Theorem : E and U for nth order linear differential equations

Chapter 3 – Theory of Linear Equations

By: Kemal Ahmed

Unrelated: call 88 if you have an on-campus medical emergency

Go to class because every teacher teaches differently even if you failed it and you think you know everything because you don’t.

[email protected]

2013−05−13

Homework #3

2 2 ' x y y

xy  

Wrong method:

2 2

1 1

'

' ' _x

x y

y

xy xy x y

y y y xy

y x

  

    

Correct method:

 

d 1 1 2 d

d 2 _{2 finish on your own} d

2 1

2 2 2

d

1 1

2 d '

,

x x v v x x x

x

x x

v v x

y y y x

y y x v y

  

    

  

3.1.1: Initial−Value Problems (IVP) & Boundary−Value Problems (BVP)

 

 

_{ }

 

_{ }

_{ }

 

 

 

 

_{ }

 

1

1 1 0

0 0

0 0

0 0

1 1

0 0

... '

IVP :

' '

" "

n n

n n

n n

a x y a x y a x y a y g x

y x y y x y y x y

y x y

 

 

     

 

 _

 _



 _

  

 



I = interval value coefficient functions:

 

, 1

 

,..., 0

   

,

n n

a x a_ x a x g x are continuous Recall from first order differential equations:

 

 

0 0

' , function continuous on is continuous on

f y

y f x y I

y x y _ I

 

 

 

=>Theorem of existence & uniqueness, then the solution exists and is unique Theorem: E and U for nth order linear differential equations

Let a_n

 

x a, _n1

 

x ,...,a x g x₀

   

, be continuous functions on I, and a_n

 

x 0for everyxI. If 0

x x I, then the solution of the IVP exists on the interval I, and is unique.

e.g. 1

         

 

 

 

1

2 0

3

3 ''' 5 '' 1 ' 7 0 1 0

' 1 0 '' 1 0

a x

a x a x g x

a x

y y y y y

y y

   

    

 



 





 

 

 

 

0 1

any 0

' 0

trivial solution: 0 '' 0

''' 0

n

I x y x y

y y

y

  

  

 _

 

 _



What happens if not all the conditions in the theorem of existence and uniqueness are not satisfied? Nothing!

e.g. 2

 

 

2

'' 2 ' 2 6 0 3

' 0 1

x y xy y y

y

   

  

 _



 

 

 

 

2 2

1 0

continuous on 2 continuous on 2 continuous on 6 continuous on

a x x a x x a x

g x I

 

       

The theorem fails if you pickx 0 I, which gives a x₂

 

x x2,  0 ,I 



0,



, which makes it not unique, so you cannot use it. Solution:

 

2

3

y x cx    x c

BVP

2nd order

   

   

 

 

 

 

2 1 0

0 1

'' '

a x y x a x y x a x y g x y a y

y b y

  

 

Note that there are 2 points, a & b where ab.

Over-determined: more than n different conditions for n order

Under-determined: less than n different conditions for n order

 

 

0 0 '

'' BVP

y a y

y b y

  

e.g. 1

 

 

 

1

 

2

 

0 0 '' 1 0,

1 0 cos 4 sin 4

x x x

x

x t c t c t

     

 

→ 2 parameter family of solutions

 

 

 

 





 

 

 

1 2 1

2 2

2 2

2 0 2

2

0 1 0 0

sin 4 sin 4

sin 2 0

sin 4

x c c c x t c t

x c c c

x t c t

 





      

 

 

  

Thus infinitely many solutions because c₂ .

e.g. 2

 

 

 

 

 

1 2

1 2

'' 1 0

cos 4 sin 4

0 0 0

1, 0 1 no solution

x x

x t c t c t

x c

x 

  

 

  

  

3.1.2: Homogeneous Equations

1. 2 '' 3 'y  y y 0is homogeneous

2. 2 '' 3 'y  y   y 1 0 is not homogeneous because you can bring the −1 to the other side and then the equation equals −1, not 0.

3. 3xy' 3 x y2 7xis not homogeneous because you have 7x is not 0.

