Week 8 [compatibility mode]

Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 200782009

KNF1023Engineering

Mathematics II

Second Order ODEs

Learning Objectives

Explain about variation of parameters

method in the inhomogeneous ODEs

Explain about Spring Mass System and

Electric Circuits

Method of Variation of Parameters

� There is a more systematic approach for

finding a particular solution of the 2nd order linear inhomogeneous ODE.

� The approach known as the method of variation of parameters makes use of two linearly independent solutions of the corresponding homogeneous ODE.

" ( ) ' ( ) ( )y f x y g x y r x+ + =


The solution of the second order inhomogeneous

ODEs is

{ {

homhom

( ) ( )p

general solution of thea particular solutioncorresponding ogeneous ODEof in ogeneous ODE

y y x Y x= +

1 2( ) ( ) ( ) ( )py u x Y x v x Y x= +


Thus, the variation parameters will involved

two equations which is

and

( )( ) ( )

∫ −−= dx

yyyy

xyxrxu

)( 2

'

1

'

21

2

( )( ) ( )

( )∫ −= dx

yyyy

xyxrxv

2

'

1

'

21

1

Example 1

Consider the inhomogeneous ODE

Notice that has two linearly

independent solutions given by

Hence

( )xyy exp3'' =−

0'' =− yy

( ) ( ) ( )( ) ( )' '

1 2 1 2 exp exp exp exp 2y y y y x x x x− = − − − − =

( ) ( )xyandxy expexp 21 =−=

Continue…

If we look for a particular solution of the

inhomogeneous ODE in the form

then u and v are given by

and

( ) ( )x x

py u x e v x e−= +

3exp( )exp( ) 3( ) exp(2 )

2 4

x xu x dx x= − = −∫

3exp( ) exp( ) 3( )

2 2

x xv x dx x

−= =∫

Continue…

Notice that we can ignore the constants of

integration, since we are only interested in getting

a particular solution.

So, a particular solution of the inhomogeneous

ODE is" 3exp( )y y x− =

( ) exp( ) ( ) exp( )

3 3exp(2 )exp( ) exp( )

4 2

3 3exp( ) exp( )

4 2

py u x x v x x

x x x x

x x x

= − +

= − − +

= − +

Continue…

The required general solution of the

inhomogeneous ODE is

Where C and D are arbitrary constants.

This can again be rewritten as

Where , ,are arbitrary constants

3 3exp( ) exp( ) exp( ) exp( )

4 2y x x x C x D x= − + + − +

3 3exp( ) exp( ) exp( )

2 4

3exp( ) exp( ) exp( ) ( )

2

y x x C x D x

x x A x B x as before

= + − + −

= + − +

3

4A C and B D= = −

APPLICATIONS OF SECOND-ORDER EQUATIONS -Spring –Mass System

Simple harmonic motion may be defined as

Motion in a straight line for which the acceleration

Is proportional to the displacement and in the

opposite direction. Example of this type of motion

are a weight on a spring.

Damped System

If we considered a mass-spring system and

modeled it by the homogeneous linear ODE,

0'" =++ kycymy


Here y(t) as a function of time t is the

displacement of the body of mass m from rest.

Forces within this system are inertia my”, the

damping force cy’ (if c>0) and the spring force ky

acting as a restoring forces.

The corresponding characteristics equation is

The roots are

2 0c k

m mλ λ+ + =

mkcmm

c4

2

1

2

2

2,1 −±−=λ


Using the abbreviated notations

and

We can write

m

c

2=α mkc

m4

2

1 2 −=β

βαλ

βαλ

−−=

+−=

2

1


The form of the solution will depend on the

damping, and,

Case I. c2> 4mk. Distinct real roots λ1, λ2

(Overdamping)

Case II. c2< 4mk. Complex conjugate

(Underdamping)

Case III. c2 = 4mk. A real double root

(Critical damping)


Case 1. Overdamping.

When the damping constant c is so large that c2 >

4mk, then λ1,λ2 are distinct real roots, the general

solution is

Case II.Underdamping.

If the damping constant c

is so small that c2< 4mk then β is pure imaginary.

The roots of the characteristics equation are

complex conjugate,

ttececty

)(

2

)(

1)( βαβα +−−− +=


The general solution is

ωαλ

ωαλ

j

j

−−=

+−=

2

1

( )tBtAetyt ωωα sincos)( += −


Case III. Critical damping.

