ChewMA1506-14 Ch1

MA1506 Mathematics II

Group A Monday 800-1000 Wed 1600-1700 UT-AUD1 Lecturer: Chew Tuan Seng

Group B Wed 800-1000 Friday 800-900 UT-AUD2

Lecturer: Quek Tong Seng

Chew T S MA1506-14 Chapter 1

follow the contents of Lecture Note but our presentation may be different

Chapter 1 Differential Equations

1.1. Introduction

In this chapter, we deal only with diff. eqs. containing derivatives, called ordinary diff. eqs. Chew T S MA1506-14 Chapter 1

A differential equation is an equation containing derivatives or partial derivatives

1st order ordinary diff. eq.

2nd order ordinary diff. eq.

2(1 ) xdy y edx

2 ( )d y dyA By R xdxdx

( ) ( 2)2

( ) ( 2) 1n n

d y d y xdx dx

−+ = + nth order ordinary diff. eq.

1.1. Introduction

Chew T S MA1506-14 Chapter 1 5

( ) ( 1) (1)1 1 0( ) ( ) ( ) ( ) ... ( ) ( ) ( ) ( ) ( )n n

n na x y x a x y x a x y x a x y x F x−−+ + + + =

An ODE of the following form is called a linear ODE

Examples: 2

2( )d y dyA By R x

dx dx+ + =

2 , yxd y dy dye xdx dxdx

Above are linear ODEs

Below are NOT linear ODEs

1.1. Introduction

In general, ODE has many solutions, e.g. siny x c= + , c is an arbitrary constant, is a solution of ' cosy x=

Such solutions, containing arbitrary constants are called general solution Giving a specific value to constant c, say c=1, we get 𝑦 = sin𝑥 + 1, which is called a particular solution

1.1. Introduction

In this Chapter, we study 1st order ordinary differential equation and its applications

2nd order ordinary differential equations

Applications of 2nd order will be given in Chapter TWO

1.1. Introduction

1.2 Separable equations

We study 1st order ODE of the following form

We shall learn how to solve separable equations by examples

( )( )

dy M xdx N y

=( ) ( )M x dx N y dy=

Example 1

We write

Then integrate both sides

1xe dx dy

+∫ ∫Chew T S MA1506-14 Chapter 1

2(1 ) xdy y edx

xe dx dyy

Example 1 (cont)

1xe dx dy

+∫ ∫

1tanxe y c−= +1tan xy e c− = −

tan( )xy e c= −

Example 2

A radioactive substance decomposes at a rate proportional to the amount present i.e.,

Radioactive Decay

dx xdt

dx kxdt

=Hence

( )x t

where k is a constant

Example 2 (cont)

dx kdtx=∫ ∫

ln x kt c= +( ) kt c c ktx t e e e+= =( ) ktx t Ae=

dx kxdt

where cA e=

What is the value A ?

( ) ktx t Ae=We have 0(0) kx Ae=

(0)A x=( ) (0) ktx t x e=

Example 2 (cont) 1.2 Separable equations

we have

Example 2 (cont)

Now we shall find the value of k

Assume the half-life of the substance is τ

ln(1/ 2) ln 2kτ τ

= = −

1( ) (0)2

x xτ =

( ) (0) ktx t x e=

The half-life is the time required for half of the substance to decay. So at time , the amount of the substance is ½ x(0)

1 (0) ( ) (0)2

kx x x e ττ= =

Hence we have ln 2

( ) (0)t

x t x e τ−

=Remark: We may use

instead of

We will get the same equation as in the above

dx kxdt

= −dx kxdt

( ) (0)t

x t x e τ−

Example 3

Temperature of an object at time 0 is it is placed in a medium of constant temperature Temperature of the object is at time (is given) Find the temperature of the object at time

Cooling(Heating) Problem

0T1( )T t

1t( )T t

Example 3 (cont)

Physical information: Rate of change dT/dt of the temperature T of an object is proportional to the difference between T and the temp T0 of the medium

Newton’s Law of Cooling (Heating)

0( )dT k T Tdt

Example 3 (cont)

0ln( )T T kt c− = +

0( ) kt c ktT t T e Ae+− = =What is the value of A?

00(0) kT T Ae− =

dT kdtT T

=−∫ ∫0( )dT k T T

dt= −

We assume

0( )T t T>cooling so

0(0)A T T= −

0 0( ) ( (0) ) ktT t T T T e− = −Now we shall find k By given, we know the value of 1( )T tSo

1 0 0( ) ( (0) ) ktT t T T T e− = −

Example 3 (cont)

( ) 1ln(0)

T t TkT T t −

= − Hence we have

0 0( ) ( (0) ) ktT t T T T e− = −where k is given above

Similarly, for heating, 0( )T t T< ,we get the same formula

Example 4

Newton’s 2nd Law

g=acceleration due to gravity =9.8m/s2

Retarded fall—air resistance

air resistance = bv2

m=weight of the man + equipment

where b is a constant

2dvm mg bvdt

We shall discuss the case when air resistance = bv in Example 9

0 Starting pt

Example 4 (cont)

2dvm mg bvdt

2 2( )dv b v kdt m

= − −

2 mgkb

=where

Example 4 (cont)

2ln v k kb t cv k m− = − + +

1 bdv dtv k m

= −−

1 1 12

bdv dtk v k v k m − = − − +

Example 4 (cont)

ptv k cev k

−−=

+2kbpm

=where

How to find c?