Homogeneous equation (get into this form): a_n

 

x y n a_n1

 

x y n1  ... a x y₁

 

'a x y₀

 

0 No-homo equation:

 

  1

 

 1 ... 1

 

' 0

 

 

n n

n n

a x y a _ x y   a x ya x yg x

 5

2 2

2 '' 3 '

7 4

2 '' 3 ' 13

1 1

y y y

y x

y y y y x

x y

  

     



The differential operator: usually in the form: d

 

2 2 dx x  x

e.g.



2



d

3 2 3

 

 

 

 

 

 

2 2 2

2 2 2

2

2

d d

3 7

d d

3

d d

3 7

d d

d d

3 7

d d

6 3 6 7 3

L x

x x f x x

L f f

x x

L f f f f x x L f x x

 

_   _

 



 

_   _

 

  

    

L:=

1

1 1 1 0

d d d '

...

d d d

n n

n n n n

a a a a

x x x

  

   

 

0

0 *** short-hand.

L y L y

 

  

The superposition principle for homogeneous equations:

Let y y₁, ₂,...y_ksolutions of the nth order homogeneous linear equationL y

 

0. Then the linear combinations of



y₁,...y_k



is again a solution of L[y]=0

i.e. y:c y_{1 1}c y_{2 2} ... c y_{k k}

In summary: a linear combination of homo solutions is a solution.

e.g.

 

_

_

2 1

2 2

3 2

ln

2 ' 4 0, 0,

y x y x x

x y xy y I

 

    

Claim that y:c y1 1c y2 2is a solution, where c1c2. 2

1

2 2

1 1

1 2 2

2 1 2

2 1 2 ' 2

0 2 2 4 0

'' 2 ''' 0

ln

' 2 ln 2 ln

'' 2 ln 2 1 ''' 2

x

x

y x

y x

x x x x

y

y

y x x

y x x x x x x

y x

y 



     

 

    

1 1 2 2

' ' '

'' ...

y c y c y y

 



Linear dependence / independence

Definition: for a given set of functions,



f x₁

 

,...,f_n

 

x



is said to be linearly dependent on an interval, I, if there exists constants, c₁,...,c_nnot all zero, such that

 

 

 

1 1 2 2 ... n n

c f x c f x  c f x a.

For everyxI, what is not linearly dependent is called linearly independent.

e.g.

 

 

 

 

1 2 3

2 4

5 5 1

f x x

f x x x

f x x

f x x

      



0,



I  

Is this linearly dependent or not?

Can you can multiply the equations by constants to get the others? If so, the set is linearly independent.

 

 





3

1 2

3 1 2

5 5

5 1

5

f

f x f x x x

f f f

 

     

   

Therefore the set is linearly independent.

e.g.

 

 

 

 

1 2

2

2 3

sin , 0, 10

1

f x x x f x x f x x



 

  

This is an example of linear independence

Wronskian can determine whether a set is linearly independent/dependent.

     

1 2

1 2

1 2

1 1 1

1 2

...

' ' ... '

'' '' ... ''

...

n n n

n n n

n

f f f

f f f

f f f

f  f  f 

e.g.

 

 

 

 

1 2

2

2 3

sin , 0, 10

1

f x x x f x x f x x



 

  



  

 









2

2

1 3 3 3

2 2

2

sin 10 1

W cos 10 2

sin 0 2

sin 10

10 1

sin 1 0 2 1

cos 10 10 2

sin 20 10 10 2 10sin 10 cos sin 10 10sin 20sin 20 cos

x x x

x x

x

x x x x

x

x x

x x x x x x

x x x x x x

 

 





      

     

     

Will it be 0? No, so independent 0

 for x

 

0,₂ , ,f f_{1 2},f₃

Let y y1, 2...yn be n solutions of the homogeneous nth order differential equation L[y] = 0 on the

interval, I. Then the set of solutions



y1,...yn



is linearly independent on I iff W



y1...yn



0 is

called a fundamental set of solutions.

