2aFE1007F1-200809

8/13/2019 2aFE1007F1-200809

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FE1007:

1

r nary eren a qua on

• Topics discussed in this course –

– Limit and derivative of multi-variable functions

– Sequence and Series

– Integral of multi-variable functions

– Ordinary differential equations (ODEs)

2

• CA and the final examination

– CA: Up to 30% (at least 3 quizzes)

– Final examination (2.5 hrs)

Part I: 60-65%, Part II: 35-40%

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Topics Of Ordinary Differential Equations :

1. First order equations

* Linear (Bernoullis Equation)

* Variable separable

* Homogeneous

* Exact differentials

-

3

coefficients

* D-Operator Method

* Homogeneous

* Non-Homogeneous

(1) variation of parameters

(2) undetermined coefficient

4. Approximation Techniques

* D-Operator Method

* Substitution method

3. Extensions to Higher-Order Equations

(a) Euler's method

* Numerical methods: ( covered in 2nd-year Maths )

4

5. Applications of ODE

* Power series method

(b) 2nd-order Runge-Kutta method(Improved Euler's Method)

(c)4th-order Runge-Kutta method

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Part I:

5

First-Order Equations

Differential Equation:

An equation that involves one or more derivatives,(dy/dx, y’ ) , or differentials, ( dx, dy ).

* Ordinary or partial

* Order

6

* Degree

x2

5

2

22

3

3e

1x

y

dx

yd

dx

yd =

++

+

Example:

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( )

( ) ( )

( )x

2

5223e1x

yyy =+++

x2

52 e1x

y)y()y( =

++′′+′′′

7

* Degree : 2

* Ordinary

* Order : 3

First Order D.E. of First Degree:

where M and N represent functions of x and y.

Mdx + Ndy = 0 (1)

A D.E. of first/higher degree and 1st order can alwaysbe put into the form

8

Separable Equations:

N : function of y alone

M(x)dx + N(y)dy = 0 (2)

M : function of x alone

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Integrating (2) gives solution:

∫ M(x)dx + ∫ N(y)dy = c (3)

Example: Solve the following D.E.

d d

9

eydx

+= xey1 2

=+

dxey1

dy x2 ∫∫ =

+ceytan x1 +=−

)cetan(y x +=∴

• One more example:

dyxy 2x 2y 4

dx= + + +

We have

dyx 2 2= + +

10

dx

Therefore,

2

2 2

dy(x 2)dx ln(y 2) x 2x c

y 2

y 2 exp(x 2x c) y exp(x 2x c) 2

= + ⇒ + = + ++

⇒ + = + + ⇒ = + + −

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• Variable separable equation (the simplest and themost basic case)

• Methods for transforming other equations intovariable separable equations:

11

= + dy

F(ax by)dx

Example:

(a) Consider the first order differential equation

where a and b are constants. Show that thechange of variable v = ax + by yields a separable

)byax(fdx

dy+=

12

.

(b) Solve the following differential eqn:

)yx(sindx

dy 2 −=

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v = ax + by(a)

ba

dxdv

b1

dxdy

dxdyba

dxdv −=⇒+=

Solution:

13

Since )byax(fdx

dy+=

∴ − =1 dv a

f(v)

b dx b

)v(bf a

dx

dv+=

)v(bf a dx

dv+=

=+

dv dx

a bf(v): variable separable eqn. (proved)

14

(b)

a = 1, b = − 1

= −2y

sin (x y)dx

)byax(f += 2f sin (v)⇒ =

v = x − y

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Using part (a) result, we get

dx)v(sin1

dv dx

2)v(bfadv =

−→=

+

dx dv vsec dx 2

vcos

dv2

=→=

15

∫ ∫ +=→= cx vtan dx dvsec2

)cx( tanv 1 += −

)cx(tan yx 1 +=− −

• Example:

2 2dyx 2xy y 2x 2y 1

dx= + + + + −

we have

2dy−

16

dxdv dy

Let v x y, we have 1dx dx

= + = +

2

2

dv dv1 v 2v 1 dx

dx v 2v− = + − ⇒ =

+

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1 1 1 vdv dx ln 2x c2 v v 2 v 2

− = ⇒ = + + +

2

dvdx

v 2v=

+

17

2x 2x1 2

v vc e 1 c e

v 2 2

−= ⇒ + = ⇒+

Complete the solution yourself.

• Variable separable equation (the simplestand the most basic case)

• Methods for transforming other equationsinto variable separable equations:

18

=dy y

Fdx x

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Homogeneous Equation:

are called homogeneous equations.

Equations that can be put into the form

=

dy yF

dx x(4)

19

Let v = y/x

y = vx

dy dv = v + xdx dx

=

y Fx

or

or

)v(Fdx

dvxv =+

(5)0)v(Fv

dv

x

dx=

−+

E uation 4 is transformed into E uation 5 which is a

20

variable separable D.E.