0(0)(0)

pv k cev k

−−=

(0)(0)

v kcv k

cev kce

Now suppose

K=4.87, c=0.345,p=4.02

cev kce

1 0.3454.871 0.345

lim ( ) 4.87t

v t→∞

Example 4 (cont)

Skydiver falling at terminal velocity

http://www.graphmatica.com

We can draw graph using graphmatica

Example 5 Mixture problem 1.2 Separable equations

air or pure water

Some substance (e.g. CO, salt) flows into a tank (room), is mixed uniformly with the contents (e.g. air , pure water) of the tanks, and flows out with the mixture (air with CO, water with salt).

CO or salt air with CO or water with salt

We want to find the amount x(t) of the substance in the tank at time t

= the rate at which the substance flows into the tank (in flow) the rate at which the substance flows out the tank (out flow)

Now we shall give an example.

Example 5 (cont)

A 2000m3 room contains air with 0.002% CO at time t=0

The ventilation system blowing in air which

contains 3% CO The system blowing in and out air at a

rate of 0.2m3/min

Mixture problem

Example 5 (cont)

Let x(t) = vol of CO in the room at time t 0.2 m3 /min

3% CO 0.2 m3 /min

Room 2000 m3

CO per m3 in the tank

= Inflow of CO - outflow of CO

0.006 0.0001x= −

0.03 0.2= × 0.22000

x− ×

Example 5 (cont)

ln(60 ) 0.0001x t c− − = +

0.006 0.0001dx xdt

= − 0.0001(60 )x= −

0.000160

dx dtx=

ln(60 ) 0.0001x t c− = − −

0.0001 0.000160 t c tx e ke− − −− = =0.000160 tx ke−= −

Now we shall find k

A 2000m3 room contains air with 0.002% CO at time t=0. Hence X(0)=2000x0.002%=2000x0.002/100=0.04

00.04 (0) 60x ke= = −

59.96k =0.000160 59.96 tx e−= −

Example 5 (cont)

0.015% CO means x(t1) = 0.00015 X 2000 = 0.3

When the air in the room containing 0.015% CO?

10.00010.3 60 59.96 te−= −

1 43.5mint ≈

0.000160 59.96 tx e−= −

What happens when ODE is not separable? For examples,

2 tricks:

• reduction to separable

• linear change of variables Chew T S MA1506-14 Chapter 1

2 22 0dyxy y xdx

− + =

(2 4 5) 2 3 0x y y x y′− + + − + =

Reduction to separable form

y v y vxx= ⇒ = y v xv′ ′= +

Now we shall give one example

' ( ) dvy f v v xdx

( )dv dx

f v v x=

Suppose

Example 6 : Reduction to separable form

2 22 0dyxy y xdx

− + =

2 2 1 12 2 2

dy y x y xdx xy x y

−= = −

y vx=Let We have

where See previous slide

( )dv dx

f v v x=

−1 1 1( )2 2

f v vv

Example 6 (cont)

v dv dxv x

= −+

1 1( 1)1

d v dxv x

+ = −+

1 1 12 2

dv dxxv v

=− −

2ln( 1) lnv x c+ = − +2ln(( 1) )v x c+ =2

1( 1) cv x e c+ = =

12 1y x cx

Linear Change of Variable

ax by cya x b y c

+ +′ =+ +

Consider ODE of the form

We shall give one example to illustrate the method

2 32 4 5

dy x ydx x y

− + −=

− +( 2 ) 3

2( 2 ) 5x yx y

− − −=

Let u=x-2y The above can be done since two st lines -x+2y-3=0 and 2x-4y+5=0 are parallel

Example 7

We may let u=-x+2y

Example 7 (cont) ( 2 ) 3'

2( 2 ) 5x yyx y

− − −=− +

Let u=x-2y Then 1 2du dydx dx

1 3(1 )2 2 5

du udx u

− −− =

+4 112 5

du udx u

subst into the above eq

2 54 11

u du dxu+

Example 7 (cont)

2 5 1 1 14 11 2 2 4 11

= −+ +

11 24 11

du dxu

− = +

1 ln(4 11) 24

u u x c− + = +

1( 2 ) ln(4( 2 ) 11) 24

x y x y x c− − − + = +

4( 2 )4( 2 ) 11 x yx y Ae − −− + =

2 54 11

u du dxu+

4( 2 ) ln(4( 2 ) 11) 8 4x y x y x c− − − + = +

4 cwhere A e=

1.3 Linear 1st Order ODEs

called Integrating factor

Std form

( ) ( )dy p x y Q xdx

( )p x dxe∫

( ) ( ) ( )( ) ( )

p x dx p x dx p x dxdy e p x ye Q x edx

∫ ∫ ∫+ =

The given ODE multiplied by integrating factor, get

We can check that the left hand side of the above

( ) ( )( )

p x dx p x dxdy e p x yedx

∫ ∫+

( )p x dxd yedx

1.3 Linear 1st Order ODE

cont. ( ) ( ) ( )

