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Module 1Module 1Introduction to Ordinary Introduction to Ordinary Differential EquationsDifferential Equations
Mr Peter Bier
Slide number 2
Ordinary Differential Ordinary Differential EquationsEquations
Where do ODEs arise? Notation and Definitions Solution methods for 1st order
ODEs
Slide number 3
Where do ODE’s ariseWhere 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 4
Example – Newton’s Law of Example – Newton’s Law of CoolingCooling
This is a model of how the temperature of an object changes as it loses heat to the surrounding atmosphere:
Temperature of the object: ObjT Room Temperature: RoomT
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”
)( RoomObjObj TT
dt
dT
Form ODE
tRoominitRoomObj eTTTT )(
SolveODE
Where is the initial temperature of the object.initT
Slide number 5
Example – Swinging of a Example – Swinging of a pendulumpendulum
mg
l
Newton’s 2nd law for a rotating object:
sin
2
22 mgl
dt
dml
0sin22
2
dt
d
This equation is very difficult to solve.
rearrange and divide through by ml 2
l
g2
where
Moment of inertia x angular acceleration = Net external torque
Slide number 6
Notation and DefinitionsNotation and Definitions
Order Linearity Homogeneity Initial Value/Boundary value
problems
Slide number 7
OrderOrder
The order of a differential equation is just the order of highest derivative used.
02
2
dt
dy
dt
yd. 2nd order
3
3
dt
xdx
dt
dx 3rd order
Slide number 8
LinearityLinearity
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 9
Linearity - ExamplesLinearity - Examples
0 ydt
dyis linear
02 xdt
dxis non-linear
02 tdt
dyis linear
02 tdt
dyy is non-linear
Slide number 10
Linearity – SummaryLinearity – Summary
y2
dt
dydt
dyy
dt
dyt
2
dt
dy
yt)sin32( yy )32( 2
Linear Non-linear
2y )sin( yor
Slide number 11
Linearity – Special PropertyLinearity – Special Property
If a linear homogeneous ODE has solutions:
)(tfy )(tgy and
then:
)()( tgbtfay where a and b are constants,
is also a solution.
Slide number 12
Linearity – Special PropertyLinearity – Special Property
Example:
02
2
ydt
yd
0sinsinsin)(sin
2
2
tttdt
td
0coscoscos)(cos
2
2
tttdt
td
0cossincossin
cossin)cos(sin
2
2
tttt
ttdt
ttd
ty sin ty coshas solutions and
Check
tty cossin Therefore is also a solution:
Check
Slide number 13
Life is mostly linear!!!Life is mostly linear!!!
Most ODEs that arise in engineering are linear with constant coefficients.
In many cases they are approximate versions of more complex nonlinear models but they are sufficiently accurate for most purposes. Often they work OK for small amplitude disturbances but for large amplitude behaviour nonlinearities start to have some effect.
For linear systems the qualitative behaviour is independent of amplitude.
The coefficients in the ODE correspond to system parameters and are usually constant.
Sometimes nonlinearities are important and there have been some important failures because nonlinearities were not understood e.g. the collapse of the Tacoma Narrows bridge.
Slide number 14
Approximately Linear – Approximately Linear – Swinging pendulum exampleSwinging pendulum example
The accurate non-linear equation for a swinging pendulum is:
0sin22
2
dt
d
But for small angles of swing this can be approximated by the linear ODE:
022
2
dt
d
Slide number 15
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)
Slide number 16
Initial Value/Boundary value Initial Value/Boundary value problemsproblems
Problems that involve time are represented by an ODE together with initial values.
Problems that involve space (just one dimension) are also governed by an ODE but what is happening at the ends of the region of interest has to be specified as well by boundary conditions.
Slide number 17
ExampleExample
1st order Linear Nonhomogeneous Initial value problem
gdt
dv
0)0( vv
wdx
Md
2
2
0)(
0)0(
and
lM
M
2nd order Linear Nonhomogeneous Boundary value
problem
Slide number 18
ExampleExample
2nd order Nonlinear Homogeneous Initial value problem
2nd order Linear Homogeneous Initial value problem
0sin22
2
dt
d
0)0( ,0 0 dt
dθ)θ(
022
2
dt
d
0)0( ,0 0 dt
dθ)θ(
Slide number 19
Solution Methods - Direct Solution Methods - Direct IntegrationIntegration This method works for equations
where the RHS does not depend on the unknown:
The general form is:
)(tfdt
dy
)(2
2
tfdt
yd
)(tfdt
ydn
n
Slide number 20
Direct IntegrationDirect 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 21
Direct Integration – ExampleDirect Integration – Example
Find the velocity of a car that is accelerating from rest at 3 ms-2:
3adt
dv
ctv 3
tv
ccv
3
00300)0(
If the car was initially at rest we have the condition:
Slide number 22
Bending of a beam - ExampleBending of a beam - Example
Beam theory gives the governing equation:
wdx
Md
2
2
with boundary conditions:
0)( and 0)0( lMM (pinned ends)
A beam under uniform load
Slide number 23
Bending of a beam - SolutionBending of a beam - Solution
wdx
Md
2
2
Step 1: Integrate
BAxwxM 2
2
1
Awxdx
dM
Step 2: Integrate again to obtain the general solution:
Slide number 24
Bending of a beam - SolutionBending of a beam - Solution
Step 3: Use the boundary conditions to obtain the particular solution.
