2-1 2. Differential Equations 2.1 ODEs of Second Order

Kreyszig by YHLee;100303; 2-1 Chapter 2. Second‐Order Linear Differential Equations

2.1 Homogeneous Linear ODEs of Second Order A second‐order linear ODE is of the form of

( ) ( ) ( )y p x y q x y r x′′ ′+ + =

• When ( ) 0r x = , it is called homogeneous.

( ) ( ) 0y p x y q x y′′ ′+ + =

Example 4 sinxy y e x−′′ + = : nonhomogeneous linear ODE

( )21 2 6 0x y xy y′′ ′− − + = : homogeneous linear ODE

( )2 2 0x y y y y y′′ ′ ′+ + = : nonlinear ODE

• A solution of a second‐order ODE has derivatives ( ) ( ) and y h x y h x′ ′ ′′ ′′= = ,

and they satisfy the ODE on the given interval. Homogeneous Linear ODE: Superposition Principle Superposition Principle

If 1 2 and y y are solutions to a homogeneous linear ODE in an open interval I,

their linear combination, 1 1 2 2c y c y+ , is also a solution to the ODE in the interval.

Proof: Insert 1 1 2 2c y c y+ into the eq.

( ) ( ) ( )1 1 2 2 1 1 2 2 1 1 2 2 " 'y py qy c y c y p c y c y q c y c y′′ ′+ + ⇒ + + + + +

( ) ( )1 1 1 1 2 2 2 2" ' " '

0

c y py qy c y py qy⇒ + + + + +

⇒

Example 1 Homogeneous Linear ODE and Superposition Principle 0y y′′ − = solution and s x xe e−⎯⎯⎯→ Try 3 8x xy e e−= − + in the ODE

→ ( ) ( )3 8 3 8 0x x x xe e e e− −− + − − + ⇒

Example 2 A nonhomogeneous linear ODE solutions1 =1+cos , =1+siny y y x y x′′ + = ⎯⎯⎯→ ( )2 1+cos and (1 cos ) (1 sin )x x x+ + + are not solutions to the ODE.

Example 3 A nonlinear ODE solutions 20 = and =1y y xy y x y′′ ′− = ⎯⎯⎯→ 2 2 and 1x x− + are not solutions

Kreyszig by YHLee;100303; 2-2 Initial Value Problem A second‐order homogeneous linear ODE has a general solution of the form of 1 1 2 2y c y c y= +

1 2 and y y are solutions that are not proportional, and 1 2and cc are constants.

1 2 and y y are also called basis of solutions.

An initial value problem for this ODE consists of two initial conditions,

( ) ( ) 1, 'o o oy x K y x K= =

A particular solution is obtained when 1 2 and c c are specified.

• 1 2 and y y are linearly independent if ( ) ( )1 1 2 2 0k y x k y x+ = can be satisfied only by 1 2= 0k k = .

1 2 and y y are linearly dependent if 1 20 and 0k k≠ ≠ .

They are proportional in this case, 21 2

1

ky y

k= − .

• A basis of solutions is a pair of linearly independent solutions. Example 4 Initial value problem 0y y′′ + = , ( ) ( )0 3.0 and 0 0.5y y′= = −

The general solution is

1 2cos siny c x c x= +

By applying the initial conditions ( ) ( ) ( )1 20 cos 0 sin 0 3.0y c c= + =

( ) ( ) ( )1 20 sin 0 cos 0 0.5y c c′ = − + = −

→ 3.0cos 0.5siny x x= − Example 5 Basis 1 2cos and siny x y x= = are solutions of 0y y′′ + = .

1 2/ cot constanty y x= ≠ → 1 2 and y y are linearly independent and a basis for all x.

Therefore, a general solution is given by 1 2cos siny c x c x= +

Example 6 Basis Solve the initial value problem 0y y′′ − = , ( ) ( )0 6 and 0 2y y′= = −

and x xe e− are two solutions that are not proportional, / constantx xe e− ≠ . Therefore, the general solution is 1 2

x xy c e c e−= +

Apply the initial conditions ( ) 0 0

1 20 6y c e c e= + = , ( ) 0 01 20 2y c e c e′ = − = −

The particular solution is 2 4x xy e e−= +

Kreyszig by YHLee;100303; 2-3 Find a Basis if One Solution Is Known. Reduction of Order Method of reduction of order One solution is found by inspection. The other solution can be obtained by solving a first‐order ODE. Reduction of order for a homogeneous linear ODE. ( ) ( ) 0y p x y q x y′′ ′+ + =

One solution 1y is known.