( ) ( )p x dx p x dx p x dxdy e p x ye Q x e

dx∫ ∫ ∫+ =

( ) ( )( )

p x dx p x dxd ye Q x edx

∫ ∫=

( ) ( )( )

p x dx p x dxye Q x e dx∫ ∫= ∫

cont. ( ) ( ) ( )

( ) ( )p x dx p x dx p x dxdy e p x ye Q x e

dx∫ ∫ ∫+ =

( ) ( )( )

p x dx p x dxd ye dx Q x e dxdx

∫ ∫= ∫ ∫

( )p x dxd yedx

( )p x dxdy edx

∫=( )p x dxdy e

dx∫+

( ) ( )( )

p x dx p x dxdy de ye p x dxdx dx

∫ ∫= + ∫( ) ( )

( )p x dx p x dxdy e ye p x

dx∫ ∫= +

( ) ( )( )

p x dx p x dxdy e p x yedx

∫ ∫+( )p x dxd ye

dx ∫=

( ) ( )d p x p xdx

=∫Use the following example to illustrate

coscos

d xdxx

dx∴ =∫

𝑑 ∫ cos𝑥𝑑𝑥𝑑𝑥

=𝑑(sin𝑥 + 𝑐)

𝑑𝑥= cos𝑥

Example 8 (i)

23xy y x′ − =13y y xx

′ − =

( ) ( )( )

Use formula

First compute integrating factor 1( 3 ) 3 lndx xxe e

− −∫ =

3ln xe−

= 3x−=

Hence, from

( ) ( )( )

we have

3 3 2 1yx xx dx x dx x c− − − −= = = − +∫ ∫

cont. 1.3 Linear 1st Order ODE

Lecture Note Example 8 (ii) Exercise

Example 9 An object of mass m dropped from rest

Newton 2nd Law

Find the position x(t) and velocity v(t) at time t.

A resistance to the object is proportional to the magnitude of the velocity of the object.

Retarded fall—air resistance

dvm mg bvdt

(Similar problem has been discussed in Example 4)

(0) 0, (0) 0x v= =x

0 starting pt

Integrating factor

dv b v gdt m

b btdtm me e∫ =

bt bt btm m mmve ge dt g e c

b= = +∫

By formula

How to find c? Use v(0)=0, from above, get mc gb

dvm mg bvdt

( 1)bt bt btm m mm m mve g e g g e

b b b= − = −

(1 )btmmv g e

(1 )btmdx mg e

dt b−

( ) (1 )btmmx t g e dt

= −∫

( ) (1 )btmmx t g e dt

= −∫( ) ( )

btmm mx t g t e d

b b−

How to find d? use x(0)=0, from above, get

00 (0) (0 )m mx g e db b

= == + +2

amt of water = constant=100 gal

3 gal/sec 100 gal water

3 gal/sec salt=0.25lb/gal

Example 10

=inflow - outflow

salt/gal

Let Q(t) be the amount of salt in the tank Q(0)=20

Mixture Problem

= 3 0.25× 3100Q

− ×amt of water

= constant

=100 gal

3 0.75100

dQ Qdt

( ) ( )( )

Use formula

310025 5

Q e−

= −get lim ( ) 25

→∞=

Similar problem has been discussed in Example 5

Example 11 A decay chain

In Example 2, we have discussed radioactive

decay. However, some radioactive elements

are transformed into unstable elements.

The product (e.g., Thorium 230) of a radioactive (e.g., Uranium 234) decay is itself a radioactive element

Let U(t) be the amount of Uranium at time t.

Let T(t) be the amount of Thorium at time t.

We assume that each decay of one Uranium atom produces one Thorium atom.

Hence Thorium atoms are being born at exactly the same rate at which Uranium atoms die.

1.3 Linear 1st Order ODE cont.

Birth rate of Thorium=(-1)Death rate of Uranium

So we have

dT K Tdt

= − UK U+

Thorium atoms are being born at exactly the same rate at which Uranium atoms die

dU K Udt

We assume U TK K≠

0(0)U U= (0) 0T =

Uranium

Thorium

We shall solve the above system of ODEs

From the 1st equation, we have

See Example 2

Hence 2nd equation becomes

dT K T K U edt

−+ =

0K tuU U e−=

dT K Tdt

= −UK U+

It is 1st order linear ODE. Hence by formula,

see Section 1.3, slide 47, we get

K dt K dtT TK tuUTe K U e e dt−∫ ∫= ∫

0K tK t K tuT T

UTe K U e e dt−= ∫

( )0( ) K K tK t U T uT

KT t e U e CK K

−= +−

Now we shall find the constant C. By given

(0) 0T =We get

KC UK K

= −−

( )0( ) (1 )K t K K tU u u T

KT t U e eK K

− −= −−

Unfortunately, we don’t know 0U

However

( )( ) (1 )( )

K K tU u T

KT t eU t K K

−= −−

0K tuU U e−=

Hence we can find the time t if we know the ratio ( )

( )T tU t

It is known that

Hence ( )( )

KT tU t K K

→− as t →∞

U TK K<

( )( ) (1 )( )

K K tU u T

KT t eU t K K

−= −−

ancient corals How old the above ancient corals?

