Introduction to Ordinary Differential Equations

Contributed by:

Sharp Tutor Wed, Jan 19, 2022 11:00 PM UTC

TOPICS:
1. Where do ODEs arise?
2. Notation and Definitions
3. Solution methods for 1st order ODEs

1. Introduction to Ordinary
Differential Equations
Asst Lect Sarmad K. Ibrahim

2. Ordinary Differential
Equations
 Where do ODEs arise?
 Notation and Definitions
 Solution methods for 1st order
ODEs
Slide number 2

3. 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)
Slide number 3

4. Example – Newton’s Law of
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: TObj Room Temperature: TRoom
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”
Form
ODE
dTObj
  (TObj  TRoom )
Solve dt
ODE TObj TRoom  (Tinit  TRoom )e  t
Where Tinit is the initial temperature of the object.
Slide number 4

5. Notation and Definitions
 Order
 Linearity
 Homogeneity
 Initial Value/Boundary value
problems
Slide number 5

6. Order
 The order of a differential
equation is just the order of
highest derivative used.
d 2 y dy
2
 0 2nd order
dt dt
dx d 3x
x 3 3rd order
dt dt
Slide number 6

7. 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.
Slide number 7

8. Linearity - Examples
dy
 y 0 is linear
dt
dx
 x 2 0 is non-linear
dt
dy 2
 t 0 is linear
dt
dy 2
y  t 0 is non-linear
dt
Slide number 8

9. Linearity – Summary
Linear Non-linear
2y y2 or sin( y )
dy dy
y
dt dt
(2  3 sin t) y (2  3 y 2 ) y
2
dy  dy 
t  
dt  dt 
Slide number 9

10. Linearity – Special Property
If a linear homogeneous ODE has solutions:
y  f (t ) and y  g (t )
then:
y a  f (t )  b g (t )
where a and b are constants,
is also a solution.
Slide number 10

11. Linearity – Special Property
Example:
d2y
2
 y 0 has solutions y sin t and y cos t
dt
2
Check d (sin t )
2
 sin t  sin t  sin t 0
dt
d 2 (cos t )
2
 cos t  cos t  cos t 0
dt
Therefore y sin t  cos t is also a solution:
Check d 2 (sin t  cos t )
2
 sin t  cos t
dt
 sin t  cos t  sin t  cos t 0
Slide number 11

12. 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)
Slide number 12

13. Example
dv  1st order
g  Linear
dt  Nonhomogeneous
v(0) v0  Initial value problem
2  2nd order
d M
2
w  Linear
dx  Nonhomogeneous
M (0) 0  Boundary value
and problem
M (l ) 0
Slide number 13

14. Example
2
 2nd order
d  2
2
  sin  0  Nonlinear
dt  Homogeneous
d
θ( 0 ) θ0 , (0) 0  Initial value problem
dt
2
 2nd order
d  2
2
   0  Linear
dt  Homogeneous
d
θ( 0 ) θ0 , (0) 0  Initial value problem
dt
Slide number 14

15. Solution Methods - Direct
Integration
 This method works for equations
where the RHS does not depend on
the unknown:
 The general form is:
dy
 f (t )
dt
d2y
2
 f (t )
dt

dny
n
 f (t )
dt
Slide number 15

16. 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.
Slide number 16

17. Direct Integration – Example
Find the velocity of a car that is
accelerating from rest at 3 ms-2:
dv
a 3
dt
 v 3t  c
If the car was initially at rest we
have the condition:
v (0) 0  0 3 0  c  c 0
 v 3t
Slide number 17

18. Solution Methods - Separation
The separation method applies only to
1st order ODEs. It can be used if the
RHS can be factored into a function of t
multiplied by a function of y:
dy
 g (t )h ( y )
dt
Slide number 18

19. Separation – General Idea
First Separate:
dy
 g (t )dt
h( y )
Then integrate LHS with
respect to y, RHS with respect
to t.
dy
h( y ) g (t )dt  C
Slide number 19

20. Separation - Example
dy
 y sin(t )
dt
Separate:
1
dy sin(t )dt
y
Now integrate:
1
y dy sin(t )dt
 ln( y )  cos(t )  c
 y e  cos( t ) c
Slide number 20
 y  Ae  cos(t )

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