Try 2 1y y uy= =

→ 2 1 1' ' ' 'y y u y uy= = + , 2 1 1 1" " " 2 ' ' "y y u y u y uy= = + +

Insert these into ODE 1 1 1 1 1 1

0

" '(2 ' ) ( " ' ) 0u y u y py u y py qy=

+ + + + + =

Change of variable 'U u≡

→ 1

1

2 '' 0

yU p U

y

⎛ ⎞+ + =⎜ ⎟⎝ ⎠

,

Separation of variables and integration

1

1

2 'ydUp dx

U y

⎛ ⎞= − +⎜ ⎟

⎝ ⎠

→ 1ln 2lnU y pdx= − − ∫

→ 21

1 pdxU e

y

−∫=

The second solution is 2 1 1y uy y Udx= = ∫

Example 7 Reduction of order

Solve ( )2 " ' 0x x y xy y− − + =

The first solution 1y x= is obtained by inspection.

Try 1 , ' ' , " " 2 'y uy ux y u x u y u x u= = = + = +

ODE becomes → Change of variable, 'v u= → → →

Kreyszig by YHLee;100303; 2-4 2.2 Homogeneous Linear ODEs with Constant Coefficients General form with constants a and b, 0y ay by′′ ′+ + = We try xy eλ= as a solution

→ xy eλ′ = λ , 2 xy eλ′′ = λ

→ ( )2 0xa b eλλ + λ + =

xy eλ= is a solution if λ satisfies

2 0a bλ + λ + = : characteristic equation

→ ( )21

14

2a a bλ = − + − , ( )2

2

14

2a a bλ = − − −

Case I. two real roots if the discriminant 2 4 0a b− > Case II. a real double root if 2 4 0a b− = Case III. complex conjugate roots if 2 4 0a b− < Case I. Two Distinct Real Roots λ λ1 2 and 1

1xy eλ= and 2

2xy eλ= constitute a basis .


1 21 2

x xy c e c eλ λ= +

Example 2 Initial value problem with distinct real roots Solve 2 0y y y′′ ′+ − = , ( ) ( )0 4, and 0 5,y y′= = −

The characteristic equation is 2 2 0λ + λ − = . → 1 21, 2λ = λ = −

→ 21 2

x xy c e c e−= +

From initial conditions ( ) 1 20 4y c c= + = and ( ) 1 20 2 5y c c′ = − = −

→ 1 21 and 3c c= =

Therefore 23x xy e e−= +

Kreyszig by YHLee;100303; 2-5 Case II. Real Double Root λ=‐a/2 When 2 4 0a b− = , roots of the characteristic equation are 1 2 / 2aλ = λ = −

( )/21 a xy e−→ =

We need a second independent solution for a basis Try 2 1y uy= (method of reduction of order)

( ) ( )1 1 1 1 1 12 0u y u y uy a u y uy buy′′ ′ ′ ′′ ′ ′+ + + + + =

( ) ( )1 1 1 1 1 1

0 0

2 0u y u y ay u y ay by= =

′′ ′ ′ ′′ ′→ + + + + + =

1 2 0, u u c x c′′→ = = +

We choose 1 21, 0c c= = to have

( )/22

a xy xe−=

1 2 and y y are not proportional.

The general equation is

( ) /21 2

axy c c x e−= +

Example 4 Initial value problem with a double root Solve 0.25 0y y y′′ ′+ + = , ( ) ( )0 3, 0 3.5y y′= = −

The characteristic equation: 2 0.25 0λ + λ + = 1 2 0.5→ λ = λ = −

The general solution : ( ) ( ) 0.51 2

xy x c c x e−= +

From the initial cond.: ( ) 10 3y c= =

( )0.5 0.52 1 2' 0.5x xy c e c c x e− −= − + → ( ) 2 10 0.5 3.5y c c′ = − = −

The particular solution : ( ) ( ) 0.53 2 xy x x e−= −

Complex Roots Complex Exponential Function A complex exponential is defined for a given complex number z r it= + as z r ite e e≡ Apply the Maclaurin series of xe for a real x to the complex exponential,

( ) ( ) ( ) ( )2 3 4 5

2 4 3 5

1 ....2! 3! 4! 5!

1 .... ...2! 4! 3! 5!

cos sin

it it it it ite it

t t t ti t

t i t

= + + + + + +

⎛ ⎞= − + − + + − + − +⎜ ⎟

⎝ ⎠= +

→ cos sinite t i t= + : Euler formula Note

( ) ( )1 1cos , sin

2 2it it it itt e e t e e

i− −= + = −

Kreyszig by YHLee;100303; 2-6 • Two complex roots When the radicand is negative, 2 4 0a b− < ,

2 2 214 ( / 4) / 4

2a b b a i b a i− = − − = − = ω

Then 1 2, 2 2a a

i iλ = − + ω λ = − − ω

Two independent solutions are

( )( )

