Introduction to Ordinary Differential Equations

Introduction to Ordinary Introduction to Ordinary

Differential Equations Differential Equations

Asst Lect Sarmad K. Ibrahim

Ordinary Differential Ordinary Differential

Equations Equations



Where do ODEs arise?



Notation and Definitions



Solution methods for 1

^st

order

ODEs

Where do ODE’s arise Where do ODE’s arise



All branches of Engineering



Economics



Biology and Medicine



Chemistry, Physics etc

Anytime you wish to find out how

something changes with time (and

sometimes space)

Example – Newton’s Law of Example – Newton’s Law of

Cooling Cooling

 This is a model of how the

temperature of an object changes as it loses heat to the surrounding

atmosphere:

Temperature of the object: T_Obj Room Temperature: T_Room

Newton’s laws states: “The rate of change in the temperature of an

object is proportional to the difference in temperature between the object and the room temperature”

) ( _{Obj Room}

Obj T T

dt

dT   

Form ODE

Notation and Definitions Notation and Definitions



Order



Linearity



Homogeneity



Initial Value/Boundary value

problems

Order Order



The order of a differential equation is just the order of highest derivative used.

2 0

2  

dt dy dt

y

. d 2^nd order

3 3

dt x x d

dt

dx  3^rd order

Linearity Linearity

 The important issue is how the

unknown y appears in the equation.

A linear equation involves the dependent variable (y) and its

derivatives by themselves. There must be no "unusual" nonlinear functions of y or its derivatives.

 A linear equation must have constant coefficients, or coefficients which

depend on the independent variable (t). If y or its derivatives appear in the coefficient the equation is non-linear.

Linearity - Examples Linearity - Examples

 0

 y dt

dy is linear

2  0

 x dt

dx is non-linear

2  0

t dt

dy is linear

2  0

 t dt

y dy is non-linear

Linearity – Summary Linearity – Summary

y 2 dt dy

dt y dy

dt t dy

2



 



 dt dy y

t) sin 3 2

(  (2  3y²)y

Linear Non-linear

y2 or sin(y)

Linearity – Special Property Linearity – Special Property

If a linear homogeneous ODE has solutions:

) (t f

y  and y  g(t)

then:

) ( )

(t b g t f

a

y    

where a and b are constants,

is also a solution.

Linearity – Special Property Linearity – Special Property

Example:

2 0

2  y 

dt y d

0 sin

sin ) sin

(sin

2

2  t   t  t 

dt t d

0 cos

cos ) cos

(cos

2

2  t   t  t 

dt t d

cos ) sin

cos

2(sin



 

t t t

t d

t

y  sin y  cost

has solutions and

Check

t t

y  sin  cos

Therefore is also a solution:

Check

Homogeniety Homogeniety

 Put all the terms of the equation

which involve the dependent variable on the LHS.

 Homogeneous: If there is nothing left on the RHS the equation is

homogeneous (unforced or free)

 Nonhomogeneous: If there are

terms involving t (or constants) - but not y - left on the RHS the equation is nonhomogeneous (forced)

Example Example

 1st order

 Linear

 Nonhomogeneous

 Initial value problem

dt g

dv 

) 0

0

( v

v 

dx w M d

² ₂



0 )

(

0 )

0 (

and



 l M M

 2nd order

 Linear

 Nonhomogeneous

 Boundary value problem

Example Example

 2nd order

 Nonlinear

 Homogeneous

 Initial value problem

 2nd order

 Linear

 Homogeneous

Initial value problem

0

2 sin

2

2    

dt d

0 ) 0 (

,

0  ₀ 

dt θ d

)

θ( 

2 0

2

2     dt

d

d

Solution Methods - Direct Solution Methods - Direct

Integration Integration

 This method works for equations

where the RHS does not depend on the unknown:

 The general form is:

) (t dt f

dy 

)

2 (

2

t dt f

y

d 



) (t y f

d ⁿ



Direct Integration Direct Integration

 y is called the unknown or dependent variable;

 t is called the independent variable;

 “solving” means finding a formula for y as a function of t;

 Mostly we use t for time as the

independent variable but in some cases we use x for distance.

Direct Integration – Example Direct Integration – Example

Find the velocity of a car that is accelerating from rest at 3 ms

^-2

:

 3

 a dt

dv

c t

v  

 3

c c

v(0)  0  0  3 0    0

If the car was initially at rest we

have the condition:

Solution Methods - Separation Solution Methods - Separation

The separation method applies only to 1^st order ODEs. It can be used if the

RHS can be factored into a function of t multiplied by a function of y:

) ( )

(t h y dt g

dy 

Separation – General Idea Separation – General Idea

First Separate:

dt t y g

h

dy ( )

)

( 

Then integrate LHS with

respect to y, RHS with respect to t.

C dt

t dy g









^{( )}

Separation - Example Separation - Example

) sin(t dt y

dy 

Separate:

dt t y dy sin( )

1 

Now integrate:

) cos(

) ln(

) 1 sin(

c t

y

dt t y dy