If



y₁,...y_n



is a fundamental set of solutions of the homogeneous equation L[y] = 0, then

 

: 1 1 2 2 ... n n

y x c y c y  c y , not all c’s zero is called a general solution.

e.g.

 

 

3 1

3 2

solutions

'' 0

x

x

y x e y x e

y gy



   

Can you deduce the general solution of y''gy0 ? First check that



y y₁, ₂



forms a fundamental set of solutions. So this equivalent with the checking that y₁and y₂are linearly independent <=>





1 2

1 2

1 2

W , 0

' '

y y y y

y y

  for

 

 x I

Note: thesymbol means any, the ϵ symbol means belongs.



1 2



33 33

W , 3 3 6 0

3 3

x x

x x

e e y y

e e





      



So 



y y1, 2



is linearly independent. So the general solution is

 

 

1 1 2 2

3 3

1 2

x x

y x c y c y y x c e c e

 

 

So the constants are c₁&c₂.

3.3: Homogeneous Linear Equations with constant coefficients

For now, let’s focus on second order.

3xy''  y' y 0←This is a homo linear equation without constant coefficients. However, we are going to work with equations that do have constant coefficients.

3 ''y   y' y 0

Homo linear constant coefficients second order

'' ' 0

a, b, c constants, 0

ay by cy

M a

  

 

Hint: assume one potential solution, M, looks like y x

 

emx

m is constant

 

 

 

2

,

' ,

'' ,

mx

mx

mx

y x e

y x me

y x m e 

  Plug into M

2

0

mx mx mx

am e bme ce 

2

2 1

2 2

0 auxiliary equation 4

2 4 2

am bm c b b ac m

a b b ac m

a

   

  



  



Cases:

1. m₁m₂is real where b24ac0 2. m₁m₂ is real where 2

4 0

b  ac 3. m m₁, ₂are complex if 2

4 0

b  ac

Case 1:

The general solution of the equation, ay''by'cy0, is

 

1 1 2 2

y x c y c y , where 1

2 1 2

m x

m x

y e y e

 

Are you surey1&y2linearly independent? Wronskian awaaay!









   





 

_

_





1 2

1 2

1 2 1 2

1 2 1 2

1 2

1 2 2 1

1 2 2 1

0 1 2

W ,

W ,

W ,

W , 0

m x m x

m x m x

m m x m m x

m m x

e e y y

m e m e y y m e m e y y e m m y y

 







 

 



Case 2:

1 2

1 2

mx

m m m y y e

 

 

General solution yc y_{1 1}c y₂



y y1,



is linearly independent

: mx

y xe

At home, check:

 

y y, is linearly independent

 

1 2

mx mx

y x c e c xe

 

 

1

2

1 2

1 2 1 1 2

complex

l.i.

a bi x m x

a bi x m x

w w

m a bi

m a bi

a b

y e e

y e e



     

  



  _



  _

Shall want to eliminate the complex powers Euler’s formula

 

1

 

2

 

cos sin

cos sin

i

ax

e i

y x e c bx c bx

 _{  }

  _  _

General solution:

 

   

1 2

a bi x a bi x

y x c e  c e 

e.g. a)

 

2

1

2

'' ' 12 0 plug-in

12 0 auxiliary equation 1 1 48 1 7

4

2 2

1 7 3 2

mx

y y y y x e m m

m

m

    

   

  

  



  

General solution y x

 

c e₁ 4xc e₂ 3x

e.g. b)





2 2

5 5

1 2

'' 10 ' 25 0 10 25 0

5 0

G.S. : x x

y y y m m

m

y c e c e

  

  

 

e.g. c)

 

 

 

 

 

2

1,2

1 2

2

1 2

'' 4 ' 7 0

4 7 0

4 16 28 2

4 12

2 4 2 3

2 3

2

cos sin

cos 3 sin 3

a

x

y y y m m m

i

i

y x e c bx c bx e c x c x

  

  

  

    

 

   

 _  _

 

 _  _

2013-05-15

Sorry, I’m late because I’m kinda sick and slept in.

e.g. 2

 





