ode ode Basic Concepts and Theorems The n th order linear ODE takes - - PowerPoint PPT Presentation

ode ode basic concepts and theorems
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ode ode Basic Concepts and Theorems The n th order linear ODE takes - - PowerPoint PPT Presentation

So Solution tion of of hi high gher er or orde der r ode ode Basic Concepts and Theorems The n th order linear ODE takes the form: n n 1 d y d y dy a a ... a a y f x ( ) n n 1 1


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SLIDE 1
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SLIDE 2

So Solution tion of

  • f

hi high gher er or

  • rde

der r

  • de
  • de
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SLIDE 3

Basic Concepts and Theorems

The nth order linear ODE takes the form:

1 1 1 1

... ( )

n n n n n n

d y d y dy a a a a y f x dx dy dx

  

    

(1) y and all its derivatives are of the first degree

1

, ,...,

n n

a a a

(2) All the coefficients are function of the independent variable x only or constants.

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SLIDE 4

Definition:

The D operator is defined to be:

d D dx 

dy Dy dx 

2 2 2

d D dx 

2 2 2

d y D y dx 

n n n

d D dx 

n n n

d y D y dx 

are n independent solutions of the homogenous DE

 

1 1 2 2

...

n n

y x c y c y c y    

1 2

, ,..., n y y y

then the general solution is

Theorem

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Complementary and particular solutions

The complementary (or homogenous) solution:

c

y

p

y

G

y

G c p

y y y  

The particular solution: is any solution satisfy the non-homogenous case is the solution of the homogenous case The general solution:

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SLIDE 6

Second Order Linear Homogenous DE with Constant Coefficients

1

y a y a y     

x

y e 

x

y e   

2 x

y e   

2 1

( )

x

a a e     

2 1

( ) a a     

This equation called the characteristic equation (c/e) (or auxiliary equation) of the DE.

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The solution is:

x x

e c e c y

2 1

2 1   

6 5       y y y

6 5

2

    

  

3 2     

1 2

2, 3       

x x

e c e c y

3 2 2 1  

 

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SLIDE 8

The solution is:

1 2

( cos sin )

x

y e c x c x

   

4 13 y y y     

2

4 13     

1,2

2 3i    

2 1 2

( cos3 sin3 )

x

y e c x c x

 

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The solution is:

1 2 x x

y c e c xe

 

 

4 4 y y y     

2

4 4     

 2

2   

1 2

2     

2 2 1 2 x x

y c e c xe  

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SLIDE 10

Example

2 y y y     

 2

1   

1 2

1      

1 2 x x

y c e c xe

 

 

Solution

2

2 1     

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SLIDE 11

Example 4 5 (0) 1 (0) y y y y y        

2

4 5     

  

5 1     

1 2

5, 1      

5 1 2 x x

y c e c e   Solution

1 2

(0) 1 1 y c c    

1 2

(0) 5 y c c     

1 2

1 5 , 6 6 c c  

5

1 5 6 6

x x

y e e  

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SLIDE 12

Example

9 y y   

2

9   

1 2

3i      

1 2

cos3 sin3 y c x c x   Solution

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SLIDE 13

Higher Orders Homogenous Linear DE with Constant Coefficients

   

1 1 1

...

n n n n

a y a y a y a y

 

     

1 1 1

...

n n n n

a a a a   

 

    

1 2

, ,..., n   

(3) According to the nature of these roots we write the linearly independent solutions. (1) Set the c/e: (2) Find the roots:

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Example 2 3 y y y      

Solution

3 2

2 3      

 

2

2 3      

  

3 1      

1 2 3

0, 1, 3       

3 1 2 3 x x x

y c e c e c e

  

3 1 2 3 x x

y c c e c e

  

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SLIDE 15

Example 4 4 y y y y       

Solution

3 2

4 4         

2

1     

4 1   

 

2

1 4     

1 2,3

1, 2i     

1 2 3

cos2 sin2

x

y c e c x c x   

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SLIDE 16

Example

 

4 2

2 1 D D y   

Solution

4 2

2 1     

2 2

( 1)   

1,2 3,4

i     

1 2 3 4

( cos sin ) ( cos sin ) y c x c x x c x c x    

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SLIDE 17

Example

 

4 2

8 16 D D y   

Solution

4 2

8 16     

2 2

( 4)   

2 2

( 2) ( 2)     

1,2 3,4

2, 2     

2 2 2 2 1 2 3 4

( ) ( )

x x x x

y c e c xe c e c xe

 

   

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SLIDE 18