Chapter 2 Differential Equations of First Order

1

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.

11

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

15

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

16

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.

19

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

20

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

21

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)

23

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

24

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.