( )( ) (1 )( )

K K tU u T

KT t eU t K K

−= −−

( )( )

T tU tIf we know

then we know the answer

Bernoulli Equations p23

Given eq multiplied by (1-n) get

( ) ( ) ny p x y q x y′ + =If n=0 or n=1, then it is 1st order linear ODE, which has been just discussed

(1 ) (1 ) ( ) (1 ) ( )n n n ny n y n y p x y n y q x y− − −′ − + − = −

Hence Bernoulli equation becomes 1st order linear ODE

1 nz y −= (1 ) nz n y y−′ ′= −

(1 ) (1 ) ( ) (1 ) ( )n n n ny n y n y p x y n y q x y− − −′ − + − = −

Let Then

So, from

we get

z′ + (1 ) ( )n p x z− = (1 ) ( )n q x−

Example (ii) Bernoulli Equation p24

2 2y y x y′ + =1 2 1z y y− −= = By formula

get 2(1 2) (1 2)z z x′ + − = −

Solve this 1st order linear ODE, get

Example (i) see Lecture note

' (1 ) ( ) (1 ) ( )Z n p x z n q x+ − = −

1 nz y −=

( ) ( ) ny p x y q x y′ + =

1 2( 2 2)x xe y e x x c− − −= + + +

2( )x xe z x e dx− −= − =∫ 2( 2 2)xe x x c− + + +Integration by parts

Review: First Order ODE • Separable

• Linear

( ) ( )M x dx N y dy=

• Bernoulli ( ) ( ) ny p x y q x y′ + =( ) ( )y p x y Q x′ + =

ax by cya x b y c

+ +′ =+ +

1st Order ODE

has many solutions. However if an initial condition y(x0)=y0, (very often x0=0) is given, then there is one and only one solution, i.e., the solution is unique.

( ) ( )y p x y Q x′ + =

1st Order ODE

Review: Applications of 1st order ODE

Radioactive Decay

Cooling (Heating)

Retarded fall-air resistance

Mixture problem

Radioactive Decay chain

1st Order ODE

1.4 Second order linear ODE

The general form of 2nd order linear ODE is

When F(x) is zero function, we have This equation is called homogeneous.

2( ) ( ) ( )d y dyp x q x y F x

dx dx+ + =

2( ) ( ) 0d y dyp x q x y

dx dx+ + =

When F(x) is not zero function, is called nonhomogenous.

2 ( ) ( ) ( )d y dyp x q x y F xdxdx

1.4 Second-order linear ODE

Examples

linear, nonhom

nonlinear

linear, hom

Superposition principle (only for homogeneous case) p27

If y1 and y2 are solutions then

c y1 + d y2 is also a solution

To prove the above result, first recall

Proof: Superposition principle

If y1 and y2 are solutions then so is c y1 + d y2

Caution

Superposition principle does not hold for nonhomogeneous ODE

are solutions but 1 cos , 1+sinxx+

1 sin 1 cosx x+ + + is Not a solution

Example 12

are solutions.

First we can check that By superposition principle,

is a solution

By initial value condition , we get

initial value condition

1 24, 1c c= =

0 01 2 1 2(0)y c e c e c c= + = +

1 2' x xy c e c e−= −So 0 0

1 2 1 2'(0)y c e c e c c= − = −

Linearly independent p29

• Two solutions u(x) and v(x) are said to be linearly dependent

if we can find a constant c such that u(x)=cv(x), for all x, otherwise they are linearly independent • For examples, sinx and cosx are linearly

indep; sinx and 2sinx are linearly dep.

Theorem: For Hom. 2nd order linear ODE

If y1 and y2 are linearly independent solutions then general solution is y = c1 y1 + c2 y2

Particular solution: Fix some values of c1 and c2

Example

two linearly indep solutions

General solution

Particular solution

Homogeneous ODE with constant coefficient p31

2 0d y dyA Bydxdx

It is clear that function is a solution

which is called a trivial solution or zero solution

My presentation is slightly diff from the L N

Now we shall look for nontrivial (nonzero) solution Recall that the general solution of first linear homogeneous ODE is

( ) 0dy p x ydx

( )p x dxy Ce

−∫=

(cont)

From this solution , we may guess that a nontrivial solution of

is of the form xy eλ=Chew T S MA1506-14 Chapter 1

Consider a special case: when p(x) is constant , say D. Then the general solution is Dxy Ce−=

2 0d y dyA Bydxdx

(cont)

Subst. these into the given ODE, get

which is called characteristic equation or auxiliary equation

xdy edx

λλ=2

xd y edx

λλ=Then

2 0x x xe A e Beλ λ λλ λ+ + =2( ) 0xA B eλλ λ+ + =

(cont)

Hence 2( ) 0A Bλ λ+ + =

When solving there are three cases: (See p32)

• Two distinct real roots • Only one real root • Two distinct complex roots

2 0A Bλ λ+ + =

(cont)

(a) Two distinct real roots Suppose that two distinct real roots are and

Then we have two distinct (linearly independent) solutions

1λ 2λ

1xy eλ= 2xy eλ=

1 21 2

x xy c e c eλ λ= +Chew T S MA1506-14 Chapter 1

General soln is

In fact , we can prove that every solution is of the form Here and are any constants. 1C 2C

1 21 2

x xy c e c eλ λ= +

(cont)

Example 13

Solution: Let Subst this y into the given ODE , get

xy eλ=

Solve 𝑦" + 𝑦′ − 2𝑦 = 0

𝑦(0) = 4,𝑦"(0) = −5

We have two distinct real roots,

1 21, 2λ λ= = −

2 2 0λ λ+ − =

(cont) Thus the general solution of the equation is we get 𝑦 = 𝑒𝑥+3𝑒−2𝑥

x xy c e c e−= +𝑦(0) = 4,𝑦′(0) = −5 By initial condition,

(b)Only one real root Suppose that the only one real root is Then we have a solution For 2nd order ODE, we can prove that we should have two distinct (linearly

indep.) solutions. What is the 2nd solution?

