differential equations review

7/31/2019 Differential Equations Review

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In the following we will BRIEFLY review the basics of solving Linear, Constant

Coefficient Differential Equations under the Homogeneous Condition

Homogeneous means the forcing function is zero

That means we are finding the zero-input response that occurs due to the effect

of the initial coniditions.

Write D.E. like this:

) ) )(...)(...)(

01

)(

01

1

1 tfbDbDbtyaDaDaD

DP

m

m

DQ

n

n

n

444 3444 2144444 344444 21==

+++=++++

.. EqDiff

)()()()( tfDPtyDQ =

m is the highest-order derivative

on the input side

n is the highest-order derivative

on the output side

We will assume: m n

)()( tyDdt

tyd kk

k

Use operational notation:


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Puty0(t) into () and use result for the derivative of an exponential:

0)...( 011

1=++++

tn

n

n

eaaac

must = 0

solutionais

solutionais

solutionais

2

1

2

1

t

n

t

t

n

ec

ec

ec

M

Then, choose c1, c2,,cn to satisfy the given ICs

t

n

tt

zinecececty

+++= ...)(:SolutionI-Z 21 21

tn

n

tn

edt

ed

=

Characteristic polynomial

Q() has at most n unique roots

(can be complex)

))...()(()(21 n

Q =

Soany linear combination

is also a solution to ()


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{ }nitie 1= Set of characteristic modes

Real Root: tii

iej

+= 0

real real t

0>i

0=i0


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To get only real-valued solutions requires the system coefficients to be

real-valued.

Complex roots of C.E. will appear in conjugate pairs:

j

j

k

i

=

+=Conjugate pair

tjt

k

tjt

i

t

k

t

i eeceecececki +=+

For some real Cj

eC

2j

eC

2

0)cos( >+ ttCet

Use Euler!


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Repeated Roots

Say there are rrepeated roots

))...()(()()( 321 pr

Q = p = n - r

ZI Solution:

:modesother...)( 111211 ++++= tr

rziietctccty

effect ofr-repeated roots

See examples on the next several pages

We can verify that: trttt etettee i 111 12 ,...,,, satisfy ()


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Find the zero-input response (i.e., homogeneous solution) for

these three Differential Equations.

Example (a)

)()(2)(3)(

)()(2

)(3

)(

2

2

2

tDftytDytyD

dt

tdfty

dt

tdy

dt

tyd

=++

=++

The zero-input form is:

0)(2)(3)(

0)(2)(

3)(

2

2

2

=++

=++

tytDytyD

tydt

tdy

dt

tyd

The Characteristic Equation is:

0)2)(1(0232 =++=++

Differential Equation Examples

w/ I.C.s


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0)2)(1(0232 =++=++

2&1 21

The Characteristic Roots are:

==

The Characteristic Modes are:

tttteeee

221 & ==

The zero-input solution is:

ttzi eCeCty

221)(

+=

The System forces this

form through its Char. Eq.

The ICs determine the

specific values of the Cis


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ttzi eCeCty

221)(

+=

and it must satisfy the ICs so:

0)0(0 210

20

1 =++==

CCeCeCyzi

The derivative of the z-s soln. must also satisfy the ICs so:

522)0(5 21

0

2

0

1=+==

CCeCeCyzi

Two Equations in Two Unknowns leads to:

5&5 21==

CC

The particular zero-input solution is:

321321modesecond

2

modefirst

55)( ttzi eety +=


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Example (b):

)(5)(3)(9)(6)(

)(5)(

3)(9)(

6)(

2

2

2

tftDftytDytyD

tfdt

tdfty

dt

tdy

dt

tyd

+=++

+=++


0)(9)(6)(

0)(9)(

6)(

2

2

2

=++

=++

tytDytyD

tydt

tdy

dt

tyd


0)3(096 22 =+=++

w/ I.C.s


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3&3 21 ==


ttttteteee

33 21 & ==


ttzi teCeCty

32

31)(

+=





0)3(096 22 =+=++

Using the rule to

handle repeated roots


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Following the same procedure (do it for yourself!!) you get


tttzi etteety

3

modesecond

3

modefirst

3 )23(23)( +=+= 321321


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0 1 2 3 4 5 6 7 8 9 100

1

2

3

FirstMode

Plots for Example 2.1 (b)

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

SecondMo

de

0 1 2 3 4 5 6 7 8 9 100

1

2

3

t (sec)Zero-InputResponse

Plots for Example (b)

Same Decay Rates

Effect of t out in front

Because the characteristic roots are real and negative

the modes and the Z-I response all .


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Example (c):

)(2)()(40)(4)(

)(2)(

)(40)(

4)(

2

2

2

tftDftytDytyD

tfdt

tdfty

dt

tdy

dt

tyd

+=++

+=++


0)(40)(4)(

0)(40)(

4)(

2

2

2

=++

=++

tytDytyD

tydt

tdy

dt

tyd


0)62)(62(04042 =+++=++ jj

w/ I.C.s


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62&62 21 jj =+=


tjtttjtteeeeee

6262 21 & + ==


tjttjtzi eeCeeCty

622

621)(

+ +=





0)62)(62(04042 =+++=++ jj

F ll i h d i h i l i f


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Following the same procedure with some manipulation of

complex exponentials into a cosine


)3/6cos(4)( 2 += tety tzi

Set by the ICs

Imag. part of root

controls oscillation

Real part of root

controls Decay

l x

Plots for Example (c)


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

Firs

tMode

Plots for Example 2.1 (c)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

SecondMode

0 0.5 1 1.5 2 2.5 3 3.5 4-4

-2

0

2

4

t (sec)Zero-InputR

esponse

CantEasilyPlot

theModes

BecauseTheyA

reComple

xPlots for Example (c)

Because the characteristic roots are complex have oscillations!

Because real part of root is negative !!!

4e-2t

-4e-2t


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Big Picture

The structure of the D.E. determines

the char. roots, which determine the

character of the response:

Decaying vs. Exploding (controlled by real part of root) Oscillating or Not (controlled by imag part of root)

The D.E. structure is determined by the

physical systems structure and

component values.

differential equations review

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