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Homogeneous Linear DE of 2

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(1)

Homogeneous Linear DE of 2

nd

Order

– Homogeneous DE

– Example of a non-homogeneous linear DE

– Example of a homogeneous linear DE

– Example of a non-linear DE

1 x

2

y  2 xy 6 y 0

     

    

y p x y q x r x

4

sin

  

x

y y e x



2

2 0

x y y y y y

    0

    

y p x y q x

(2)

Homogeneous Linear DE of 2

nd

Order

– Example of a non-linear DE

• Superposition or Linearity Principle

– DE

– Solution

– Linear combination of the known solutions y1, y2

,

,

x

x

y e y e

1 1 2 2

 

y c y c y

  0 y y

2

1

   

y y

3 2

5

 

x

x

y e e

(3)

Superposition Principle

--- Any linear combination of two solutions is again a solution on I.

Sums and constant multiples of solutions are again solutions.

Does not hold for non-homogeneous and nonlinear DE – Example of non-homogeneous DE

– Solution

– But, not a solution

1 cos , 1 sin

   

y x y x

   1 y y

     

2 1 cos  x , 1 cos  x   1 sin x

(4)

Superposition Principle

– Example of non-linear DE

– Solution

– But, not a solution

2

, 1

 

y x y

2 2

, 1

x x

    0

y y xy

(5)

Initial Value Problem (1)

– Solution of 2nd order homogeneous linear DE

– Two IC’s

– Example of IVP

– Two known solutions ex, e-x – Particular solution

   

0, 0 4, 0 2

      

y y y y

1 1 2 2

 

y c y c y

 

0

0

,   

0

1

y x K y x K

1 2

x

x

y c e c e

3

x

x

y e e

(6)

Initial Value Problem (2)

– Wrong selection of known solutions ex, e-x

Cannot solve the DE

1 2

x

x

y c e c le

(7)

Basis (Fundamental System)

– General sol.of homogeneous linear DE

– Not proportional to the known solutions

– Particular solution can be obtained by assigning c1, c2

– Linearly independent y1, y2 on I

   

1 1

2 2

 0

k y x k y x

1 1 2 2

 

y c y c y

1

,

2

y y

Basis

1 1 2 2

 

y c y c y

1

 0,

2

 0

k k

(8)

Basis (Fundamental System)

– Linearly dependent

2

1 2

1

  k

y y

k

(9)

Basis (Fundamental System)

– General sol.of homogeneous linear DE

--- a pair y1, y2 on I of linearly independent sol.

– Example – General sol.

– Example

– General sol.

  0 y y

1 2

x

x

y c e c e

  0 y y

1

cos

2

sin

 

y c x c x

(10)

Method of Reduction of Order (1)

– How to obtain basis if one solution is known – set

– Differentiating

– Original DE

– Rearranging in u

2 1 1

2 1

2

1 1

    

       

y u y uy

y u y u y uy

2

1

y uy

 

1

2

1 1 1 1 1

0

            u y u y uy p u y uy quy

   

1

2

1 1 1 1 1

0

           u y u y py u y py qy

0

(11)

Method of Reduction of Order (2)

– Divide by y1 and set

– Then

– Separation

1 1

2 0

  

     

 

U y p U

y

1 1

 2  

    

 

dU y

U y p dx

ln U   2 ln y

1

  pdx

  ,    u U u U

1 1

1

2   0

   y pyu u

y

(12)

Method of Reduction of Order (3)

– Taking exponentials

– Desired sol. y2

--- where

2

1

1

y uy y Udx

2 1

1

pdx

U e

y

2 1

   

y u Udx

y const.

(13)

– 2nd order DE – Standard form

– One known sol.

– Formulation of the other sol.

ln 2

1 1

x

U e

x x

2

     0 x y xy y

2

1 1

     0

y y y

x x

1

y x

2

    ln

y ux x Udx x x

Example of Reduction of Order

(14)

– 2nd order DE

– 1st order DE and its solution

– Substitution

k

y e x

     0 y ay by

,

2

    y e x

y e x y e x

 

   0 y ky

2

a b e

x

0

Characteristic equation

2

nd

Order DE with Const. Coeff. (1)

(15)

– Roots of the characteristic eqn.

– Bases of the original DE

– Three cases

(Case I) two real roots if

(Case II) a real double root if

(Case I) complex conjugate roots if

• Two Distinct Real Roots

– Bases of the original DE

1 2

1

 ,

2

y e x y

e x

2

nd

Order DE with Const. Coeff. (2)

2

 

2

1 2

1 1

4 , 4

2 2

   a ab    a ab

 

2

 4  0

a b

2

 4  0

a b

2

 4  0

a b

1 2

1

 ,

2

y e x y

e x

(16)

– General sol.

– Example

– Characteristic Eqn., its roots

– General sol.

  0 y y

Case I. Two Distinct Real Roots (1)

1 2

1 2

x

x

y c e

c e

2

1 2

1 0,

1, 1

 

  

 

1 2

x

x

y c e c e

(17)

– Initial Value Problem

– Characteristic Eqn., its roots

– General sol.

– Particular sol.

