# Homogeneous Differential Equation (Maths)

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```					Homogeneous Differential Equation

Non-Homogeneous Differential Equation

Presented by : M. Nauman Zubair
Presented to : Sir Atif
Semester     :2nd
Roll #       : 9021
Definitions:
A differential equation is called a homogeneous
differential equation if it can be written in the
form
M(x,y)dx + N(x,y)dy = 0
where M and N are of the same degree.

First order ‘ordinary differential equation’ is
homogeneous if it has the following form.
dy/dx =F(y/x)
It is simply an equation where both coefficients of the
differentials dx and dy are homogeneous.
A linear homogeneous differential equation having form:
L(y) = 0
Where L is differential oprator.
A linear nth order differential equation of the form
an(x) d^n(y)/d^n(x)+an-1(x)d^n-1y/dx^n-1+…+a0(x)y=0
is said to be homogenous differential equation.
Standard Form:
A linear homogeneous ordinary differential equation
with constant coefficients has the general form of

where             are all constants.

Homogeneous function:
In mathematics, a homogeneous function is a
function with multiplicative scaling behavior: if the
argument is multiplied by a factor, then the result is
multiplied by some power of this factor.
Formal definition:
 Homogeneous functions are functions where the sum of the
powers of every term are the same. So the first function below is
homogeneous of degree 3, the second and third are not
homogeneous.

 This method of recognizing a homogeneous function does not
work all of the time, but is useful for many examples. A more
formal definition states that if a function is homogeneous it can
be written in the following form:

 OR:
Second order:
The Euler-Cauchy equation crops up in a number of
engineering applications. It is given by the equation:

We assume a trial solution given by

Differentiating, we have:

and

Substituting into the original equation, we have:

Or rearranging gives:
We then can solve for m. There are three particular
cases of interest:
Case #1: Two distinct roots, m1 and m2
Case #2: One real repeated root, m
Case #3: Complex roots, α ± iβ
In case #1, the solution is given by:

In case #2, the solution is given by:

To get to this solution, the method of reduction of
order must be applied after having found one
solution y = xm.
In case #3, the solution is given by:

This equation also can be solved with x = et
transformation.
Example:
Given

we substitute the simple solution xα:

For this to indeed be a solution, either x=0 giving the
trivial solution, or the coefficient of xα is zero, so
solving that quadratic, we get α=1,3. So, the general
solution is
1st Order Homogeneous DE :
A first order homogeneous differential equation involves only
the first derivative of a function and the function itself, with
constants only as multipliers. The equation is of the form

and can be solved by the substitution

The solution which fits a specific physical situation is
obtained by substituting the solution into the equation and
evaluating the various constants by forcing the solution to
fit the physical boundary conditions of the problem at
hand. Substituting gives
2nd Order Homogeneous DE :
A linear second order homogeneous differential
equation involves terms up to the second derivative
of a function. For the case of
constant multipliers, The equation is of the form

and can be solved by the substitution

The solution which fits a specific physical situation is
obtained by substituting the solution into the
equation and evaluating the various constants by
forcing the solution to fit the physical boundary
conditions of the problem at hand. Substituting
gives

which leads to a variety of solutions, depending on the values
of a and b.
Applications :
An analog computer study of Vander Pol’s equation and
its resonse to write Gaussian Noice.
In 1st Order Homogeneous Differential
Equations
The general form of the solution of the homogeneous
differential equation can be applied to a large number
of physical problems
Barometric pressure variation with altitude:
ii)-Discharge of a capacitor:

Differential Equation Applications:
i)-Simple harmonic motion:

ii)-Simple pendulum:

iii)-Azimuthal equation, hydrogen atom:

iv)-Velocity profile in fluid flow.
Charging a Capacitor :
An application of non-homogeneous differential equations
A first order non-homogeneous differential equation

has a solution of the form :

For the process of charging a capacitor from zero charge
with a battery, the equation is
Using the boundary condition Q=0 at t=0 and identifying
the terms corresponding to the general solution, the
solutions for the charge on the capacitor and the
current are:

In this example the constant B in the general solution had
the value zero, but if the charge on the capacitor had
not been initially zero, the general solution would still
give an accurate description of the change of charge
with time. The discharge of the capacitor is an
example of application of the homogeneous
differential equation
Capacitor Discharge:
An application of homogeneous differential equations
A first order homogeneous differential equation

has a solution of the form :

For the process of discharging a capacitor which is
initially charged to the voltage of a battery, the
equation is
Using the boundary condition and identifying the terms
corresponding to the general solution, the solutions for
the charge on the capacitor and the current are:

Since the voltage on the capacitor during the discharge is
strictly determined by the charge on the capacitor, it
follows the same pattern.

