N ( x, y) 
x
y
F ( x, y)   M ( x, y)dx  g ( y) and
F ( x, y)   N ( x, y)dy  h( x)
Example 1
sin ydx  ( x cos y  2 y)dy  0
39
2.5.2 Integrating factors
Even if M and N fail to satisfy Eq. (10), so that the equation
M ( x, y)dx  N ( x, y)dy  0
is not exact, it may be possible to find a multiplicative factor
s(x,y) so that
s ( x, y)M ( x, y)dx  s ( x, y) N ( x, y)dy  0
is exact, then we call it an integrating factor, and it satisfied


(s M )  (s N )
y
x
40
How to find s(x,y)
s y M  s M y  s x N  s Nx
(23)
Perhaps an integrating factor s can be found that is a
function of x alone. That is sy=0. Then Eq. (23) can be
reduced to the differential equation
s
sMy 
N  s N x or
x
M y  Nx
s
 s ( x, y )(
)
x
N
if
M y  Nx
N
s ( x)  e

 function of x alone, then
M y  Nx
N
dx
(24)
(25)
(26)
41
If (My-Nx)/N is not a function of x alone, then an integrating
factor s(x) does not exist, but we can try to find s as a function
of y alone: s(y). The Eq. (23) reduces to
s
M  s M y  s N x or
y
M y  Nx
s
 s ( x, y )(
)
y
M
if -
M y  Nx
M

s ( y)  e

 function of y alone, then
M y  Nx
M
dy
(27)
(28)
42
M N

( Exact )
y
x
M N

( Non  exact )
y
x
Try Integrating factor
 ( IM )  ( IN )

y
x
43
Example 1
Example 1’
3 ydx  2 xdy  0
2xydx  ( y2  x2 )dy  0
44
Example 2
(3xy  2 y 2 )dx  (2x2  3xy)dy  0
Eq : M ( x, y )dx  N ( x, y )dy  0
Multiply Eq. by I ( x, y )  xy
2 2
3
3
2 2
(3
x
y

2
xy
)
dx

(2
x
y

3
x
y )dy  0

M

 y  3x  4 y
  ( IM )
2
2

6
x
y

6
xy

 y

N


 4x  3y
 x
  ( IN )  6 x 2 y  6 xy 2
 x
M N

( x  y )
1
y x


 f ( xy )
yN  xM  xy ( x  y ) xy
 I ( x, y )  e

1
d ( xy )
xy
 xy
45
Using part known integrating factor to
determine the integrating factor of the Eq.
Example 3
2
xy
  y  dx  xdy  0
46
Example 4
xy  y
2



 1 dx  xy  x 2  1 dy  0
47
2.5.3 Integrating factor of a homogeneous
differential Eq.
For the homogeneous differential equation
M(x,y)dx+N(x,y)dy=0
with the same degree of M and N, if
1
Mx  Ny  0, 
is an integral factor, and if
Mx  Ny
1
Mx  Ny  0, 
is an integral factor
xy
4
4
3
(
x

y
)
dx

xy
dy  0
Example 6
48
Example 1
Example 1’
3 ydx  2 xdy  0
2xydx  ( y2  x2 )dy  0
49
Example 2
(3xy  2 y 2 )dx  (2x2  3xy)dy  0
Eq : M ( x, y )dx  N ( x, y )dy  0
Multiply Eq. by I ( x, y )  xy
2 2
3
3
2 2
(3
x
y

2
xy
)
dx

(2
x
y

3
x
y )dy  0

M

 y  3x  4 y
  ( IM )
2
2

6
x
y

6
xy

 y

N


 4x  3y
 x
  ( IN )  6 x 2 y  6 xy 2
 x
M N

( x  y )
1
y x


 f ( xy )
yN  xM  xy ( x  y ) xy
 I ( x, y )  e

1
d ( xy )
xy
 xy
50
Using part known integrating factor to
determine the integrating factor of the Eq.
Example 3
2
xy
  y  dx  xdy  0
xy 2 dx   ydx  xdy   0
 xy 2 dx  d  xy   0
1
1
dx d  xy 

 2 2 0
2 2
x y
x
x y
1
 ln x 
 C or  x  e
xy

1
xy
B
51
Example 4
xy  y
2



 1 dx  xy  x 2  1 dy  0
2
2
xy

y
dx

xy

x

 
 dy   dx  dy   0
 y  x  y  dx  x  x  y  dy  d  x  y   0
 x  y  d  xy   d  x  y   0 1
1  d xy  d  x  y   0
 
x y
 x  y
 xy  ln x  y  C or e xy x  y  K
52
Problems for Chapter 2
Exercise 2.2
2. (b) 、(f)
3. (b) 、(d)
9.
10. (b)、(f) 、(g)、(h)
12. (b)、(f)
13. (a)、(c)
Exercise 2.3
2. (a)
12. (a)
15. (a)
Exercise 2.4
1. (a)、(f)、(g)、(m)
6. (b)、(e)
7. (c)、(f)
8. (b)
10. (c)
11. (b)
Exercise 2.5
1. (b)、(f)、(i)
2. (b)
5. (b)、(h)、(n)
8. (b)
9. (c)
11.
53