2

   2 0,

 

2

1 2

x

x

y c e c e

Case I. Two Distinct Real Roots (2)

   

2 0, 0 4, 0 5

        

y y y y y

   

1 1

1 1

1 9 1, 1 9 2

2 2

        

 

3

2

x

x

y e e

(18)

– Only one sol.

– To obtain a second independent solution y2, use reduction of order.

– Set

– Substituting and rearranging

Since

Case II. Real Double Root (1)

 2

1

a x

y e

2

1

y uy

2 1 1

2 1

2

1 1

    

       

y u y uy

y u y u y uy

   

1

2

1 1 1 1 1

0

           u y u y py u y py qy

0 0

2

1 1

2 y    ae

ax

  ay

(19)

– Thus

– Double integrating

– Select an independent sol.

– Corresponding general solution

Case II. Real Double Root (2)

1

0, 0

   

u y u

1 2

 

u c x c

2

1

, 

y uy u x

1 2

2

 

ax

y c xc e

(20)

– DE

– Characteristic Eq.

– Double root

– Corresponding general solution

Example of Real Double Root (1)

8 16 0

    

y y y

2

 8   16 0

 

  4

1 2

4

 

x

y c xc e

(21)

– Initial Value Problem – Characteristic Eq.

– Double root

– Corresponding general solution

– Particular solution

Example of Real Double Root (2)

   

4 4 0, 0 3, 0 1

       

y y y y y

2

 4   4 0

 

 2

1 2

2

 

x

y c xc e

3 5

2

 

x

y x e

(22)

Case III. Complex Conjugate Roots (1)

– 2nd order DE

– Characteristic Eq.

– Roots of the characteristic eqn.

– (Case III) complex conjugate roots if

     0 y ay by

2

a   b 0

 

2

 

2

1 2

1 1

4 , 4

2 2

   a ab    a ab

 

2

 4  0

a b

(23)

Fundamental of Complex Roots (1)

– 2nd order DE

– Characteristic Eq.

– Roots of the characteristic eqn.

– Two complex bases

0 y   y

2

1 0

  

1 i

     

ix

,

ix

e e

(24)

Fundamental of Complex Roots (2)

– Complex exponential function

– Adding

– Subtracting

– The other bases:

cos sin e

ix

x ix

cos ,sin x x cos sin e

ix

x ix

 

cos 1

2

ix ix

xee

Euler formula

 

sin 1

2

ix ix

x e e

i

 

(25)

Complex Exponential Function

– Real case:

– Maclaurin series

cos t i sin t

 

       

2 3 4 5

1 2! 3! 4! 5!

it

it it it it

e    it    

cos sin

z s it s it s

ee

e ee t it zs

2 4 3 5

1 2! 4! 3! 5!

t t t t

i t  

           

 

(26)

Case III. Complex Conjugate Roots (2)

– One root of the characteristic eqn.

– Roots of the characteristic eqn.

where

1 2

1 1

2 a i , 2 a i

         

2 2

1

1 1 1 1

2 a 2 a 4 b 2 a i b 4 a

        

1

2

b 4 a

  

(27)

Case III. Complex Conjugate Roots (3)

– Using complex exponential function

– Adding and divide by 2 y1

– Subtracting and divide by 2i y2

– General solution

2

1 ax

cos

ye

x

   

 

   

 

1

2

2 2

2 2

cos sin

cos sin

a x i x a x

x

a x i x a x

x

e e e x i x

e e e x i x

 

 

  

  

2

2 ax

sin

ye

x

 

2

cos sin

ye

ax

Ax B   x

New bases

(28)

Example of Complex Conjugate Roots (1)

– Initial value problem

– Characteristic eqn.

– Roots of the characteristic eqn.

– General solution

0.1 2i

   

   

0.2 4.01 0, 0 0, 0 2

y   y   yyy  

 

0.1x

cos 2 sin 2 ye

A xB x

2

0.2 4.01 0

    

(29)

Example of Complex Conjugate Roots (2)

– First IC

– Then, remaining and its derivative

– Second IC

– Particular solution

  0 2 2, 1

y   BB

  0 0

y   A

0.1x

sin 2 ye

x

0.1x

sin 2 yBe

x

0.1

0.1x

sin 2 2

0.1x

cos 2

y   Be

xe

x

(30)

Example of Complex Conjugate Roots (3)

e

0.1x

e

0.1x

Damped vibrations

(31)

– DE

– General solution

2

0

y    y

cos sin

yAx B   x

Example of Complex Conjugate Roots (4)

(32)

Summary

Case

Roots of

Characteristic Eq.

Basis of DE General sol.

I

Distinct real

II

Real double root

III

Complex conjugate

1

,

2

  e

1x

, e

2x

y c e

1 1x

c e

2 2x

1

  2 a

e

ax 2

, xe

ax 2

1 2

2

 

ax

y c c x e

1,2

1

  2 a i

 

2 2

cos , sin

ax ax

e x

e x

  

2

cos sin

y e

ax

Ax Bx

(33)

Boundary Value Problem

– DE

– Example

– General sol.

– Particular sol.

 

1

1

,  

2

2

y P k y P k

Boundary

conditons

   

0, 0 3, 3

     

y y y y

  

1

cos 

2

sin

y x c x c x

   3cos 

2

sin

y x x c x

Not determined yet

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