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Chapter 2 Differential Equations of First Order 2.1 IntroductionThe general first-order equation is given by
where x and y are independent and dependent variables, respectively.
0)',,( yyxF )1(
)()(')( 10 xfyxayxa )1(
)2(
2.2 The Linear Equation
where p(x) and q(x) are continuous over the x interval of interest.
)()(' xqyxpy
2
2.2.1. Homogeneous case.
In the case where q(x) is zero, Eq. (2) can be reduced to
Which is called the homogeneous version of (2).
,0)(' yxpy)3(
0p(x)dx + y
dy )4(
Cdxxpy )(ln
dxxp
Bexy)(
)()6(
dxxp
Aexy)(
)(
)7(
x
a- p( )d
y(x) = Ae
When subjected to an initial condition y(a)=bx
a- p( )d
y(x) = eb
A can be positive, zero, or negative
3
Example 2 p.22
'( 2) 0x y xy (0) 3y
4
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 (x).
qpyy ' )16(
The idea is to seek (x) so that
And the left-hand side of (16) is a derivative:
( )d
y qdx
)(' ydx
dpyy )17(
σ(x) is called an integrating factor.
)18(
5
How to find (x)? Writing out the right-hand side of (17) gives
yypyy '''
dxxp
ex)(
)(
)19(
)20(
)()()()(
xqeyedx
d dxxpdxxp
If we choose (x) so that
)'17(
Eq. (17’) is satisfied identically
Putting Eq. (20) into (18), we have
' p
6
Cdxxqeyedxxpdxxp
)()()(
)21(( ) ( )
( ) ( ) ( )
( ) ( ( ) )
= ( )
=
p x dx p x dx
p x dx p x dx p x dx
h p
y x e e q x dx C
Ce e e q x dx
y y
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).
( ) ( )( ) ( ( )
x
a axp d p d
ay x e e q d b
)24(
7
Example 2
32xy y x
8
Some special differential equations1. Bernoulli’s equation
2. Riccati’s equation
nyxqyxpy )()('
)()1()()1(' xqnvxpnv
)()()(' 2 xryxqyxpy
)1.9(
)2.9(
)1.11(
9
3. d’Alembert-Lagrange equation
( ) ( )y xf p g p (13.1)
( ) ( )
( ) ( ) ( )
because
( ) ( ) ( )
( ) [ ( ) ( )]
( ) '( )
( ) ( )
y xf y g y
y f y xf y y g y y
y p
p f p xf p p g p p
p f p xf p g p p
dx f p g px
dp p f p p f p
(13.2)
(13.3)
10
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 given direction if it is the flow of positive charge in that direction.
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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:
R is called the resistance and is measured in ohms; (1) is called Ohm’s law.
( ) ( )E t Ri t (1)
For an inductor, the voltage drop is proportional to the time
rate of change of current through it:
L is called the inductance and is measured in henrys.
(2)dt
tdiLtE
)()(
12
For a capacitor, the voltage drop is proportional to the charge Q(t) on the capacitor:
Where C is called the capacitance and is measured in farads.
(3)
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:
dt
tdQti
)()( (4)
)(1
)( tQC
tE
13
dttiC
tE )(1
)( (5)
From Eqs. (3) and (4) it follows that the desired voltage/currentrelation for a capacitor can be expressed as
According to Kirchhoff’s voltage law, we have
a d b a c b d c(V V ) (V V ) (V V ) (V V ) 0
gives
1( ) - - 0
diE t Ri L idt
dt C (7)
(6)
2
2
1 ( )d i di dE tL R idt dt C dt
(8)
14
Example 1 RL Circuit.