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Example: Show that the equation

is homogeneous and solve it.

(x2 + y2) dx + 2xydy = 0

21

where v = y/x .or

y x +y + y x y = = = F

dx 2xy 2( y x) x− −

21+vF(v) =

2v−

Equation (5) becomes:

Integration gives:

2

dx dv + = 0

x 1+vv+

2v

or2

dx 1 dv + = 0

x 3 1+3v

6v

21ln x + ln(1+3v ) = c

3

22

If the initial condition is y = 1 when x = 1, then

c' = 4 and

x3(1 + 3v2) = e3c = c'

23

2

yx 1+3 = 4

x

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One more example:

( )+= −

3

3

dy x y1

dx x

Let x y

v x

+=

dy dvx y vx v x 1

dx dx⇒ + = ⇒ = + −

3 3dv dvv x 1 v 1 x v v+ − = − ⇒ = −

23

3 2

22

1

dv dx v 1 dx dv

x v xv v v 1

1 v 1Hence, ln(v 1) lnv ln x c c x

2 v

= ⇒ − =

− −

−− − = + ⇒ =

Complete the solution yourself.

• Variable separable equation (the simplest andthe most basic case)

•

24

variable separable equations.

• Linear equations and Bernoulli’s equations

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Linear Equation:

Each term of the equation is of degree 1 or 0.

Degree: Add the exponents of the dependent variable

and that of any of its derivatives that occur.

25

: 1st Degree

: 2nd Degree

dx

yd2

2

dxdyy

22

2 : 1st de

d ye

dxgrex

22

2 : 3rd ded y e

dxgrey

A linear D.E. of 1st order can be expressed as:

where P and Q are functions of x only.

Integrating factor:

dyP y Q (6)

dx+ =

26

Multiply Equation (6) by ρ gives

)7(e Pdx∫=ρ

dyρ ρPy ρQ (8)

dx+ =

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L.H.S. of Equation (8) is =

This is because

d(ρy)

dx

Pdxdy d

d dy dρ(ρy) = ρ + y

dx dx dx

27

dx dx

=

∫= +Pdxdy

ρ y e Pdx

= +dy ρ y ρ Pdx

Therefore: d (ρy) = ρQdx

∫ρy = ρQ dx

= ∫1

y ρQ dx (9)ρ

28

ummary:

dy + Py = Qdx

PdxLet ρ = e∫

∫1

then y = ρQ dxρ

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Example: xeydx

dy=+

xdx1 e e ==ρ ∫

dxeee

1 y xxx ∫=

P = 1 Q = xe

29

+= ce

2

1

e

1 x2

x

xx ce e2

1 −+=

Example:

0> xxy3dxdyx 2=−

x=Q x

3P x

x

y3

dx

dy −==−

3 −

30

x

eρ = 3e x

−

= =

dxx

1xdxx

x

1xy

23

33

∫∫ ==1

3 xx c

1

− = + −

2 3x cx= − +

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Bernoullis Equation:

* Divide both sides by yn :

ndyPy Qy (10)

dx+ =

P, Q: functions of x or constants.

31

* Let:

y Py Q (11)dx

− −+ =

ndz dy (1 n)y (13)dx dx

−⇒ = −

1 ny (12)−=z

* Multiply (11) by (1−n)

n 1 ndy(1 n)y (1 n)Py (1 n)Q (14)

dx

− −− + − = −

Combine eqns (12), (13) and (14):

(15)Q)n1(Pz)n1(dz

−=−+

32

which is a linear first-order equation.

Summary:

Let

Bernoulli Equation is transformed into the lineareqn. (15).

1 ny (12)−=z

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Example: Solve the following D.E.

3x4eydx

dy32y =−

Re-arranging the terms, we get

33

− −∴ − = −3x

4 3dy 2 ey y (i)

dx 3 3

3

eyy

3

2

dx

dy 3x4=−

Let

dx

dyy3

dx

dz 4−−=∴

341 yyz −− ==

Multi l in throu h b −3 , the E n i becomes :

34

i.e.

x334 ey2dx

dyy3 =+− −−

x3ez2dx

dz=+

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Letx2dx2 ee ==ρ ∫

dxeeez x3x2x2∫

−=∴

5x2x e

ze c= +

35

2x 5x

3

e e a

5 5y∴ = +

2x3

5x

5ey

e a⇒ =

+

• Variable separable equation (the simplestand the most basic case)

• Methods for transforming other equationsinto variable separable equations.