/2 /2

/2 /2

= cos sin

= cos sin

ax i x ax

ax i x ax

e e e x i x

e e e x i x

− ω −

− − ω −

ω + ω

ω − ω : A basis

The linear combination of these solutions

/2 /2 /2

/2 /2 /2

1( + )= cos

21( ‐ )= sin

2

ax i x ax i x ax

ax i x ax i x ax

e e e e e x

e e e e e xi

− ω − − ω −

− ω − − ω −

ω

ω : Another basis

Since we are dealing with a real function, we take the latter basis

( )/2 cos sinaxy e A x B x−= ω + ω

Example 5 Initial value problem with complex roots Solve 0.4 9.04 0y y y′′ ′+ + = , ( ) ( )0 0, 0 3y y′= =

The characteristic equation 2 0.4 9.04 0λ + λ + = 0.2 3i→ λ = − ± The general solution 0.2 ( cos3 sin3 )xy e A x B x−= + The initial condition ( )0 0y A= =

0.2 0.2' ( 0.2 sin3 3 cos3 )x xy B e x e x− −= − +

( )' 0 3 3y B= =

The particular solution 0.2 sin3xy e x−=

Kreyszig by YHLee;100303; 2-7 2.4 Modeling : Free Oscillations Setting up the Model Assumptions: 1. Vertical motion only. 2. Downward is + direction. 3. Spring mass is negligible. • System in static equilibrium 1. Gravitational force : 1F mg= m , mass of the body

(weight) g , acceleration of gravity (980cm/sec2) 2. Restoring force: 2 oF ks= − k , spring constant

(Hooke's law) os , vertical displacement

The restoring force balances the weight 1 2 0oF F mg ks+ ⇒ − =

• System in motion Force on the body: 1 ( )"oF m s y= + force = mass X acceleration

(Newton’s second law) Restoring force of spring: 2 ( )oF k s y= − + Hooke’s law

Undamped System The net force on the body ( )( )o om y s mg k y s′′+ = − +

0

omy mg ks ky=

′′→ = − −

The equation of motion, → 0my ky′′ + = The general solution ( ) cos sino oy t A t B t= ω + ω : Harmonic oscillation

/o k mω =

( ) ( ) cos oy t C t→ = ω −δ , : 2 2 , tan /C A B B A= + δ =

/ 2oω π , natural frequency.

Kreyszig by YHLee;100303; 2-8 Damped System The damping force is proportional to the velocity 3F cv= − : c, damping constant

The net force on the body ( ) ( )( ) 'o o om y s mg k y s c y s′′+ = − + − +

The equation of motion 0my cy ky′′ ′+ + = The characteristic equation

2 0c km m

λ + λ + =

→ λ = −α ±β where 2 4

, 2 2c c mkm m

−α = β =

Case I. Overdamping ( 2 4c mk> ) 1 2, λ = −α +β λ = −α −β The general solution

( ) ( ) ( )1 2

t ty t c e c e− α−β − α+β= +

Note 0α >β >

→ 1 2 and 0λ λ <

→ 0 as y t→ →∞

Kreyszig by YHLee;100303; 2-9 Case II. Critical Damping ( 2 4c mk= )

1 2λ = λ = −α

The general solution ( ) ( )1 2

ty t c c t e−α= +

0te−α > , ( )1 2 0c c t+ = once only for 1 2 0c c < .

→ The mass can cross the equilibrium point only once. The difference from case I is that this is on the border and ready to oscillate with slight change of conditions.

Case III. Underdamping ( 2 4c mk< ) β is pure imaginary.

1 2, i iλ = −α + ω λ = −α − ω : 2cm

α = , 2

24k cm m

ω= −

The general solution

( ) ( )cos sin

cos( )

t

t

y t e A t B t

C e t

−α

−α

= ω + ω

⇒ ω −δ : Damped oscillation

2 2 2C A B= + , 1tan /B A−δ =

As 0c→ , the frequency ω approaches to the natural frequency /o k mω = .

Example 2 Three cases of Damped Motion (1) 210 , 90 / , 100 / sm Kg k Kg s c Kg= = =

The equation of motion: ( ) ( )10 " 100 ' 90 0 0 0.16, ' 0 0y y y y y+ + = = =

The general solution: 9 91 2 1 2, ' 9t t t ty c e c e y c e c e− − − −= + = − −

The initial condition: 1 2 1 20.16, 9 0c c c c+ = − − =

The particular solution: 90.02 0.18t ty e e− −= − + (2) 60 / sc Kg=

The characteristic equation: 210 60 90 0λ + λ + = The general solution: ( ) 3

1 2ty c c t e−= +

The particular solution: ( ) 30.16 0.48 ty t e−= +

(3) 10 / sc Kg=

The characteristic equation: 210 10 90 0λ + λ + = The general solution: ( )0.5 cos2.96 sin2.96ty e A t B t−= +

The particular solution: ( )0.50.162 cos 2.96 0.17ty e t−= −

Kreyszig by YHLee;100303; 2-10 2.5 Euler‐Cauchy Equations Euler‐Cauchy equation is of the form,

2 0x y axy by′′ ′+ + = Try my x=

→ ( )2 2 11 0m m mx m m x axmx bx− −− + + =

( )2 1 0m a m b+ − + = : auxiliary equation

Two roots

( ) ( )21

1 11 1

2 4m a a b= − + − − , ( ) ( )21

1 11 1

2 4m a a b= − − − − (4)

my x= is a solution if and only if m is given by (4).