In the civil engineering:
On the structural side of civil engineering, beam theory is
based on a 4th order differential equation.
Solutions/approximations of the solutions of this are the
basis behind all structural engineering topics. Beam
way.
Physical Problem for Civil Engineering:
A physical problem of finding how much time it would take a lake to
have safe levels of pollutant. To find the time, the problem is
modeled as an ordinary differential equation.

Mass of pollutant = Mass of pollutant entering – Mass of pollutant
leaving
which also gives
Rate of change of mass of pollutant = Rate of change of mass of pollutant
entering – Rate of change of mass of pollutant leaving.

In fluids :
Anything that involves rates of flow can be modeled using diff EW,
such as water treatment plants, and many environmental
applications (typically grouped with civil topics).

Physical Problem for Ordinary Differential Equations:
A physical problem of finding how much time it would take a
trunnion to cool down in a refrigerated chamber. To find the
time, the problem would be modeled as a ordinary differential
equation.
Helicopter design can greatly change the flight
patterns of the helicopter

•   Wing length and width have
large effects on rotational and
linear velocities

•   Larger helicopters suffer from
deformity due to strength of
paper

•   Current model provides for the
longest drop time
– Same model used for all drops
[1]

17
At equilibrium, the drag force deflects the
rotors upward

•   Rotor deflection angle (wing
flex), φ, decreases as linear
velocity increases

•   At terminal linear velocity, φ ≈
70˚

φ

18
Turbulence due to the wake of rotors was
ignored
thickness (100 μm)

•   Angular velocity is relatively
[4]

19
Due to rotation of the rotors, the drag force
acts on a larger effective area
•   As the helicopter falls, it
spins quickly, and the drag
force acts on an area
proportional to the circle
traced by the wings as they
rotate

•   Effective rotor area is a
linear combination of the
stationary wing area and
the disc traced as the
wings rotate

20
The rotational ‘blur’ effectively adds to the
area the drag force acts on

A  t2 2 
  LW W
ω, 2 Δ
φ    
ω si
L  
φ                    


approximately 25% to the
area of the rotors for small
Δt      [1]

21
Determination of angular velocity depended
largely on the capabilities of the sensor

•    Our first experiment was designed to determine whether
our data collection method was possible

•    Linear position was held constant, and rotor rotation was
controlled

– By obtaining a set of static rotational speeds, we could
compare the results with the results of the actual drops

22
While the results were encouraging, it proved
difficult to apply this technique to our drops

0.09 m/s

0.65
Distance from sensor (m)

0.6

0.55

0.5

0.45

0.4
0   0.5          1       1.5   2
Time (s)

0.24 m/s

0.7
Distance from sensor (m)

0.65
0.6
0.55
0.5
0.45
0.4
0   0.5          1       1.5   2
Time (s)
23
To reduce crosswinds and eliminate reflective
noise, a large ‘box’ was constructed
•   At just over 3 meters tall,
and 1 meter wide to
account for the 15˚ arc
swept out by the sensor

•   Only interested in the
initial drop (1-2 seconds)

24
While the data gave good linear data, the
angular resolution was poor
Linear Position vs Time
•   Data sensor operates at
~20 Hz, too low for
1.2

credible angular rotation
1

0.8
Distance (m)

0.6                                                                                 data analysis
0.4

•   If helicopter path is not
0.2

0
0       0.2          0.4            0.6         0.8         1
directly lined up with sensor
during the entire fall, we
Time (s)

Linear Position vs Time
lose angular data
1.2
1                                                                   •   Chose not to use this
technique for angular
Distance (m)

0.8

rotation
0.6
0.4
0.2

•   vt = 0.60 m/s ± 0.02 m/s
0
0     0.2          0.4                0.6         0.8       1
Time (s)