1xy eλ=

The 2nd solution is We can verify that (see p34)

is also a solution (superposition principle) In fact , we can prove that every solution is

of the form

1xy xeλ=

1 11 2

x xy c e c xeλ λ= +

1 11 2

x xy c e c xeλ λ= +

(cont)

Example14 (ii): Solve The auxiliary equation is We have only one solution Hence the general solution is (Example 14 (i) Exercise )

2 4 4 0d y dy ydxdx

− + =

2 21 2

x xy c e c xe= +

2 4 4 0λ λ− + =1 2λ =

(c) Two distinct complex roots Suppose that we have two distinct complex roots, namely and Then we have two distinct (linearly indep) complex-valued solutions and Suppose that Then

1λ 2λ

1xy eλ= 2xy eλ=

1 a ibλ = +

2 a ibλ = −Chew T S MA1506-14 Chapter 1

Note that these two solutions are complex-valued . However we want real-valued solutions. How to get real-valued solutions ?

We shall look at the real part and imaginary part of the complex-valued solution

(cont)

1x ax ibxy e e eλ= =

cosaxe bx= i+ sinaxe bx

(cont) 1.4 Second-order linear ODE

axe= (cos sin )bx i bx+

We can verify that the real part and the imaginary part are two (real-valued) solutions

cosaxy e bx=

sinaxy e bx=

(cont) cos sinax axy e bx ie bx= +

we can prove that every solution is of the form

cos sin

( cos sin )

y c e bx c e bx

e c bx c bx

Do we need to consider ANS: NO, since it induces the same general solution.

2xy eλ=

(cont)

Example 15 (i) Solve y" + 2y′ + 5y = 0 The complex roots of the auxiliary equation are Hence the general solution is. Example 15 (ii) Sovle (i) with From (i) and initial condition, we get

1 21 2 , 1 2i iλ λ= − + = − −

1 2( cos 2 sin 2 )xy e c x c x−= +

(0) 1, y'(0)=5y =

(cos 2 3sin 2 )xy e x x−= +

(0) 1, y'(0)=5y =

cos sin

( cos sin )

y c e bx c e bx

e c bx c bx

2nd-order nonhomogeneous linear ODE

From now onwards till the end of this chapter, pp 38-54, my presentation is different from the L. N. . However, the content remains unchanged.

We deal only ODEs with constant coefficients

2nd-order nonhomogeneous linear ODE with constant coefficients

The general form is Solving this equation can be reduced to three steps 1.Find the general solution to the

homogeneous equation

, say the solution is

2 0d y dyA Bydxdx

hyChew T S MA1506-14 Chapter 1

2. Find a particular solution to the nonhomogeneous equation

3. Add the solutions from step 1 and step 2 , get + ,which is the general solution to (see Appendix 1)

We have learnt step 1. There are two methods for step 2. Method 1. The method of undetermined coefficients. Method 2. The method of variation of parameters.

Now we shall use examples to illustrate method 1

3 2 10y y′′ + =

Method 1. The method of undetermined coefficients

Example 1

Guess a solution? If the function R(x) on the right hand side is constant, then we can guess that

py A=This is always true except some special but important cases, see Examples 3-6

2'' ' 2 4y y y x− − =

2py Ax=

2py A Bx Cx= + +

No, no

The correct form is

Example 2 Solve

Can we guess a solution?

( ) ' 2py B Cx= + ( ) '' 2py C=

2'' ' 2 4y y y x− − =

get So

2 22 2 2 2 2 4C B Cx A Bx Cx x− − − − − =

2, 2, 3C B A= − = = −

Now shall find A, B, C

Subst above into

(cont)

2C-B-2A=0 -2C-2B=0 -2C=4 Hence

(cont) Hence is a particular solution of

23 2 2py x x= − + −

2'' ' 2 4y y y x− − =On the other hand is the general solution of

x xhy C e C e−= +

'' ' 2 0y y y− − =Chew T S MA1506-14 Chapter 1

Therefore is the general solution of the nonhomogeneous ODE Here and can be any constant

2 21 2 3 2 2x x

h py y C e C e x x−+ = + − + −

2'' ' 2 4y y y x− − =

(cont)