4

4 3 2

2 2

1 2

2

3

1

''' '' 0 0 1 0 0

1 0 3

1 1 3 1 3

2 2 2

1 1 3 1 3

2 2 2

y y y m m m m m m m m m m

m i

m i

  

  

  

 

        

   

 

   

 

 

 

 

1

1 2 2

1 2

x 3 3

1 2 3 2 4 2

1 2

cos sin

after subbing in m's ...

m m x

y x c e c xe e c x c x

y c c x e 



 

   _  _

   

3.1.3 Homogeneous Equations

 

 

2 2

' 16

4 constant & particular solution.

x x x t

   

General Solution for no-homo equations

Let y_pbe a particular solution of the no-homo equation,a_m

 

x y n  ... a x y₁

 

'a x y₀

 

g x

 

And let



y y₁, ₂,...,y_n



be a fundamental set of solutions of the associated homo equation,

 

 

_{ }

_{ }

1 0

... ' 0

n m

a x y  a x ya x y . Then the general solution for the homo equation is

 

 

1 1 ... n n p

yc y x  c y x y , the solution of the homo equation. The solution of a homo linear (with constant coefficients) differential equation is also called the complementary solution.

e.g.

11 1 12 2

''' 6 '' 11 ' 6 3

p

y y y y x

y x

   

  

What is the general solution?

Observation 1: Non-homo equation, constant coefficients

Obs 2: Look at associated homo equation: y''' 6 '' 11 ' 6 y  y  y0

3 2

6 11 6 0

m  m  m  Synthetic Division:

3 m 1

2 m −6

1 m 11

0 m −6 Factors:



   1, 2, 3, 6



1 1

1 1

−5 −4

−6 −2

0

2 1 −5

2

6 −6

0

1 −3 0 → so (m−3)









1 2

1 2 3

1 2 3

1 2 3 0

1 2, 3

x x x

c

m m m m ym m

y c e c e c e

   

  

  

Then the general solution is y y_cy_p

2 3 11 1

1 2 3 12 2

c p

x x x

y y

yc e c e c e   x

Go here for tests https://sites.google.com/site/macengfifteen/eng-docs/engineering-ii/term-1/math-2z03.

Let

1, 2,... k

p p p

y y y for k particular solutions of the no-homo linear (nth) order differential equation (*) on an interval, i corresponding to k distinct functions: g x g₁

   

, ₂ x ,...,g_k

 

x , then

 

 

1 ... k

p p p

y  y x  y x is a particular solution of (*), where g(x) from (*) is g x

 

g₁ ... g_k

1. y''' 3 ' 4 y y 16x224x 8 g₁

2. ''' 3 ' 4 2 2 2

x

y  y y e g

3. y''' 3 ' 4 y y2xex ex g₃

1. 1

2 4

p

y   x 2.

2 2x p

y e

3. 3

x p

y xe

The superposition principle states that if you look at the no-homo equation,

 

 

 

1 2 3

''' 3 ' 4

y  y yg x g x g x , then a particular solution of it is

1 2 3

p p p p

y  y y y

3.4 – Undetermined Coefficients

Used for no-homo linear differential equations, y y_n  n  a_n y n1  ... a y₁ 'y y₀ g x

 

. It works well if g(x) is:

 A polynomial, like 2

7 1 x  x .

 An exponential, like e3x7,ex,e2x.

 sin(x), cos(x), sin(3x)

 Or a combination of any of the above, like exsinx

 Does not work for:

o polynomial fractions

 

2 3 3 1

x x or

1

x

o tan(x)

o ln(x)

o inverse functions

o Everything else

e.g.

2

'' 4 ' 2 2 3 6

 

   

2

4 2 6 2

2 6 2 6

1 2

4 2 0

4 16 4 2 2 4 24

2

2 6

x x

c

m m m

m m m

y c e c e

 

   

  

  

      

  

  









 



2

2 2

2 2

2 2

Guess: ' 2 '' 2

2 4 2 2 2 3 6

2 8 4 2 2 2 2 3 6

2 8 2 2 4 2 2 3 6

p

p p

y ax bx c

y ax b

y a

a ax b ax bx c x x

a ax b ax bx c x x

ax a b x a b c x x

    



       

       

        

a= −1

8(−1) − 2b = −3 b = −5/2

2(−1) + 4(−5/2) − 2c = 6 −2 – 10 − 2c = 6

c = −9 2 5

2 9

p

y   x x

e.g.