Case I. Distinct real roots 1 2 and m m are real and distinct.


1 21 2

m my c x c x= +

Case II. Double root

The radicand ( )211 0

4a b− − = and ( )1 / 2m a= −

Euler‐Cauchy equation, in this case, and its first solution

( )2 211 0

4x y axy a y′′ ′+ + − = , ( )1 /2

1ay x −=

Use the reduction of order, 2 1y uy=

( ) ( )2 11

1 1 1exp exp

a

aU pdx dx

x xy x −

⎛ ⎞= − ⇒ − ⇒⎜ ⎟⎝ ⎠∫ ∫

lnu Udx x= ⇒∫


( ) ( )1 /21 2 ln

ay c c x x −= +

Case III. Complex roots Two roots are given by complex conjugate 1 2, m i m i= μ + ν = μ − ν Two solutions could be

( ) ( )ln ln, i ii x i xx x e x x eν − νμ+ ν μ μ− ν μ= =


( ) ( )cos ln sin lny x A x B xμ= ν + ν⎡ ⎤⎣ ⎦

Kreyszig by YHLee;100303; 2-11 Example 1 Different real roots Solve the Euler‐Cauchy equation, 2 1.5 0.5 0x y xy y′′ ′+ − = The auxiliary equation, 2 0.5 0.5 0m m+ − = → 0.5, 1m = − The general solution

21

cy c x

x= +

Example 2 A double root Solve 2 5 9 0x y xy y′′ ′− + = The auxiliary equation 2 6 9 0m m− + = → A double root m=3 The general solution ( ) 3

1 2 lny c c x x= +

Example 3 Complex roots Solve 2 0.6 16.04 0x y xy y′′ ′+ + = The auxiliary equation 2 0.4 16.04 0m m− + = Two complex roots 1,2 0.2 4m i= ±

Two solutions are

( ) ( ) ( )

( ) ( ) ( )

1

2

40.2 4 0.2 ln 0.2 (4ln ) 0.2

40.2 4 0.2 ln 0.2 (4ln ) 0.2

cos 4ln sin 4ln

cos 4ln sin 4ln

im i x i x

im i x i x

x x x e x e x x i x

x x x e x e x x i x

+

−− −

= ⇒ ⇒ ⇒ +⎡ ⎤⎣ ⎦

= ⇒ ⇒ ⇒ −⎡ ⎤⎣ ⎦

By linear combinations of these, the general solution is

( ) ( )0.2 cos 4ln sin 4lny x A x B x= +⎡ ⎤⎣ ⎦

Example 4 Electric Potential between Two Concentric Spheres v(r) is given by a solution of " 2 ' 0rv v+ = ( ) ( )5 110 , 10 0v r cm V v r cm V= = = =

The auxiliary equation, 2 0m m+ = Roots are 0, 1m m= = −

The general solution, ( ) 1 2 /v r c c r= +

The initial condition, 2 21 1110, 0

5 10

c cc c+ = + =

The particular solution, ( ) 110 1100 /v r r= − +

Kreyszig by YHLee;100303; 2-12

2.6 Existence and Uniqueness of Solutions. Wronskian A second‐order homogeneous linear ODE

( ) ( ) 0y p x y q x y′′ ′+ + =

An initial condition

( ) ( ) 1, o o oy x K y x K′= =

A general solution 1 1 2 2y c y c y= +

Theorem 1. Existence and Uniqueness Theorem for Initial Value Problem

If ( ) ( ) and p x q x are continuous functions on some open interval I, and ox is in I,

then the initial value problem has a unique solution ( )y x on the interval.

Proof: It will not be presented here. Linear Independence of Solutions Two solutions 1 2, y y are linearly independent on I if

( ) ( )1 1 2 2 0k y x k y x+ =

is satisfied only by 1 2 0k k= = on I.

1 2, y y are linearly dependent if it is also satisfied by other than 1 2 0k k= = .

In this case 1 2, y y are proportional,

1 2 2 1 or y ky y ly= =

• Wronskian of two solutions 1 2 and y y

( ) 1 21 2 1 2 2 1

1 2

,y y

W y y y y y yy y

′ ′= = −′ ′

Theorem 2 Linear Dependence and Independent solutions

The coefficients ( ) ( ) and p x q x are continuous on some open interval I.

Two solutions 1 2 and y y are linearly dependent if and only if W=0 at some ox on I.

If W=0 at an ox x= on I, W=0 on entire I.