Now we shall consider two special but important cases

'' ' 0y Ay+ = '' 0y =The general solution of '' ' 0y Ay+ =is

1 2 1 2ox Ax Ax

hy c e c e c c e− −= + = +

Hence constant functions are solutions of

'' ' 0y Ay+ =

This result will be used in Examples 3-4

The general solution of '' 0y =is 1 2hy c c x= +

Hence constant functions and functions are solutions of

1c2c x '' 0y =

Example 3 A particular solution of '' ' 10y y+ =

is of the form py xA=Why we have extra term x

Now consider 2nd case

This result will be used in Examples 5-6

The guiding principle is: Need to ensure that no term in a particular soln is a soln of the corresponding homogeneous ODE If we let py A=then we can’t find such A, since constant A is a soln of '' ' 0y y+ =The correct form is py xA=Subst this into '' ' 10y y+ = get 10A =So 10py x=

Example 4 '' 'y y x+ =

A correct form of ( )py x A Bx= +

Suppose we let py A Bx= +

Then we can’t find such A and B, since in this particular solution , there is one term, namely, A, is a solution of '' ' 0y y+ =Subst. ( )py x A Bx= + into '' 'y y x+ =

get 1, 1/ 2A B= − = So 21

2py x x= − +

Example 5 '' 10y =

The correct form of 2

py x A=We can’t let py A= py xA=Since they are solutions of the corresponding homogeneous ODE '' 0y =Subst. 2

py x A= into '' 10y =get A=5 so

25py x=In fact, we can get this result by integrating the ODE '' 10y = twice

Example 6 3''y x=

The correct form of

2 2 3( )py x A Bx Cx Dx= + + +

We can’t let 2 3( )py x A Bx Cx Dx= + + +

2 3py A Bx Cx Dx= + + +

Subst.

2 2 3( )py x A Bx Cx Dx= + + +

into 3''y x=get A=B=C=0, D=1/20 So 5(1/ 20)py x=In fact, we can get this result by integrating the ODE 3''y x= twice

See Example 5

Since A, xA are solutions of the corresponding homogeneous ODE '' 0y =

Important Remark: From the above examples, we know that a particular solution of

not only depend on R(x) but also depend on the general solution of

2 0d y dyA Bydxdx

So when we want to find a particular solution, first we need to find the general solution of

2 0d y dyA Bydxdx

(A)The general solution of is (1)A particular solution of is of the form (2)A particular solution of is of the form

Example 7 '' 3 ' 4 0y y y− − =

x xc e c e−+2'' 3 ' 4 xy y y e− − =

2 xAe4'' 3 ' 4 xy y y e− − =

4 xxAe

(cont) (B)The general solution of is So a particular soln of is of the form

'' 2 ' 0y y y+ + =

'' 2 ' xy y y e−+ + =

2 xx Ae−

1 2x xc e c xe− −+

Why we have extra term 2x

are solutions of

,x xAe xAe− −

2 xx Ae−

'' 2 ' 0y y y+ + =

First the general solution of is

'' 3 ' 4 0y y y− − =4

1 2x x

py c e c e−= +

Example 8 Solve '' 3 ' 4 2siny y y x− − =

We guess a particular solution is of the form

py = cosA x + sinB x

123 Chew T S MA1506-14 Chapter 1

A particular solution

As in Example 2, we can find the values of A and B

3 /17, 5 /17A B= = −

py = sinB xcosA x +

So the general solution of is Chew T S MA1506-14 Chapter 1

'' 3 ' 4 2siny y y x− − =

h py y+

Hence a particular solution is (3 /17)cos ( 5 /17)sinpy x x= + −

Find a particular soln of First, the general soln of is

Example 9

'' siny y x+ =

'' 0y y+ =

1 2sin cosc x c x+

A particular soln is of the form

since Asinx and Bcosx are solutions of

with the extra term x

( sin cos )py x A x B x= +

'' 0y y+ =

(cont) We can check that Hence a particular soln is The general soln of is

'' siny y x+ =

0, 1/ 2A B= = −

( 1/ 2)cospy x x= −

1 2sin cos (1/ 2) cosc x c x x x+ −

2nd order nonhomogeneous linear ODE

The general solution is

hy py+ The general soln for homogeneous ODE

A particular soln for nonhomogeneous ODE

How to find

2 ( )d y dyA By R xdx dx

Review

pyHow to find The method of undetermined coefficients.

Notation used in the following slides

0 1( ) , 0nn n nP x a a x a x a= + + + ≠

0 1( ) , 0nn n nQ x A A x A x A= + + + ≠

(1) ( )R x C=

spy x A=

( ) ( )nR x P x= ( )sp ny x Q x=

s xpy x Aeα=(3)

( sin cos )spy x A bx B bx= +

( ) xR x eαβ=

( ) sinR x bxβ=

( ) cosR x bxβ=

S=0,1,2 Four basic and important cases

What is s? s is the smallest nonnegative integer such that no term in

is a solution of the corresponding homogeneous ODE (or corresponding homo. Complex-valued ODE)

What is the meaning of term used above? 3 2 2

3 2 2 2 2 2

x x x x

Ax Bx Cx D e

Ax e Bx e Cxe De

= + + +In the above, there are four terms

Consider We can guess that a particular soln is By the method used in previous examples, we can find A, B, C, D. However the computation is very involved. We will use the following method to simplify the computation

3 2 2( ) xAx Bx Cx D e+ + +

Example 10 3 2'' 4 ' 2 2 xy y y x e− + =

Four more Examples 10-13, computations are very involved

(cont) Let Then a particular soln is

Subst the above into the given ODE, get,

We have

2 xy ue=2 22x xy u e ue′ ′= +

2 2 24 4x x xy u e u e ue′′ ′′ ′= + +

3" 2 2u u x− =

3 2( )u x Ax Bx Cx D= + + +

Thus a particular soln is

Now subst

We can find A,B, C, D, we get

into 3" 2 2u u x− =

3( ) 3u x x x= − −

3 2( 3 ) xpy x x e= − −

3 2( )u x Ax Bx Cx D= + + +(cont)

Remark: When we subst.

into the given nonhom. ODE, we do not compute the derivative of the right-hand side of

until the last step

2 xy ue=

3 2( )u x Ax Bx Cx D= + + +

Example 11 See Appendix 4

Why we have extra term x in the above?

