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Introduction

INTRODUCTION TO DIFFERENTIAL EQUATIONS

A differential equation is an equation involving some hypothetical function and its derivatives.

Example

xyy =

′

+

′

′

2 is an differential equation. As such, the differential equation is a description of some function

(exists or not).

A solution to a differential equation is a function that satisfies the differential equation.

Example

xxy 5

3+= is a solution to

3

xyxy −+=

′′ .

Some differential equations are famous/important:

• yy =

′

,

x

ey =.

• ayy =

′

,

ax

ey =.

• 0=+

′

′

yy ,

x

y

cos

=

.

• 0=+

′

′

ayy , xaycos=.

Recall that a differential equation describes a phenomenon in terms of changes. For example, if mvP

=

, then

vdmFdt⋅= or

F

dt

dP=

.

Example

A pool contains V liters of water which contains M kg of salt. Pure water enters the pool at a constant rate of v

liters per minute, and after mixing, exits at the same rate. Write a differential equation that describes the

density of salt in the pool at an arbitrary time t.

• Let

(

)

t

ρ

be the density at time t. Then

( )

(

)

V

tM

t=

ρ

.

• To model change in

(

)

t

ρ

, let

(

)

11 t

ρρ

= and

(

)

22 t

ρρ

=. Then

(

)

(

)

12112 ttvV−−≈−

ρρρ

, so

V

v

V

tt 1

12

12

ρ

ρρ

−≈

−

−

or

(

)

(

)

( )

V

v

tV

ttt

ttt

ρ

ρρ

−≈

−∆+

−∆+

.

• Now, as 0

→

t,

( )

V

v

t

dt

d

ρ

ρ

−=

or V

v

ρρ

−=

′

.

• Without solving this equation, we can predict facts about this system. As

∞

→

t

, 0

→

ρ

.

• To solve this differential equation, write

dt

V

v

d−=

ρ

ρ

. Integrating both sides, we get

( ) ( )

t

V

v

t

V

v

C

Ct

V

v

AeeeetCt

V

v

t

−−+−

===+−=

ρρ

ln

.

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• If we add

(

)

ρρ

=0 to V

v

ρρ

−=

′

, then we have an IVP (initial value problem).

ISSUES ABOUT THE USE OF DIFFERENTIAL EQUATIONS

1) How to translate a real problem to a differential equation. Keep your eyes open!

2) Some patterns of nature are ill-defined. Use different points of view and different mental differential equation

models to reformulate them.

3) Differential equations have infinitely many solutions. Which one is yours? The initial value are extremely

important.

4) There may be no analytic solution found.

• Is there a solution?

• Is this solution unique?

• If a numeric answer is required, i.e. the value of the solution at one particular point, then use numerical

approximations. It does not give any feelings for the pattern, nor does it give elbow room.

• Use theoretical analysis if you need the behavior of the solution. This does not give any values.

• To know the behavior locally/in a neighborhood, solve in series.

5) The data does not fit you solution. You need to repeat (as in a feedback/controlled system).

NOTATIONS WITH REGARD TO THE INPUT/OUTPUT SYSTEMS

Example

xyxyyx tansin2 =⋅−

′

+

′

′

can be written as

[

]

xyLtan=. Solve it, and the answer is the output.

•

[

]

yL is the “black box system”.

• xtan is the ‘input”.

For theoretical purposes, mathematicians use these equivalents: x

y

y

x

x

x

x

y

′

−+=

′′′ 2

sintan is the same as

(

)

yyxfy

′

=

′

′

′

,, or

(

)

0,,,, =

′

′

′

′

′

′

yyyyxF.

LINEAR VS. NON-LINEAR DIFFERENTIAL EQUATIONS

• x

yey

x

yx x

tan1

11

tan+

=+

′

+

′′

⋅ is linear.

• 0

2=+

′

+

′′′ yyy is non-linear.

First Order Differential Equations

LINEAR EQUATIONS

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First order linear equations have the form

(

)

(

)

tgytpy=+

′

.

Derivation

• Suppose I can find

(

)

t

µ

so that

(

)

(

)

(

)

ttpt

µµ

′

=.

• Multiply both sides of

(

)

(

)

tgytpy=+

′

by

(

)

t

µ

:

(

)

(

)

(

)

(

)

(

)

tgtytptyt

µµµ

=+

′

which is

(

)

(

)

(

)

(

)

tgtytyt

µµµ

=

′

+

′

, i.e.

( )( ) ( ) ( )

tgtyt

µµ

=

′

.

• Integrate both sides:

( ) ( ) ( )

Cdttgtyt+=

µµ

, so

( ) ( ) ( ) ( )

Cdttgt

t

ty+=

µ

µ

1.

• But what is

(

)

t

µ

? Since

( ) ( )

(

)

( ) ( )

tp

t

t

tpt=

′

⇔

′

=

µ

µ

µµ

, therefore

( ) ( )

=dttpt

µ

ln. So

(

)

(

)

=dttp

et

µ

.

General Solution

To solve

(

)

(

)

tgytpy=+

′

,

1) Let

(

)

(

)

=dttp

et

µ

(no constant needed).

2) The solution is

( ) ( ) ( ) ( )

[

]

Cdttgt

t

ty+=

µ

µ

1.

Example

Solve 2

22 t

tetyy −

=+

′.

• Here,

(

)

ttp2=,

( )

2

2t

tetg−

=.

• Let

( )

2

2t

tdt

eet==

µ

.

• The solution is

( )

[

]

[

]

(

)

22

22

22

2

22

1

2

1

2

1tt

tt

tt

tCeetCt

e

Ctdt

e

Cdttee

e

ty−−− +=+=+=+= .

• Note that as

∞

→

t

, 0

→

y.

Importance of Analysis

One needs to have an understand of the solution before (even after) solving it, with respect to:

1) Behavior of the solution as

∞

→

t

.

2) The nature and behavior of the solution within a family (depends on y0).

Variation of Parameter

Recall that the family of solutions of a first order linear differential equation

(

)

(

)

tgytpy=+

′

,

(

)

(

)

=dttp

et

µ

,

( ) ( ) ( ) ( )

[

]

( ) ( ) ( ) ( )

t

C

dttgt

t

Cdttgt

t

ty

µ

µ

µ

µ

µ

+=+= 11 . Notice that the family of solutions is generated by

(

)

t

µ

. This leads to the technique of variation of parameter.

Recall a differential equation

[

]

(

)

xgyL=. If

(

)

0=xg, then we have a zero-input system, or a homogeneous

differential equation

[

]

0=yL which describes the solutions to a great extent.

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