Two solutions are linearly independent on I if 0W ≠ at some 1x on I.

Proof:

(a) If 1y and 2y are linearly dependent on the interval I,

then 1 2y ky= and

( ) ( ) 2 21 2 2 2 2 2 2 2

2 2

, , 0ky y

W y y W ky y ky y y kyky y

′ ′⇒ ⇒ ⇒ − ⇒′ ′

Kreyszig by YHLee;100303; 2-13 (b) If ( )1 2, 0W y y = for some ox x= on I,

Consider the equation with unknowns 1 2and k k

( ) ( )1 1 2 2 0o ok y x k y x+ =

Its derivative is ( ) ( )1 1 2 2 0o ok y x k y x′ ′+ = . (7)

Eliminate 2k and combine

( ) ( ) ( ) ( )1 1 2 1 1 2' ' 0o o o ok y x y x k y x y x− =

→ ( )1 1 2 ( ), ( ) 0o ok W y x y x = : 1Either 0 or 0k W= =

Similarly ( )2 1 2 ( ), ( ) 0o ok W y x y x = : 2Either 0 or 0k W= =

( )1 2 ( ), ( ) 0o oW y x y x = means that 1 2and k k can be other than 1 2 0k k= = .

Using these 1 2 and k k , define a function y(x)

( ) ( ) ( )1 1 2 2y x k y x k y x= + : Note x is used not xo

Its initial condition is, from (7), ( ) ( )0, ' 0o oy x y x= =

A solution with the initial condition can be y=0. This is a unique solution since ( ) ( ) and p x q x are continuous.

→ ( ) ( ) ( )1 1 2 2 0y x k y x k y x= + =

since 1 2 and k k can be other than 1 2 0k k= = , ( ) ( )1 2 and y x y x are linearly dependent in whole I.

(c) If ( ) 0oW x = at ox in I, ( ) ( )1 2 and y x y x are linearly dependent in whole I. [proof (b)]

If ( ) ( )1 2 and y x y x are linearly dependent in I, ( )1 2, 0W y y = in whole I. [proof (a)]

Example 2 A double root ( )1 2

xy c c x e= + is the general solution of 2 0y y y′′ ′− + = .

The Wronskian

( ) 2, 0x x xW e xe e= ≠

Two solutions and x xe xe are linearly independent.

Kreyszig by YHLee;100303; 2-14 A General Solution Theorem 3 Existence of a general solution

If ( ) ( ) and p x q x are continuous on some open interval I,

then the ODE has a general solution on I. Proof: This can be skipped By theorem 1, the ODE has a unique solution ( )y x , a particular solution.

Let’s assume ( ) ( ) ( )1 2y x y x y x= + with ( ) ( )( ) ( )

1 2

1 2 1

, 0

' 0, 'o o o

o o

y x K y x

y x y x K

= =⎡⎢

= =⎢⎣

Their Wronskian

( ) ( )( )1 2 1, 0o o oW y x y x K K= ≠

By theorem 2, ( ) ( )1 2 and yy x x are linearly independent.

Their linear combination is also a solution, 1 1 2 2y c y c y= +

Theorem 4 A general solution includes all solutions

If ( ) ( ) and p x q x are continuous on some open interval I,

then every solution of the ODE, ( )Y x , is of the form

( ) ( ) ( )1 1 2 2Y x C y x C y x= +

Hence there is no solution not obtainable from a general solution, no singular solution.

Proof: This can be skipped By Theorem 3 the ODE has a general solution ( ) ( ) ( )1 1 2 2Y x c y x c y x= +

One can always find 1 2and c c that produce ( )Y x .

At fixed ox in I

( ) ( ) ( )1 1 2 2o o oc y x c y x Y x+ =

( ) ( ) ( )1 1 2 2o o oc y x c y x Y x′ ′ ′+ =

Solve these. Since the Wronskian 1 2 1 2' ' 0W y y y y= − ≠ , non‐trivial solution of 1 2and c c exists.

( ) ( )

2 2 1 11 2

1 2 1 2 1 2 1 2

' ' ' ',

' ' ' '

Yy y Y Y y y YC C

y y y y y y y y

− −= =

− −

A particular solution satisfying the initial conditions ( ) ( ), 'o oY x Y x

( ) ( ) ( )1 1 2 2y x C y x C y x= +

The uniqueness theorem 1 implies that ( ) ( )y x Y x=

Kreyszig by YHLee;100303; 2-15 2.7 Nonhomogeneous ODEs A nonhomogeneous linear ODE

( ) ( ) ( )y p x y q x y r x′′ ′+ + = (1)

The corresponding homogeneous ODE ( ) ( ) 0y p x y q x y′′ ′+ + = (2)

Definition General Solution, Particular Solution

A general solution of the nonhomogeneous ODE is of the form ( ) ( ) ( )h py x y x y x= +

1 1 2 2hy c y c y= + is a general solution of the homogeneous ODE.

py is a specific solution with no arbitrary constants.