Example 12

A particular soln is of the form

Since Bsin2x and Dcos2x are solns of

[( )sin 2 ( )cos2 ]x Ax B x Cx D x+ + +

'' 4 0y y+ =

'' 4 16 sin 2y y x x+ =

Again, it is not easy to find A,B,C, D.

We shall use the method in Example

10 to find a particular soln.

Furthermore, we need to use the corresponding complex -valued ODE to help us

first notice that

To find a particular soln of

2 cos(2 ) sin(2 )ixe x i x= +So now we consider the corresponding complex-valued ODE

where z is a complex–valued function ,

say z (x)=w(x)+iy(x)

'' 4 16 sin 2y y x x+ =

2'' 4 16 i xz z xe+ =

(cont)

Remark (*) : The imaginary part of is So if z(x) is a complex soln of Then the imaginary part Im(z) of z is a soln of

2'' 4 16 i xz z xe+ =

'' 4 16 sin 2y y x x+ =

16 sin 2x x

(cont)

216 16 cos(2 ) 16 sin(2 )ixxe x x i x x= +

A, B are complex numbers

As in Example 10, we assume a particular soln is

( ) ( )u x x Ax B= +

Why we have extra term x in u

2i xpz ue=

(cont)

Since 2i xBeis a solution of the corresponding hom. ODE '' 4 0z z+ =

Now from 2i xpz ue=

We compute ( ) ',( ) ''p pz zSubst. them into 2'' 4 16 i xz z xe+ =

we get '' 4 ' 16u iu x+ =

(cont)

4 2 1 2iλ = ± − = ± − = ±

2 4 0λ + =

Subst into the

equation , we get

(cont)

( ) ( )u x x Ax B= +

2 , 1A i B= − =

22u ix x= − +

'' 4 ' 16u iu x+ =

From the remark (*), we have

Now look at what we have done

'' 4 16 sin 2y y x x+ =2'' 4 16 i xz z xe+ =

2 2( 2 ) i xpz ix x e= − + 2( 2 )(cos2 sin 2 )ix x x i x= − + +

2Im( ) sin2 2 cos2p py z x x x x= = −

(cont)

Example 13 See Appendix 5

( 2 )cos 2 ( 2 )( sin 2 ) cos 2 sin 2 )[2 sin 2 cos 2 ] [ sin 2 2 cos 2 ]

ix x ix i x x x i x xx x x x i x x x x

= − + − + +

= + + −

We remark that so far we only consider nonhomogeneous ODE with one term on the right hand side, e.g.,

How to solve the case when right hand side has more than one term, see next example

'' 3 ' 4 2siny y y x− − =

'' 3 ' 4 4y y y− − =

Example 14

We known that a particular soln of is A particular soln of is Then a particular soln of is

'' 3 ' 4 2siny y y x− − =

'' 3 ' 4 4y y y− − =

'' 3 ' 4 2sin 4y y y x− − = +

(1/17)(3cos 5sin )py x x= −

ˆ 1py = −

ˆp py y+

Remark: Method of undetermined coeff only works for the following

• Polynomials

• Exponentials

• Sine/Cosine

• Sum or product of the above functions

constant

'' ' ( )y Ay By R x+ + =

Method 2: Method of Variation of Parameters

is 1 2( ) ( ) ( ) ( )py u x y x v x y x= +Chew T S MA1506-14 Chapter 1

Suppose that the general solution of homogeneous ODE '' ' 0y Ay By+ + =

Then a particular solution of the corresponding nonhomogeneous ODE

'' ' ( )y Ay By r x+ + =

1 1 2 2( ) ( )hy c y x c y x= +

How to find u(x) and v(x)? We can find u and v by using the following two equations

1 2' ' 0u y v y+ =

1 2' ' ' ' ( )u y v y r x+ =Solving these two equations, get

1 2 1 2

' y ruy y y y

= −′ ′−

1 2 1 2

' y rvy y y y

=′ ′−

(cont)

1 2 1 2

y rv dxy y y y

=′ ′−∫

1 2 1 2

y ru dxy y y y

= −′ ′−∫

(cont)