• ’

36

• Exact Equation

M NM(x,y)dx N(x,y)dy 0 where

y x

∂ ∂+ = =

∂ ∂

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Exact Differential Equations:

The total differential of f, df, is given by :

f(x,y) = CConsider an equation:

0dyf

dxf

df =∂

+∂

=

37

We have an exact ODE : 0dy)y,x(Ndx)y,x(M =+

)y,x(Ny

f and )y,x(M

x

f=

∂

∂=

∂

∂In general :

Example: x yf 5xy e xe 0= + + =

x yf5y e e M(x,y)

x

∂= + + =

∂

yf5x xe N(x,y)

y

∂= + =

∂

38

2 2y yf M f N

5+ ; 5y x y x y x

e e∂ ∂ ∂ ∂

= = = = +∂ ∂ ∂ ∂ ∂ ∂

i.e.x

N

y

M

∂

∂=

∂

∂

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If a D.E. is of the form:

M(x,y)dx + N(x,y)dy = 0 (16)

M(x,y) + N(x,y)dy/dx = 0or

M(x,y) N(x,y)

∂ ∂=

39

then Equation (16) is said to be exact, and

y x∂ ∂

is the solution of Equation (16)

f(x,y) = c (18)

Example : (x2 + y2)dx + (2xy + cos y)dy = 0

(19)with )y,x(Mx

)y,x(f =∂

∂

(20))y,x(Ny

)y,x(f=

∂

∂

40

M(x,y) = (x2 + y2) , y2yM =

∂∂

N(x,y) = 2xy + cos y, y2x

N=

∂

∂

therefore x

N

y

M

∂

∂=

∂

∂

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Next, find a function f(x, y) = C such that :

yxx

f 22 +=∂

∂

∫ ∫ +=∂

∂dx)yx(dx

x

f 22

∂= +

∂

f2xy cos(y)

yand (a)

41

32x

f(x,y) xy g(y)3

= + +

∂

= +∂

f dg(y)

2xy (b)y dy

( We need to fine g(y) )

Comparing (a) and (b), we get :

ycosdy

)y(dg=

or 1g(y) siny c= +

42

3 21

xf(x,y) xy sin y c c

3= + + + =

or3

22

xxy sin y c

3+ + =

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is exact and find the solution such that y (-1) = 0.

(2x + y3)dx + (3xy2 − e-2y)dy = 0

Example: Show that

∂

43

The equation is therefore exact.

yyxy

=+∂

2y22 y3)exy3(x

=−∂

∂ −

There exists a function f(x,y) = C such that :

and f(x,y) = x2 + xy3 + g(y)

,yx2x

f 3+=∂

∂ −= −∂

∂ 2 2yf3xy e

y

44

∂ = +∂

2f dg(y)3xyy dy

-2ydg(y) e

dy⇒ = −

2y1

1g(y) e c

2

−= +

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The solution is then :

2 3 2y1

1x xy e c c

2

−+ + + =

2 3 2y2

1x xy e c

2

−+ + =

45

22

1y( 1) 0 gives ( 1) 0 c

2− = − + + =

23

i.e. c

2

=

Summary:

For ODE

M(x,y)dx + N(x,y)dy = 0

ifM(x,y) N(x,y)

,y x

∂ ∂=

∂ ∂then the solution is

46

( , ) ( , ) ( ) ( , ) ( ) ,( )

'( ) ( , )

F x y M x y dx g y G x y g y cwhere g y can be calculated from

F Gg y N x y

y y

= + = + =∫

∂ ∂= + =

∂ ∂

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• Variable separable equation (the simplest and themost basic case)

• Methods for transforming other equations intovariable separable equations

• Linear Equation and Bernoulli’s Equation

• Exact Differential Equations

47

• Applications of the first-order ODE

(Differential equations characterize the rates of change of certainphysical parameters. To achieve ODE formulation for anapplication problem, the central issue is to identify what ischanging and how the rate of change is decided by various

factors).

A tank with a 1000-litre capacity contains 500 litres ofbrine containing 50 grams of salt. At time t = 0, pure wateris added at a rate of 20 litre/min and the mixed solution isdrained off at a rate of 10 litre/min. How much salt is inthe tank when it reaches the point of overflowing?

Application Example (Mixing Problem):

48

Water : 20 litre/min ( salt input rate to tank = 0 gm/min )

-------------------

------

At time t = 0, 50 grams of salt

in 500 litres of solution

Mixed soln : 10 litre/min

( Salt outpute rate = G gm/litre x 10 litre/min )

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At time t, the amount of solution in the tank:

500 (litre) + (20 − 10) (litre/min) t (min) = 500 + 10t (litre)

Let y(t) be the amount of salt at time t.

dy y y 0 1

gm litre gm( ) ( ) ( )0

−∴ = − =

49

dy y 0 and y(0) 50

dt 50 t+ = =

+

dy 1 dt

y 50 t

= −

+

1t 0, y 50 c 2500= = ⇒ =

2500

lny ln(50 t) c= − + + 1cor lny ln50 t

= +

1c y

50 t⇒ =

+

50

t50 +The tank will overflow when :

500 + 10 t = 1000, i.e., t = 50 min

2500y 25 gram

50 50∴ = =

+

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The time rate of change of the temperature of a body isproportional to the temperature difference between thebody and its surrounding medium.