If we omit the capacitor from the circuit, then (7) reduces to the first-order linear equation
)(tERidt
diL )10(
)11(/ ( ) /
0 0
1( ) ( )
tRt L R t Li t i e e E dL
)12(/ /00
/0 00
( ) (1 )
( ) ( )
Rt L Rt L
Rt L
Ei t i e e
RE E
i t i eR R
or
)13(
Case 1: If E(T) is a constant, then Eq.(11) gives
0 0
00
( )( )
t R Rtd d
L LE
Li t e e d i
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Case 2: If E(T)=Eosint and i0 =o, then Eq.(11) gives
/02 2
( ) ( sin cos )( )
Rt LE L Ri t e t t
R L L
)14()(tERi
dt
diL
0 0
0
0
0
sin
sin
0
sint t
h
RT
Lh
R Rd d
L Lp
diL Ri E tdt
Edi Ri t
dt L Ldi R
find i let idt L
i Ce
Ei e e tdt
L
0 0
0
2 20
2
02 2 2
( )
( ) sin
sin
( sin cos )
( sin cos )
t
Rt
Lp
Rd
L
Rt
L
Rt Rt
L L
Rt Rt
L L
i t e
Et e tdt
L
E Lt de
L R
E t Lt e t e
R R R
E L Rt e t e
R L L
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Example 2 Radioactive decay
The disintegration of a given nucleus, within the mass, isindependent 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:
kNdt
dN )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
kmdt
dm )22(
m(t) is the total mass. Solve Eq.(22), we havektemtm 0)( )23(
17
kTemm 0
0
2
Ttmtm 2)( 0)24(
Thus, if t =T, 2T, 3T, …, then m(t) = mm00,, 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:
κ is the net birth/death rate.
dNN
dt (25)
18
teNtN 0)(
Solving Eq. (25), we have
(26)
We expect that κ will not really be a constant but will varywith 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 Ndt
(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 (lb/gal)c1(t):Constant concentration of solute at inlet (lb/gal)x(t): Instantaneous mass of solute (lb)V: Volume of the tank
x dx dcc x cV V
V dt dt
Rate of increase of mass of solute within V=Rate in – Rate out
1 1 1( ) ( ) ( ) ( )dx dc dc Q Q
Q t c t Q t c t V Qc Qc c cdt dt dt V V
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2.4 Separable Equations2.4.1. Separable equations (page 46-48)
)3(( ) ( )y X x Y y
If f(x,y) can be expressed as a function of x times a function of y, that is
then we say that the differential equation is separable.
1
( )
( )
y dx X x dxY y
dyX x dx
Y y
(5)
Example 12y y
Direct
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Example 2 Solve the initial-value problem
4 y(0)=1
1 2 y
xy
e
xy y xdxdye
01421
Observe that if we use the definite integrals.
22
)1(
)2('
yx
yyy
x
dxdy
yy
y
)2(
1
Cx
yy2
)2(ln
2
)17(
)18(
)21(
Example 3 Solve the equation listed in the following
The solution can be expressed as
B is nonnegative. Thus,
02 22 Axyy 211 Axxy
2C2
( - 2)so =e B (0 B< )
y y
x
(22)
2
( - 2)= B A (- <A< )
y y
x (23)
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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
2
2)( t
gtx )26(
'''
'''
'' dxxdtdt
dxxdt
dt
dx
dt
dxdx
dt
dxdxx )27(
gdxdxx '' )28(
Agxx 2'2
1
212' xgx 2)2(4
1)( Ctgtx
)29(
)30( a
'' , (0 t< ) subjected to B.C.
(0) 0 and '(0) 0
x g
x x
(25a)
(25b&c)
Let us multiply Eq. (25a) by dx and integrate on x
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Example 5 Verhulst Population Model.
Naba
NaNba
NbNbNa
1111
)(
111
Ct
a
e
ba
N
N
1
ataCat Bee
ba
N
N
CtNab
aN
a ln
1ln
1
atat AeBe
ba
N
N
00
0
)()(
bNebNa
aNtN
at
)48(
)49(
)50(
)51( )52(
( )
dNdt
a bN N
0( ) ( ) ; N(0)=NN t a bN N
25
Example 6
2 2
2 2
2 2
2 2
2 2
'
1 1
2 2
x y
y x
y x
y x
x y
yy xe e
ye dy xe dx
ye dy xe dx
e e C
e e A
2 2( )' x yyy xe Sol:
Example 7
' 4 /(1 2 )yy x e
2
' 4 /(1 2 )
4
1 2
4 (1 2 ) (1)
2 2
y
y
y
y
y x e
dy x
dx e
xdx e dy
x y e C
26
Indirect
(a) Homogeneous of degree zero
' ( ) ,y y
y f let u dy xdu udxx x
Hence
' ( )
( )
[ ( ) ]
[ ( ) ]
dy xdu udxy f u
dx dxxdu udx f u dx
xdu f u u dx
du dx
f u u x
27
' 3y x
yx y
udxxdudysoux
yset ,
1
2
1
2
1
2
3
2
3
2
3
1 1
3
1 1
3
1 2ln
3 3
2ln ( )
9
xdu u dx
dx u dux
dx u dux
x C u
yx C
x
Example 8
28
)1.11(
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.