A particular solution of nonhomogeneous ODE is obtained with specific values of 1 2 and c c .

Theorem 1 Relation between solutions of (1) and (2)

(a) The sum of solutions of (1) and (2) on I is a solution of (1) on I. (b) The difference of two solutions of (1) on I is a solution of (2) on I.

Proof: The left side of (1) [ ]L y≡ . Let and y y be two solutions of (1), y be a solution of (2).

(a) Insert ( )y y+ into (1)

[ ] [ ] [ ] ( ) ( )+ 0L y y L y L y r x r x= + = + =

(b) Insert ( )y y− into (1)

[ ] [ ] [ ] ( ) ( )‐ 0L y y L y L y r x r x= − = − =

Theorem 2 A General Solution of a Nonhomogeneous ODE includes All Solutions

If ( ) ( ) ( ), and p x q x r x are continuous on some open interval I,

every solution on I is obtained from the general solution by assigning values to 1 2and c c .

Proof: skipped here

Kreyszig by YHLee;100303; 2-16 Method of Undetermined Coefficients The specific solution py can be obtained by the method of undetermined coefficients

when ODE is of the form

( )y ay by r x′′ ′+ + = : a and b are constants

Choice Rules for the Method of Undetermined Coefficients

(A) Basic Rule ( )Term in r x Choice for py

xkeγ xCeγ nkx (n=0,1,2..) 1 1

1 1 0..n nn nK x K x K x K−

−+ + + +

cos

sin

k x

k x

ω ⎫⎬ω ⎭ cos sinK x M xω + ω

cos

sin

ax

ax

ke x

ke x

⎫ω ⎪⎬

ω ⎪⎭ ( )cos sinaxe K x M xω + ω

(B) Modification Rule If the choice for py happens to be a solution of the homogeneous equation, try pxy .

If the choice happens to be a double root of the characteristic equation, try 2px y .

(C) Sum Rule If ( )r x is a sum of the functions in the first column,

try the sum of the corresponding choice functions. Example 1 Application of the basic rule Solve 20.001y y x′′ + = ( ) ( )0 0, ' 0 1.5y y= =

The general solution of 0y y′′ + = cos sinhy A x B x= +

Try 2

py Kx= , " 2py K=

→ 2 22 0.001K Kx x+ = → 22 ( 0.001) 0K x K+ − = → 0 and 0.001K K= = (contradiction !!) Try 2

2 1p oy K x K x K= + +

→ 2 22 2 12 0.001oK K x K x K x+ + + =

→ 22 1 2( 0.001) ( 2 ) 0ox K K x K K− + + + =

→ 2 10.001, 0, 0.002oK K K= = = −

→ 20.001 0.002py x= −

The general solution is 2cos sin 0.001 0.002h py y y A x B x x= + = + + −

A=0.002, B=1.5 are obtained from the initial conditions.

Kreyszig by YHLee;100303; 2-17 Example 2 Application of the modification rule Solve 1.53 2.25 10 xy y y e−′′ ′+ + = − , ( ) ( )0 1, ' 0 0y y= =

The characteristic equation : 2 3 2.25 0λ + λ + = The general solution : 1.5

1 2( ) xhy c c x e−= +

The choice of py would be 1.5xCe− , but it is a double root of the characteristic equation.

Try 2 1.5xpy Cx e−= in the ODE,

( ) ( )2 2 22 6 2.25 3 2 1.5 2.25 10C x x C x x Cx− + + − + = − → C=‐5

The general solution of nonhomo. ODE : 1.5 2 1.5

1 2( ) 5x xh py y y c c x e x e− −= + = + −

The initial condition ( ) 10 1y c= = , ( ) 1.5 1.5 2 1.5

2 1 2' 1.5 1.5 10 7.5x x xy c c c x e xe x e− − −= − − − + → 2 1.5c =

The particular solution : ( ) ( )1.5 2 1.5 2 1.51 1.5 5 1 1.5 5x x xy x e x e x x e− − −= + − = + −

Example 3 Application of the sum rule Solve ( ) ( )0.5" 2 ' 5 40cos10 190sin10 , 0 0.16, ' 0 40.08xy y y e x x y y+ + = + − = =

The characteristic equation : ( )( )2 2 5 1 2 1 2 0i iλ + λ + = λ + + λ + − =

The general solution : ( )cos2 sin2xhy e A x B x−= +

The specific solution : 0.5 cos10 sin10xpy Ce K x M x= + +

Insert hy into the nonhomogeneous ODE

( ) ( ) ( ) ( )0.5

0.5

0.25 2 0.5 5 100 cos100 100 sin10 2 10 sin10 10 cos10 5 cos10 sin10

40cos10 190sin10

x

x

Ce K x M x K x M x K x M x

e x x

+ + + − − + − + + +

= + −

i

From the exponential term: C=1/6.25 From the cosine term: ‐100K+2∙10M+5K=40 From the sine term: ‐100M‐2∙10K+5M=‐190 → K=0, M=2 The general solution of nonhomogeneous ODE ( ) 0.5cos2 sin2 0.16 2sin10x x

h py y y e A x B x e x−= + = + + +

The initial condition y(0) = A+0.16 ( ) 0.5' cos2 sin2 2 sin2 2 cos2 0.08 20cos10x xy e A x B x A x B x e x−= − − − + + +