BAxwxM 2
2
1
Step 4: Substitute back the values for A and B
0002
10 2 BBAw
BlAlw 2
2
10
)(2
1xlwxM wlxwxM
2
1
2
1 2
0)0( M
0)( lMwlA
2
1
Slide number 25
Solution Methods - SeparationSolution 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:
)()( yhtgdt
dy
Slide number 26
Separation – General IdeaSeparation – General Idea
First Separate:
dttgyh
dy)(
)(
Then integrate LHS with respect to y, RHS with respect to t.
Cdttgyh
dy )(
)(
Slide number 27
Separation - ExampleSeparation - Example
)sin(tydt
dy
Separate:
dttdyy
)sin(1
Now integrate:
)cos(
)cos(
)cos()ln(
)sin(1
t
ct
Aey
ey
cty
dttdyy
Slide number 28
Cooling of a cup of coffeeCooling of a cup of coffee
Amount of heat in a cup of coffee:
cTVQ heat volume specific heat
density temperature
Heat balance equation (words): Rate of change of heat = heat lost to surrounding air
Slide number 29
Cooling of a cup of coffeeCooling of a cup of coffee
Heat lost to the surrounding air is proportional to temperature difference between the object and the air
The proportionality constant involves the surface area multiplied by a heat tranfer coefficient
Newton’s law of cooling:
Heat balance equation (maths) :
)( RoomTThAdt
dQ
Slide number 30
Cooling of a cup of coffeeCooling of a cup of coffee
)( RoomTThAdt
dQ
)( RoomTThAdt
dTcV
)( RoomTTdt
dT
cV
hA
whererearrange
cTVQ Substitute in
Now we solve the equation together with the initial condition:
InitialTT )0(
Slide number 31
Cooling of a cup of coffee - Cooling of a cup of coffee - SolutionSolution
)( RoomTTdt
dT
Step 1: Separate
dtTT
dT
Room
)(
ctTT Room )ln( Step 2: Integrate
ctRoom eTT
tRoom AeTT ceA
where
Make explicit in unknown T
Slide number 32
Cooling of a cup of coffee - Cooling of a cup of coffee - SolutionSolution
Step 3: Use Initial
Condition
Step 4: Substitute
back to obtain final answer
tRoom AeTT
InitialTT )0(
0 AeTT RoomInitial
)( RoomInitial TTA
tRoomInitialRoom eTTTT )(
Slide number 33
Solution Methods - Integrating Solution Methods - Integrating FactorFactor
The integrating factor method is used for nonhomogeneous linear 1st order equations
The basic ideas are: Collect all the terms involving y and
on the left hand side of the equation. dt
dy
Combine them together as the derivative of a single function of y and t.
Solve by direct integration.
The cunning trick is that step 2 cannot usually be done unless you first multiply the whole equation by an integrating factor.
Slide number 34
Integrating Factor – ExampleIntegrating Factor – Example
1 ydt
dy 2)0( y
There are several ways to solve this problem but we will use it to demonstrate the integrating factor method. To understand the integrating factor method you must be very familiar with the formula for the derivative of a product:
Slide number 35
Integrating Factor – ExampleIntegrating Factor – Example
Product Rule:
ydt
df
dt
dyf
dt
yfd
).(
The basic idea is that if we multiply the ODE by the correct function (an integrating factor) we can make the LHS of the ODE look like the RHS of the product rule.
Slide number 36
Integrating Factor – ExampleIntegrating Factor – Example
We will look ahead, use the answer and show how it is derived later. For our ODE the integrating factor is te
Thus the ODE becomes:
ttt eyedt
dye
Now the LHS of this equation looks like the RHS of the Product Rule with:
tef
Slide number 37
Integrating Factor – ExampleIntegrating Factor – Example
We can rewrite the equation as:
tt
t eydt
ed
dt
dye
)(
or t
t
edt
yed
)(
Now we can use direct integration
Ceye tt
Do not forget C, the constant of integration!