Example1 Solve '' tany y x+ =First note that 1 2cos sinhy c x c x= +is the general soln of '' 0y y+ =Hence a particular soln of '' tany y x+ =

is ( )cos ( )sinpy u x x v x x= +

Need to use

1 2 1 2

y ru dxy y y y

= −′ ′−∫ 1

1 2 1 2

y rv dxy y y y

=′ ′−∫

( )cos ( )sinpy u x x v x x= +

1 2cos , siny x y x= = 1 2sin , cosy x y x′ ′= − =

1 2 1 2 1y y y y′ ′− =

From get

get sin tanu x xdx= −∫

cos tanv x xdx= ∫

2sincos

= −∫(cos sec )x x dx= −∫ sin ln sec tanx x x= − +

2cos 1cos

x dxx−= ∫

sin cosxdx x= = −∫

1 2 1 2

y ru dxy y y y

= −′ ′−

1 2 1 2

y rv dxy y y y

=′ ′−∫

'' tany y x+ = 1 2cos , siny x y x= =

A general soln of '' tany y x+ =is

h py y+1 2cos sin cos sinc x c x u x v x= + + +

1 2cos sin cos ln sec tanc x c x x x x= + − +

Example 2

'' sin cosx x xy y e e e− − −− = +

1 2x x

hy c e c e−= +

The general soln of '' 0y y− =is

Hence a particular soln of '' sin cosx x xy y e e e− − −− = +

is ( ) ( )x xpy u x e v x e−= +

1 ( sin cos )2

x x x xu e e e e dx− − − −= +∫1 ( sin cos )2

x x x xv e e e e dx− − −= − +∫

(cont) 2

1 2 1 2

y ru dxy y y y

= −′ ′−∫

1 2 1 2

y rv dxy y y y

=′ ′−

'' sin cosx x xy y e e e− − −− = +

( ) ( )x xpy u x e v x e−= +

1 2 1 2 ( ) 1 1 2x x x xy y y y e e e e− −′ ′− = − − = − − = −

See Appendix 2

See Appendix 3

1 (2sin cos )2

x x xu e e e− − −= − −

1 cos2

x xv e e−= −

sinx x x xpy ue ve e e− −= + = −

1 2 sinx x x xh py y y c e c e e e− −= + = + −

(cont)

Appendix 1 Optional General soln of nonhomogeneous ODE

General Solution

We can check that

'' ' ( )y Ay By R x+ + =

h py y y= +

0h h hy Ay By′′ ′+ + =where

( )p p py Ay By R x′′ ′+ + =

( ) ( ) ( ) ( )h p h p h py y A y y B y y R x′′ ′+ + + + + =

21 1sin cos2 2

x x x xu e e dx e e dx− − − −= +∫ ∫2 sin cosx x x xe e dx e d e− − − −=∫ ∫

cos cosx x x xe e e de− − − −= − ∫cos sinx x xe e e− − −= −

Appendix 2

cos cosx x x xe e dx e de− − − −= −∫ ∫sin xe−= −

1 ( cos 2sin )2

x x xu e e e− − −= −

Appendix 2 (cont)

1 ( sin cos2

x x xv e dx e e dx− −= − +∫ ∫sin cosx x xe dx e d e− −=∫ ∫

cos cosx x x xe e e de− −= − ∫Hence 1 cos

2x xv e e−= −

Appendix 3

Appendix 4 Example 11 Consider First note that has only one root the general soln of is

3 2'' 4 ' 4 20 xy y y x e− + =

2 4 4 0λ λ− + =2λ =

'' 4 ' 4 0y y y− + =2 2

1 1x xc e c xe+

So a particular soln of is of the form Note that we have extra term above By method used in Example 10 , we can get A=1,B=C=D=0

3 2'' 4 ' 4 20 xy y y x e− + =2 3 2 2( ) xx Ax Bx Cx D e+ + +

(cont)

2xWhy we have extra term

Suppose we use 3 2 2( ) xAx Bx Cx D e+ + +

Then 2 xDe are solns of

Suppose we use

2 xCxeand

3 2 2( ) xx Ax Bx Cx D e+ + +

Then 2 xDxe is a soln of

(cont)

'' 4 ' 4 0y y y− + =

Here we use

to ensure that no term in the above is a soln of

We remark that in Example 10, we use 3 2 2( ) xAx Bx Cx D e+ + +

since no term in the above is a soln of '' 4 ' 2 0y y y− + =

(cont)

'' 4 ' 4 0y y y− + =

2 3 2 2( ) xx Ax Bx Cx D e+ + +

The meaning of “term”

There are four terms in the above particular solution We have to ensure that no term in a particular solution of nonhom. ODE is a solution of the corresponding hom. ODE

(cont)

3 2 2( ) xAx Bx Cx D e+ + +3 2 2 2 2 2x x x xAx e Bx e Cxe De= + + +

Appendix 5 Example 13

As in Example 12, we consider 2'' 2 ' 5 16 x ixz z z xe e−+ + =

A particular soln is of the form where

The real part of z(x) is a particular soln of y-equation .

As in example 12, we can find A and B. Why we have extra term x in u(x)?

'' 2 ' 5 16 cos2xy y y xe x−+ + =

( 1 2 )( ) ( ) i xpz x u x e − +=

( ) ( )u x x Ax B= +

Appendix 6 Example one of the most short-lived animals in the world Radioactive Decay Chain

0( ) ( )K t K tU u T

KT t U e eK K

− −= −−

U(t)=# of eggs T(t)=# of mayflies

𝐾𝑈=ln2/2

𝐾𝑇=ln2/1 Chew T S MA1506-14 Chapter 1 168

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