 

'' ' 2sin 3

y   y y x ← if right side was 2cos(3x), use the same yp

Guess :

 

 

 

 

 

 

sin 3 cos 3 ' 3 cos 3 3 sin 3 " 9 sin 3 9 cos 3

p

p

p

y A x B x

y A x B x

y A x B x

 

 

  

 

 

 

 

 

 



   

  

9 sin 3 9 cos 3 3 cos 3 3 sin 3 sin 3 cos 3

9 3 sin 3 9 3 cos 3

A x B x A x B x A x B x

A B A x B A B x

     

 

 

 

 

8 64 9

3 3 3

13 6

3 73

8 8 6 16

3 3 73 73

16 6

73 73

9 3 2 8 3 2 8 3 2 2

9 3 0 8 3 0 8 3 2

sin 3 cos 3

B

B

p

A B A A B B B B

B A B B A B A B B

A

y x x



  



              

              

  

 

e.g.



1



2

1

2 2

2 " 2 ' 3 4 5 6

p

p

x

g g

x p

y _y

y y y x xe y Ax B Cx D e

    

   

Sometimes, after you tried to find A, B, C, D and you weren’t able to do so, try to use a polynomial of higher degree.

e.g.



Cx2DxE e



2x OR

e.g. Ax3Bx2Cx D



Ex2Fx2GxH e



2x

Variation of Parameters

Used to find particular solutions for non-homo linear (nth) order equation, and it is a bit more general than undetermined coefficient (reacts nicely with lnx, 1/x, tanx, etc.) it is very

computational so for now, I will solve 2nd order linear non-homo equation

   

   

   

 

2 " 1 ' 0

a x y x a x y x a x y x g x and a₂

 

x 0. Standard form:

 

 

 

 

 

 

 

 

 

1 0

2 2 2

"

P x Q x f x

a x a x g x

y y y

a x a x a x

  

The method…

 

1

_{ }

4 " 36 csc 3

sin 3

y y x

x

  

Step 1: solve for yc

2 2

3 3

1 2

4 36 0

9 0 3

x x

c

m m m

y c e c e

     

 

 

 

 

 

 

3 3

3 3

1 3 3 2 3 3

3 3 3 3

3

0 0

1 1

3 3

4 sin 3 4 sin 3

' , '

3 3 3 3

... 6 4 sin 3

x x

x x

x x x x

x x x x

x

e e

e e

x x

x x

e e e e

e e e e

e x                   

Plug back into differential equations

 

3

 

3

1 2

' ... " ...

x x

y x e x e

y y           _ 

 

 

 

 

 

 

1 1 _{3 3}

1 2 2 2 ' plug into: ' x x x x

y x e x e

x x          _     _





 

 

 

 

2 1 2 1 1 2 1 2

1 2 1 2

1 2 1 2

0 0 ' ' W W ' , ' W W ' ' ' ' y y

f x y y f x

x x

y y y y

y y y y

      e.g.









 

 

 













 





2 2

2 2 2

1 2

2 2

1 2

2

2 2 2 4

2 1

1 2 2 4

2 2 2

2 2

4

2 2 2

2

2 2 2 4

2 2 2

" 4 ' 4 1

, 4 4 0 2 0

0

1 2 1

W ' 1 W 2 2 1 2 2 W ' W 2 2 x x x c x x x

x x x x

x

x x

x x x

x x

x

x x x

x

x x

x x x

y y y x e

y c e c xe m m m

y x e x xe

xe

x e xe e x x e

x x x x x

e

e xe

e xe e

e xe

x e

e xe e

x x

e

e xe

e xe e

                                       



1



 

 

 

 









3 2

2

3 2 2

1 1 3 2

2 2 2

2 2

3 2 2

' '

x x

x

x x

x x x

p

x x

x x x

y e x xe

                