If c2 = 4mk, then β=0, λ1 = λ2 = , and the

general solution is

Undamped System

If the damping of the system is so small and can be

disregarded, then

The general solution is

tectcty

α−+= )()( 21

0" =+kymy

tBtAty 00 sincos)( ωω +=

α−

Example 2

In testing the characteristics of a particular

type of spring, it is found that a weight of 4.90 N

stretches the spring 0.49 m when the weight and

spring are placed in a fluid which resists the motion

with a force equal to twice the velocity. If the

weight is brought to rest and then given a velocity

of 12 m/s, find the equation of motion.

Continue…

Using Newton’s second law, we have

The weight is 4.90 N and the acceleration due to

gravity is 9.80 m/s2. Thus, the mass, m is

m=4.90/9.80 = 0.500 kg

The constant k is found from the fact that the

spring stretches 0.49 m for a force of 4.9 N, Thus,

using Hooke’s Law,

0'" =++ kycymy

kyymy

kyymy

−−=

=++

'00.2"

0'00.2"

Continue…

4.90 = k (0.490)

k = 10.0 N/m

This means that the differential equation to be

solved is

Solving this equation, we have

00.20'00.4"00.1

010'00.2"500.0

=++

=++

yyy

yyy

Continue…

)00.4sin00.4cos(

00.400.2

00.2

)00.1)(0.20(40.1600.4

00.2000.400.1

21

00.2

2

tctcey

jm

m

mm

t +=

±−=

−±−=

=++

−

Continue…

Since the weight started from the equilibrium

position with a velocity of 12.0 m/s, we know that

y= 0 and y’ = 12.0 for t = 0. Thus,

Thus, since c1=0, we have

0

1 2

1

0 ( 0 )

0

e c c

c

= +

=

00.3

)00.2)()(0()00.4)(1(0.12

)00.2)((00.4sin)00.4)(00.4(cos'

00.4sin

2

0

2

0

2

00.2

2

00.2

2

00.2

2

=

−+=

−+=

=−−

−

c

ecec

etctecy

tecy

tt

t

Continue…

This means that the equation of motion is

2.003.00 sin 4.00ty e t

−=

Application on Electric Circuit

Given the RLC circuit as below,

(a) Show that the circuit can be modeled as

(b) Given that Solve the differential equation for the initial conditions, when

and

( )2

'

2

d i di iL R E t

dt dt C+ + =

( ) 0.01 tE t e

−= −

0, 0t i= = 0di

dt=

Solution

(a)

differential with respect

to t.

(b) Given hence

( ):L R C

KVL V V V E t+ + =

( )1di

L iR idt E tdt C

+ + =∫

( )2

'

2

d i di iL R E t

dt dt C+ + =

( ) 0.01 ,tE t e

−= − ( )' 0.01 tE t e

−=

2

20.001 0.005 0.01

250

td i di ie

dt dt

−+ + =

2

25 4 10 td i di

i edt dt

−+ + =

Continue…

� For Particular Solution

'' '

2

4

5 4 0

5 4 0

4@ 1

t t

i i i

I Ae Be

λ λ

λ− −

+ + =

+ + =

= − −

= +

( )

'

'' 2

t

p

t t

p

t t t t t

p

i te

i e te

i e e te e te

α

α α

α α α α α

−

− −

− − − − −

=

= −

= − − − = − +

Continue…

Compare to

Thus

( ) ( ) ( )'' '5 4 2 5 4t t t t ti i i e te e te teα α α α α− − − − −+ + = − + + − +

3 teα −=

10 te−

3 10

10

3

α

α

=

=

10

3

t

pi te

−=

Continue…

Thus, the general solution is

Given and

4 10

3

t t t

pi i I Ae Be te

− − −= + = + +

( )0 0i = ( )' 0 0i =

( )

( ) ( )

4 0 0 0

' 4

4 0 0 0 0

100 0 , 0 (1)

3

10 104

3 3

10 10 100 4 0 , 4 (2)

3 3 3

t t t t

Ae Be e A B

i Ae Be e te

Ae Be e e A B

− − −

− − − −

− − − −

= + + + = − − −

= − − + −

= − − + − + = − − −

Continue…

�Try to use variation parameter to solve this application question and check whether the answer is the same or not…

( ) 4

10 10,

9 9

10 10 10

9 9 3

t t t

A B

i t e e te− − −

= = −

= − +

Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 2007/2008

Week 8 [compatibility mode]

Education

Transcript of Week 8 [compatibility mode]