( )' 0 2 0.08 20 40.08y A B= − + + + = → B=10

The particular solution 0.510 sin2 0.16 2sin10x xy e x e x−= + + Stability If all roots of the characteristic equation of ODE , ( )y ay by r x′′ ′+ + = , are negative or have a negative real part,

then, 0 as 0.hy x→ →

In this case the transient solution h py y y= + approaches the steady‐state solution py .

The system is called stable.

Kreyszig by YHLee;100303; 2-18 2.8 Modeling : Forced Oscillations In the presence of the external forces, the equation of motion is given by ( )my cy ky r t′′ ′+ + = : ( )r t , forcing function, input, or driving force

( )y t , output, or response of the system

For a periodic external force, ODE becomes cosomy cy ky F t′′ ′+ + = ω my′′ : force of inertia cy′ : damping force ky : restoring force cosoF tω : driving force

At each t, the force of inertia, damping force and restoring force are in equilibrium with ( )r t .

Solving the Nonhomogeneous ODE Using the method of undetermined coefficient ( ) cos sinpy t a t b t= ω + ω (3)

By inserting py and its derivatives

( ) ( )2 2cos sin cosok m a cb t ca k m b t F t⎡ ⎤ ⎡ ⎤− ω +ω ω + −ω + − ω ω = ω⎣ ⎦ ⎣ ⎦

By collecting coefficients of cosine and sine

( ) ( )− ω +ω = −ω + − ω =2 2, 0ok m a cb F ca k m b

→ ( )

( ) ( )

2 2

2 22 2 2 2 2 2 2 2 2 2, =

o

o o

o o

m ca F b F

m c m c

ω −ω ω=

ω −ω +ω ω −ω +ω : /o k mω =

( )

( ) ( )⎡ ⎤ ⎡ ⎤ω −ω ω⎢ ⎥ ⎢ ⎥= ω + ω⎢ ⎥ ⎢ ⎥ω −ω +ω ω −ω +ω⎣ ⎦ ⎣ ⎦

2 2

2 22 2 2 2 2 2 2 2 2 2cos sin

o

p o o

o o

m cy F t F t

m c m c

Case 1 Undamped Forced Oscillations. Resonance When the damping is neglected, c=0

→ ( ) ( )2 2coso

p

o

Fy t t

m= ω

ω −ω

Using the previous result of the homogeneous solution (Section 2.4) ( ) ( )cosh oy t C t= ω −δ


( ) ( ) ( )2 2cos coso

o

o

Fy t C t t

m= ω −δ + ω

ω −ω :

ωω

, natural frequency

, frequency of driving forceo

Superposition of two harmonic oscillations.

Kreyszig by YHLee;100303; 2-19 Resonance As oω→ω , the oscillation amplitude goes to infinity.

This is called resonance. At resonance, the nonhomogeneous ODE becomes

2 cosoo o

Fy y t

m′′ + ω = ω

By the modification rule ( ) ( )cos sinp o oy t t a t b t= ω + ω

Insert this into ODE and find 0, / 2o oa b F m= = ω

( ) sin2

op o

o

Fy t t t

m= ω

ω

Beats

When ω ω≈ o with initial conditions ( ) ( )0 0, 0 0y y′= =

( ) ( ) ( )2 20

cos cosoo

Fy t t t

m= ω − ω

ω −ω

→ ( ) ( )2 2

2sin sin

2 2o o o

o

Fy t t t

m

ω +ω ω −ω⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟ω −ω ⎝ ⎠ ⎝ ⎠

Case 2 Damped Forced Oscillations When c > 0 but negligibly small, ( ) ( )−α≈ ω + ωcos sint

h o oy t e A t B t → ( ) = 0hy t

↑ ↑ / 2c mα = →∞As t ( )py t ⇒ Steady state solution.