(b) Almost-homogeneous equation
1 1 2( , ,...., constants)a b c
1 1 1 2 2 2( ) ( ) 0a x b y c dx a x b y c dy
Steps for changing the Almost-homogeneous Eq. to homogeneous Eq.
1 1
2 2
( ) a b
aa b
1 1 1
2 2 2
'a x b y c
ya x b y c
29
dydvdxdu
yvxulet
,
)(),(
3.
0)()(
'
2211
22
11
dvvbuaduvbuaor
vbua
vbua
du
dv
dx
dyy
2.
tduudtdvtu
vlet
It turns to be a homogeneous of degree zero.
4.
1. Find the intersection (α ,β)
30
Sol:
642
352
yx
yx)1,1(),(
vy
ux
1
1;
dvdy
dudx
so
)2(,
)1(0)42()52(
tduudtdvsotu
vset
dvvuduvu
0))(42()52(
0)42()52(
tduudttdut
dvtdut
Example 10
(2 5 3) (2 4 6) 0x y dx x y dy
31
P.S. tbtbata 4224
22
44
ba
ba
3
4
3
2 ba
2
2
1
(2 5 2 4 ) (2 4 )
2 4( )2 7 4
2 4ln ( )
(1 4 )(2 ) 2 1 4
t t t du t udt
du tdt
u t tt a b
u C dt dtt t t t
1
2
2 1ln ln 2 ln 1 4
3 3
( 4 3)( 2 3)
u C t t
x y y x C
32
1 1 1
2 2 2
( ) a b c
b ma b c
)()()('
,
2
1
222
111
22
zfcz
cmzf
cybxa
cybxafy
sozybxaset
)('
),(
2
2
2
2
zfdxb
dxadz
dx
dyy
thusb
dxadzdy
33
Example 11
( ) (3 3 - 4) 0x y dx x y dy
( ) , set x y z so dy dz dx
(3 - 4)( - ) 0
(3 - 4) ( -3 4) 0
zdx z dz dx
z dz z z dx
,
3 4 3 2 3 1( ) ( ) ( )2 4 2 2 4 2 2
Therefore
zdx dz dz dz
z z z
Integration
3ln 2
23 3
( ) ln 2 ( ) ln 22 2 2
x c z z
xx c x y x y y x y c
34
(c)
' ( )
( ) ( )
set ax by c t adx bdy dt
dt adx dt adxbdy dt adx dy y f t
b bdxdt adx bf t dx dt bf t a dx
' ( )y f ax by c
Example 12 2' tan ( )y x y
2
2
2
( )
tan
(tan 1)
cos
let x y z so dy dz dx
dy dz dxz
dx dx
dz z dx
dx zdz
35
1 cos 2( )
2sin 2
2 4sin(2 2 )
2 42 2 sin(2 2 )
zx c dz
z z
x y x y
y x x y k
36
2
3
2
2
3
' 1 2 1 3
22 3
3
mlet y x v
x
y m
xy y m m m
m m
2 1 2
3 3 32, ' '
3set y x v y x v x v
(c) Isobaric Eqs.
