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SECOND ORDER LINEARORDINARY DIFFERENTIAL

EQUATIONS WITHCONSTANT COEFFICIENT

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Given ordinary differential equation

2

2 ( )

d y dy

a b c y f xdx dx+ + =

The differential equation is in order 2, linear, and the

coefficients a, bandcareconstants.Such differential equation

is known as second order linear ordinary differentialequation with constant coefficient.

f(x) = 0: homogenous

2

2 0d y dya b c ydx dx+ + =

2

2 ( )d y dya b c y f xdx dx+ + =

F (x) = 0 : nonhomogenous0)( xf

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I. Homogenous differential equations2

2 0

d y dya b c y

dx dx+ + =

Method of solving homogenous differential equation (*)

Step 1 From

change to

2

2 0

d y dya b c y

dx dx+ + =

2

0am bm c+ + =

characteristic equation

Step 2 Find roots of the characteristic equation

Step 3

Step 4 If given initial condition(s), substitute into thesolution to obtain values of AandB.

Find a general solution 3 cases arise according as

the roots are real and distinct (CASE 1), real andequal (CASE 2) or complex (CASE 3)

2

2

2

mdx

yd

mmdx

dy

1

10 my

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CASE 1: b2> 4ac CASE 2: b2= 4ac CASE 3: b2< 4ac

Distinct real roots Equal real roots Complex roots

Suppose the rootsare m1and m2

Then the generalsolution is

1 2m x m xy Ae Be= +

Suppose the rootsare m1and m2,

real and equal

such that

Then the generalsolution is

( ) mx

y A Bx e= +

1 2m m m= =

Suppose the rootsare complexnumber

Then the generalsolution is

( cos sin )x

y e A x B xa

b b= +

1

2

m im i

a b

a b

= +

= -

Example: CASE 1

Example: CASE 2 Example: CASE 3

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II. Nonhomogenous differential equations

2

2 ( )d y dya b c y f x

dx dx+ + =1.

)(xfcyybya 2.

3. )(xfcyybya

or

or

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1. Undetermined coefficients

2

2 ( )d y dya b c y f xdx dx+ + =

Consider

The class off(x)to which this method applies is actuallyquite restricted.f(x)defined by one of the following:

1. ,where nis a positive integer or zero.2.3.4. Linear combination of function 1-3

5. Product of function 1-3

nxmxe

bxbx sinorcos

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Step 1

Step 2 Find particular integral, py

Step of solving nonhomogenousdifferential equation

by Undetermined coefficients

Step 3

Step 4 If given initial condition(s), substitute into thesolution to obtain values of AandB.

Find complementary function, use method (*)to find general solution of homogenous DE

cy

General solution of nonhomogenous differentialequations:

or h py y y= +pc yyy

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How to solve nonhomogenous differential equationusing undetermined coefficients method?

2 In Step 2, use undetermined coefficients method to findparticular integral. Do the following steps:

pyStep 1uc Choose a correct according to the type of .Refer the Table of particular integral

( )f x

0r= py

py

If no? Do Step 3uc

Ifyes? Substitute with and check again1r=If no? Do Step 3uc

Ifyes? Substitute with and go to Step 3uc2r=

Step 2uc Substitute into the chosen and checkeither the term in corresponds to any term in cy

1 Do Step 1: Find refer step of solving homogenous DE (*)cy

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Step 3uc

Step 4uc Equating the coefficients to find the unknown

coefficients

3 Do Step 3and Step 4

To be easier, we will classify the 5 applicable f (x)for this

method to 5 casesand will discuss thoroughly each case.

Example: Undetermined coefficients

Find and then substitute into thedifferential equation given

'

py''

py

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How to solve nonhomogenous differentialEquation UsingVariation of parametersmethod?

2 Refer step of solving homogenous DE (*)

Rewrite in formcy 21 ByAyyc

1 Determine the values of aand f (x). Make sure coefficient ofsecond order derivative is 1.

In Step 3, use variation of parameters method to findparticular integral. Do the following steps:

Step 1vp Calculate the Wronskian,

1221

21

21yyyy

yy

yyW

3

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Obtain the general solution using the formula

1 2y uy vy= +

4

Example:Variation of parameters

Step 2vp Calculate

and dxWxfyu )(2

dx

W

xfyv

)(1

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Application

Let the spring have unstretched length l.

The mass mis attached to its lower

End and come to rest in its equilibriumposition, thereby stretching the springan amount dso that its stretched length

is l+d

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Pull the mass down and let it go, its startsoscillating

--------------------------------------------------------------------------------The differential equation for the motion of mass on the spring is

2

2 ( )

d y dym a ky F t

dt dt + + =

where mis the mass,yis the displacement of stretched springwhen it pulled down, ais the damping constant, kis the springconstant andF (t)is external forces.

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We will consider 2 cases

Free oscillation Forced oscillation

Undampedmotion

Dampedmotion

Cases when

a= 0 andF (t) = 0then the diff eqnbecomes

2

2 0

d ym ky

dt

+ =

Cases when

a> 0 andF (t) = 0then the diff eqnbecomes

2

2 0

d y dym a ky

dt dt

+ + =

Cases that considereffect of dampingupon the mass onthe string and anexternal forces.Thusthe diff eqn is

2

2 ( )

d y dym a ky F t

dt dt + + =

Example: Free-u Example: Free-dExample: Forced

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