Kreyszig ; 2-1

Chapter 2. Second-Order Linear Differential

Equations

2.1 Homogeneous Linear ODEs of Second Order 2.2 Homogeneous Linear ODEs with Constant Coefficients 2.4 Modeling : Free Oscillations 2.5 Euler-Cauchy Equations 2.6 Existence and Uniqueness of Solutions. Wronskian 2.7 Nonhomogeneous ODEs 2.8 Modeling : Forced Oscillations 2.9 Modeling: Electric Circuits 2.10 Solution by Variation of Parameters

Kreyszig ; 2-2 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

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" ' " '0c 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 ; 2-3 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 2 and 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 k k1 2 0= = .

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 ; 2-4 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" '(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= − − ∫

→ 1

1 pdxU e

y−∫= ∫=

− pdxe

yU 2

1

1(**1/y1^2)

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

Example 7 Reduction of order Solve The first solution 1y x= is obtained by inspection. Try ODE becomes → Change of variable, 'v u= → → →

Kreyszig ; 2-5 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

21

42

a 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 21 and λλ 1

1xy eλ= and 2

2xy eλ= constitute a basis .

The general solution is 1 2

1 2x 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 ; 2-6 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−→ = , 0''' =++ byayy

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 ( ) /2

1 2axy 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 4 4 0λ − λ + = 1 2 2→ 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,

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

Kreyszig ; 2-7 • 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

( - )= sin2

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 ; 2-8 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 =

(cos(a-b)=cosacosb+sinasinb) / 2oω π , natural frequency.

Kreyszig ; 2-9 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 2t ty t c e c e− α−β − α+β= +

Note 0α >β > → 1 2 and 0λ λ < → 0 as y t→ →∞

Kreyszig ; 2-10 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

Ce t

−α

−α

= ω + ω

⇒ ω −δ : Damped oscillation

2 2 2C A B= + , tan /B Aδ = As 0c → , the frequency ω approaches to the natural frequency /o k mω = .

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= − + − − , ( ) ( )2

11 1

1 12 4

m 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. The general solution is 1 2

1 2m my c x c x= +

Kreyszig ; 2-11 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= ( ( ) ( ) 0y p x y q x y′′ ′+ + = )

( ) ( )2 11

1 1 1exp exp

a

aU pdx dx

x xy x −

= − ⇒ − ⇒ ∫ ∫

lnu Udx x= ⇒∫

The general solution is

( ) ( )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= =

The general solution

( ) ( )cos ln sin lny x A x B xµ= ν + ν

Example 1 Different real roots ( 2 0x y axy by′′ ′+ + = , ( )2 1 0m a m b+ − + = ) 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

Kreyszig ; 2-12 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= +