Harmonic oscillation with frequency equal to that of the input. Amplitude of py

Rewriting the above equation (3) ( ) ( )* cospy t C t= ω −η

where

( )

* 2 2

22 2 2 2 2

o

o

FC a b

m c= + =

ω −ω +ω : Amplitude

( )2 2

tano

b ca m

ωη= =

ω −ω : Phase angle

C * has a maximum at =ω

*

0dCd

→ ( )2

2 2 2 2 2 22

2 or 2o o

cc m

m= ω −ω ω = ω −

( )*max 2 2 2

2

4o

o

mFC

c m cω =

ω −

Kreyszig by YHLee;100303; 2-20 2.9 Modeling: Electric Circuits

Different physical systems can be analyzed by the same mathematical model. Equivalent electrical model is used to explain complicated systems. Setting up the Model of RLC circuit Using Kirchhoff’s voltage law for a current I,

′+ + = ω∫1

sinoRI Idt LI E tC

/ cosoLI RI I C E t′′ ′→ + + = ω ω

Solving the Equation By the method of undetermined coeff. ( ) cos sinpI t a t b t= ω + ω

→ ( ) ( )' sin cospI t a t b t= ω − ω + ω

→ ( ) ( )2" cos sinpI t a t b t= ω − ω − ω

Insert these into ODE ( )2 cos sinL a t b tω − ω − ω + ( )sin cosR a t b tω − ω + ω + ω + ω(1/ )( cos sin )C a t b t cosoE t= ω ω

→ ( ) ( )2 2/ cos / sin 0oaL bR a C E t bL aR b C t− ω + ω+ − ω ω + − ω − ω+ ω =

Each parenthesis should be zero,

2 2

oE Sa

R S

−=

+,

2 2oE R

bR S

=+

, where 1

S LC

= ω −ω

, reactance

Rewrite the solution

( ) ( )sinp oI t I t= ω −θ , where 2 2oI a b= + , tan /a bθ = −

The characteristic equation

2 10

RL LC

λ + λ + = → 1,2λ = −α ±β , where / 2R Lα = , 21 42

LR

L Cβ = −

The homogeneous solution 1 2

1 2t t

hI c e c eλ λ= +

As t→∞ , 0hI → and pI becomes the steady state solution,

which is a harmonic oscillation with the same frequency as the input.

Kreyszig by YHLee;100303; 2-21 Example 1 RLC Circuit Find the current in an RLC circuit with 211 , 0.1 , 10R L H C F−= Ω = = with zero current and charge at t=0.

The ODE

The characteristic equation

The general solution of the homogeneous ODE

The specific solution from the method of undetermined coefficients

The coefficients

The general solution of nonhomogeneous ODE

The initial condition

From 1

sinoLI RI Idt E tC

′ + + = ω∫ → → ( )' 0 0I =

The particular solution

Analogy of Electrical and Mechanical Quantities / cosoLI RI I C E t′′ ′+ + = ω ω and cosomy cy ky F t′′ ′+ + = ω

There is analogy: L m− , R C− , 1 /C k− , ( ) ( )I t y t−

Kreyszig by YHLee;100303; 2-22 2.10 Solution by Variation of Parameters The standard form of nonhomogeneous ODE

( ) ( ) ( )y p x y q x y r x′′ ′+ + =

When p, q, and r are continuous on I

( ) 2 11 2p

y r y ry x y dx y dx

W W= − +∫ ∫

where 1 2 and y y form a basis of solutions of homogeneous ODE, 1 2 1 2W y y y y′ ′= − .

Derivation Continuous p and q gives a general solution of the homogeneous ODE ( ) ( ) ( )1 1 2 2hy x c y x c y x= +

In the method of variation of parameters we try

( ) ( ) ( ) ( ) ( )1 2py x u x y x v x y x= + → 1 1 2 2py u y uy v y vy′ ′ ′ ′ ′= + + +

Two unknowns and u v require two conditions.

we assume 1 2 0u y v y′ ′+ = . (a)

Then, 1 2py uy vy′ ′ ′= +

Inserting , ', "p p py y y into ODE

( ) ( )1 1 1 2 2 2 1 2u y py qy v y py qy u y v y r′′ ′ ′′ ′ ′ ′ ′ ′+ + + + + + + = → 1 2u y v y r′ ′ ′ ′+ = . (b)

Combine (a) and (b), and cancel v’

( )1 2 1 2 2u y y y y y r′ ′ ′− = − 2 u W y r′→ = − → 2y ru dx

W= −∫

Similarly, cancel in (a) and (b)

( )1 2 1 2 1v y y y y y r′ ′ ′− = 1 v W y r′→ = → 1y rv dx

W= ∫

Using the results of and u v , we can check 1 2 0u y v y′ ′+ =

Example 1 Method of Variation of Parameter Solve secy y x′′ + = From the homogeneous ODE, a basis of solution is 1 2cos , siny x y x= =

The Wronskian ( )cos cos sin sin 1W x x x x= − − =

From variation of parameters

cos sin sec sin cos sec cos ln cos sinpy x x xdx x x xdx x x x x= − + ⇒ +∫ ∫

The general solution 1 2cos sin cos ln cos siny c x c x x x x x= + + +

2-1 2. Differential Equations 2.1 ODEs of Second Order

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Transcript of 2-1 2. Differential Equations 2.1 ODEs of Second Order