Slide number 38
Integrating Factor – ExampleIntegrating Factor – Example
Rearrange to make this explicit in y
tCey 1
Now use the initial condition to calculate C
1
21
212)0( 0
C
C
Cey
Substitute back to obtain the final solution
tey 1
Slide number 39
How do we calculate the How do we calculate the integrating factor?integrating factor? Let us now pretend we do not know what
the integrating factor should be Call it Φ and use it to multiply the ODE
from the previous example
ydt
dy
To make the LHS of this equation look like the RHS of the Product Rule we must choose
dt
d
Slide number 40
How do we calculate the How do we calculate the integrating factor?integrating factor? Then the ODE becomes
ydt
d
dt
dy
Now using the product rule in reverse the LHS can be written as a single term (a very clever trick)
dt
yd )(
Slide number 41
How do we calculate the How do we calculate the integrating factor?integrating factor? Now we can integrate once we know Φ
We can separate to find Φ
t
ct
Ae
ect
dtd
dt
d
ln
The convention is to put A = 1. It appears in every term of the ODE, and therefore can be divided out. This gives the integrating factor:
te
Slide number 42
Finding the integrating Finding the integrating factor in generalfactor in general
Given the general form of a nonhomogeneous 1st order equation:
)()( tfytgdt
dy
0)0( yy
How do we use the integrating factor method to find a y?
Slide number 43
Finding the integrating Finding the integrating factor in generalfactor in general Step 1: Multiply by Φ:
)()( tfytgdt
dy
Step 2: Compare with the RHS of the Product Rule and set up equation for Φ:
)(tgdt
d
Step 3: Use separation to solve for Φ:
dttge
dttg
dttgd
)(
)(ln
)(
Slide number 44
Finding the integrating Finding the integrating factor in generalfactor in general Step 4: Combine terms on the LHS:
)(] [
tfdt
yd
Step 5: Integrate:
Cfdty Step 6: Divide by to make explicit in y :
C
fdty 1
Step 7: Use the initial conditions to evaluate C :
Slide number 45
Finding the integrating Finding the integrating factor in generalfactor in general
Notes: After you have been through the process a
few times then skip some of the steps. For example you can remember the formula for the integrating factor, you do not have to re-derive it every time.
dttg
e)(
In principle this process can be used to
solve any linear nonhomogeneous 1st order ODE but some of the details may be tricky or impossible. Both the integrals
and may be impossible to evaluate.
dttg )( dttf )(
Slide number 46
Solving an example using the Solving an example using the integrating factor methodintegrating factor method
ttydt
dy 0)0( y
)()( tfytgdt
dy
Step 1: Put the ODE into the general form:
The ODE is already in that form!
Step 2: Find the integrating factor:
221
)(t
tdt
e
ettg
Slide number 47
Solving an example using the Solving an example using the integrating factor methodintegrating factor method
Step 5: Integrate and make explicit in y:
Step 3: Multiply by the integrating factor:2
212
212
21 ttt
tetyedt
dye
Step 4: Use the reverse Product Rule:
221
221
][ tt
tedt
yed
221
2212
212
21
1t
ttt
Cey
CeCdtteye
Slide number 48
Solving an example using the Solving an example using the integrating factor methodintegrating factor method
Step 7: Substitute back into the original equation:
Step 6: Use the initial conditions to find the exact solution:
1
01
010)0( 0
C
C
Cey
221
1t
ey
Slide number 49
Exponential substitutionExponential substitution
The exponential trial or guessing method can be used for solving linear constant coefficient homogeneous differential equations.
tCey
The basic trial for the solution of the ODE is:
Slide number 50
Characteristic EquationsCharacteristic Equations
gives the differentialstCey
tnn
t
t
t
Cey
Cey
Cey
Cey
)(
3)3(
2
An algebraic characteristic equation comes from substituting in for y and its derivatives and cancelling out tCe
Slide number 51
Exponential Trial - ExampleExponential Trial - Example
05 yy
05 tt AeAe
Try tAey
Cancelling out gives the characteristic equation
tAe
5 05
Substituting this back into givestAey
tAey 5
Slide number 52
Solving GuideSolving Guide
General Form Description Solving Method
)(tfdt
ydn
n
)()(),( yhtgytfdt
dy
)()( tfytgdt
dy
Direct Integration
Separation
Integrating Factor Method
1st order only
1st order nonhomogeneous linear equation
1st or higher order RHS does not
depend on the unknown y
Slide number 53
Solving GuideSolving Guide
General Form Description Solving MethodExponential trial. See Module 2
Problems like this can be solved for some types of f.
Covered in Modules 3 and 5
001
1
1
1
yadt
dya
dt
yda
dt
ydn
n
nn
n
)(01
1
1
1
tfyadt
dya
dt
yda
dt
ydn
n
nn
n
2nd order or higher homogeneous linear equation constant
coefficients
2nd order or higher nonhomogeneous linear equation constant
coefficients
Slide number 54
Solving GuideSolving Guide
General Form Description Solving MethodCan only solve a few ‘special’ problems
Not Covered in MM2
Generally can’t be solved analytically
(see Module 4 for numerical methods)
2nd order or higher nonhomogeneous linear equation variable
coefficients
2nd order Function f
contains t, y and y’ terms all mixed up
)()()(
)(
01
1
1
1
tfytgdt
dytg
dt
ydtg
dt
ydn
n
nn
n
dt
dyytf
dt
yd,,
2
2