2 2 33 ' 2 0xy y x y Example 13
37
3
3
2
4 1 22 2 2 33 3 3
2 3 3 2 2 2 3
2
2
3
2
3
23 ( )( ') 2 0
3
2 3 ' 2 0
3 ' 1 0
3
ln
( )
v
y
x
x x v x v x v x x v
x v x v v x x v
xv v
dxv dv
x
v x c
e kx
v yx
e kx
38
2.5 Exact equations and integrating factors2
( ) ( )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
( , ) ( , ) ( , ) 0dF x y M x y dx N x y dy
Considering
( , ) 0F F
dF x y dx dyx y
and
( )M F
y y x
( )
N F
x x y
39
Therefore, if
( ) ( )M F F N
y y x x y x
Then Mdx+Ndy is an exact differential equation, and according to the definitions
( , ) ( , )( , ) and ( , )
F x y F x yM x y N x y
x y
( , ) ( , ) ( ) and
( , ) ( , ) ( )
F x y M x y dx g y
F x y N x y dy h x
Example 1
sin ( cos 2 ) 0ydx x y y dy
(10)
40
2.5.2 Integrating factors
Even if M and N fail to satisfy Eq. (10), so that the equation
( , ) ( , ) 0M x y dx N x y dy
is not exact, it may be possible to find a multiplicative factor (x,y) so that
( , ) ( , ) ( , ) ( , ) 0x y M x y dx x y N x y dy
is exact, then we call it an integrating factor, and it satisfied
( ) ( )M Ny x
41
y y x xM M N N
How to find (x,y)
(23)
Perhaps an integrating factor can be found that is a function of x alone. That is y=0. Then Eq. (23) can be reduced to the differential equation
or
( , )( )
y x
y x
M N Nx
M Nx y
x N
(24)
if function of x alone, theny xM N
N
( )y xM N
dxNx e
(25)
(26)
42
or
( , )( )
y x
y x
M M Ny
M Nx y
y M
(27)if - function of y alone, theny xM N
M
( )y xM N
dyMy e
(28)
If (My-Nx)/N is not a function of x alone, then an integrating factor (x) does not exist, but we can try to find s as a function of y alone: (y). The Eq. (23) reduces to
43
( )
( )
( ) ( )
M NExact
y x
M NNon exact
y x
Try Integrating factor
IM IN
y x
44
Example 1 3 2 0ydx xdy
Example 1’ 2 22 ( ) 0xydx y x dy
45
Example 22 2(3 2 ) (2 3 ) 0xy y dx x xy dy
1( )
: ( , ) ( , ) 0
3 4
4 3
( ) 1( )
( )
( , )d xy
xy
Eq M x y dx N x y dy
Mx y
y
Nx y
xM N
x yy xf xy
yN xM xy x y xy
I x y e xy
2 2 3 3 2 2
2 2
2 2
. ( , )
(3 2 ) (2 3 ) 0
( )6 6
( )6 6
Multiply Eq by I x y xy
x y xy dx x y x y dy
IMx y xy
y
INx y xy
x
46
Example 6
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
0, is an integral factor, and if Mx NyMx Ny
10, is an integral factorMx Ny
xy
4 4 3( ) 0x y dx xy dy
47
Example 32 2(3 2 ) (2 3 ) 0xy y dx x xy dy
1( )
: ( , ) ( , ) 0
3 4
4 3
( ) 1( )
( )
( , )d xy
xy
Eq M x y dx N x y dy
Mx y
y
Nx y
xM N
x yy xf xy
yN xM xy x y xy
I x y e xy
2 2 3 3 2 2
2 2
2 2
. ( , )
(3 2 ) (2 3 ) 0
( )6 6
( )6 6
Multiply Eq by I x y xy
x y xy dx x y x y dy
IMx y xy
y
INx y xy
x
48
Example 4 2 0xy y dx xdy
Using part known integrating factor to determine the integrating factor of the Eq.
2
2
2 2 2 2
1
0
0 1
10
1ln xy
xy dx ydx xdy
xy dx d xy
d xydx
x y x x y
x C or x e Bxy
49
2 2 0
0
0 1
10
ln xy
xy y dx xy x dy dx dy
y x y dx x x y dy d x y
x y d xy d x y
d x yd xy
x y x y
xy x y C or e x y K
011 22 dyxxydxyxyExample 5
50
Problems for Chapter 2Exercise 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.