T : temperature of the body

Application Example (Newton’s Law of Cooling)

51

m .

mdT

k(T T )dt

= − − mdT

i.e., kT kTdt

+ =

k : positive constant T > Tm : cooling

T < Tm : heating

After 20 mins, metal bar temp : 50 0F

Find (a) time taken for the bar to cool down to 25 0F.

b tem erature of the bar when t = 10 mins.

Given metal bar temp T = 100 0F ; room temp = 0 0FmT

Example:

52

mdT

kT kT 0dt

+ = =

ktT ce−=

Assume room temp remainsunchanged.

Two unknowns : c and k

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(i) Given T = 100 0F at t = 0.

−∴ = =k(0)100 ce or c 100

ii Given T = 50 0F at t = 20.

ktT 100 e−=

53

k(20)50 100 e−=

k = 0.035

0.035tT 100 e−=

(a) Determine t when T = 25 0F.

Find (a) time taken for the bar to cool down to 25 0F.

(b) temperature of the bar when t = 10 mins.

54

0.035t25 100e

−=

t = 39.6 min

(b) Answer: 70.5 0F. (Work it out yourself.)

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A quantity is said to have an exponential growth (decay) model if at each instant of time, its rate ofincrease (decrease) is proportional to the amount ofthe quantity present.

Application Example (Exponential Growth):

55

dy k y

dt=

,

where k : constant of proportionality. =

dy

dtky

= =dydt

dyyk

y dt

dy k dt

y=

: relative growth rate

kt c e=

56

kt0y(t) y e=

Initial condition: y(0) = y0

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Example:

The population of a certain country is known to increase ata rate proportional to the number of people presently livingin the country. If after 2 years the population has doubled

and after 3 years the population is 20,000, find the numberof people initially living in the country.

57

Let x(t) : No. of people in the country at time t

g : growth rate

dx gx

dt

= gt0 0x(t) x e , x(0) x= =

= =0 02g 1

x e , g ln(2) =2x 0.3472

0.347t0x(t) x e∴ =

= = 0t , x 2x2 ,(i) Given that when i.e.,

58

= , = , . .,

3347.00 ex 000,20 ×=

1.0410x 20,000 e−= 7062=

The number of people initially living in the country = 7062.

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Doubling and Halving Time

Suppose y has an exponential growth model, i.e.,

y = y0ekt (k > 0) (1)

At time t1,

59

1 1 0y y(t ) y e (2)= =

Let T denote the amount of time required for y todouble in size. We have,

1k(t T)

1 1 0 2y y(t T) y e

+

= + =

ln(2) kT=

1kt kT1 0 2y y e e⇒ =

kT1 1i.e., 2y y e=

kT2 e=

(3)1

T = ln(2)∴

60

Note: T does not depend on y0 or t1.

At time t1 , we have y1 . At time t1+T, we have 2y1.

At time t2 , we have y2 . At time t2+T, we have 2y2.

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1k t

1 1 0y y ( t ) y e= =

Suppose y has an exponential decay model, i.e.,

y = y0ekt (k < 0)

At time t1,

Let T be the half time of the decay model.

61

We then have : 1k(t T)1 1 0

1y y(t T) y e

2

+= + =

kT1 e2

=

1k kT kT1 0 1

1 ty = y e e = y e2

(4)1 T = ln(2)k−∴

1 ln 2= − −=

1k ln 2= −

Example:

The radioactive element carbon-14 has a half-life of5750 years. If 100 grams of this element are presentinitially, how much will be left after 1000 years?

62

5750

.

0y y(0) 100= =

0.00012(1000) y(1000) 100e−∴ =

0.12 100e−= 88.692 (gm)=

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Summary: First-Order ODE (1)

• Variable Separable Equation (the simplestand the most basic case)

• Methods for transforming other equationsinto variable separable equations

dyF ax b= +• First order differential equation:

63

xv ax by= +

dy yF

dx x

=

y

v ,

x

= then dy dvv x

dx dx

= +

Let

• Homogeneous Equation:

Let

Summary: First-Order ODE (2)

• Linear First-Order Equation:

Let

• Bernoulli’s Equation:

Pdx,e ∫ρ = then

dyPy Q

dx+ =

1y Qdx= ρ

ρ ∫ndy

Py Qydx

+ =

• Exact Differential Equations

M(x,y)dx+N(x,y)dy =0, where

Let z y ,−= en

dz(1 n)Pz (1 n)Q

dx+ − = −

M N

y x

∂ ∂=

∂ ∂

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2aFE1007F1-200809