25
An understanding of the forces involved allow
us to create differential equations

v  Dra
(ω
d, φ   Fo
F ,)Linea


(φ
ω  Drag
Force
F,)Angular
d

F  m g
26
The resistance experienced by an object
moving through a fluid is dependent on its
velocity
• Generally                     • Incorrect assumption
– b is constant, and depends
on viscosity of the fluid,
for bodies falling in
and the size and shape of    Earth’s atmosphere
the object
– α depends on shape of
object and size of the
F  bv
D
α
object
FD  v               • Experimentally, the
•   For small objects,             drag force has been
– Raindrop falling in air     approximated by
– Aerosols falling in air
[5]
– Objects passing through
heavy oil                       FD  v   2

27
Applying this linearly, we obtain our linear drag
force
•   Dependent on shape, b is
not constant for our
purposes
– Incorporates effective
F  2
D  bv
rotor area A(ω, φ)‫‏‬
– Also dependent on density
of air [7]
v 1ω D
F  ρ A ,φc
1 2
•   We also introduce the           D
2
coefficient of drag, cD
– Surface/Shape of object
– Viscosity of air (fluid)‫‏‬

  A c
m  ρ 1 ,)
1 v ω
vmg 2 ( φ
   2     D

28
We find the rotational drag force is dependent
only on the angular velocity

•   Area of the helicopter does
not change with increased
linear or angular velocities,
only wing deflection angle      F  2
D  bv
(assumed constant)
•   Recall v = rω
 ω2 2 φc
F  ρr  A  D
1
•   Torque                           D
2
τI  
ω r F

ω 2ω  r 2γ
 1 φ  ( D
 ()  2 φ
I  v ,β1 A
1 A c 2 ω c
ρ   D ρ
2         )

29
Our two coupled differential equations provide
insight into the flight of the helicopter
•   Linear equation incorporates
the summation of the falling
body and the drag              m  mg12ρv2 A (ω φ)cD
v               1  ,
associated with the fall,
which varies with both         Iω 12ρv2 A (ω φ)βcD 12ρrω A (φ)γcD
2
              ,
linear and angular                          1                     2

velocities
[1]

•   Angular equation
experiences a forcing from
the liner motion, and a
separate drag force which
varies only with the angular    1  ,     
A ω φ  2 LWωΔ  L W2 sin( )
t 2       
φ
velocity

•   β and γ are scaling factors    A φ  A  A cos()
2       body rotor φ
with units of length

30
With some careful rearrangement we arrive at
our differential equations
•   Linear equation
– v = linear velocity
K        K                       – ω = angular velocity
  g  v  v2  vω  v2  ω
v                                        – g = acceleration due to gravity
m        m                           (9.81m/s2)
 Kv 2 Kvω 2  γ  Kω 2– m = mass of helicopter
ω β   v 
                      v  ω       ω – Kv & Kvω = Linear drag force
 I         I            I           coefficients
•    Angular equation
– β = linear drag scaling factor
– I = moment of inertia
    
Kv  ρ LW cD sin( )φ                     – γ = angular drag scaling factor
– Kω = angular drag force coefficients
 
Kvω  ρ Δtr cD sin( )
2
φ             •    K – equations
– ρ = density of dry air
Kω  12 ρr2 A cD
2                          – L = rotor length
– W = rotor width
– cD = coefficient of drag
– Δt = change in time (~0.10 s)

31
Graphical solutions to our equations reflect the
physical results
1.2
Li near Veloci ty
•   When dropped, the helicopter
accelerates uniformly as there is
0.96
no rotational motion
Velocity (m/s)

0.72

•   When the rotation begins, the
linear velocity decreases, and
0.48

0.24                                                              the two settle into an equilibrium
– the terminal velocities
0
0       0.4     0.8          1.2    1.6    2
Ti me (s)

25
Ang ular Vel ocity
•   vt = 0.62 m/s
– vt experimental = 0.60 m/s ±
0.02 m/s
20

15

10                                                         •   ωt = 21.49 rad/s
5

0                                                      •   Numerical Solutions found using ODE Architect,
0    0.4     0.8          1.2    1.6       2       implementing a fourth order Runge-Kutta Method
Ti me (s)                                                                         32
Analytical solutions to our system of equations
provides insight into the terminal velocities
•   Obtained by setting
K    K
v g v v v  2ω
   2 ωv

m

m
                  v  ω 0
 
• In turn, we can solve for
           γ
  ωω β and γ in terms of the
K   K         K 2
 β 2
     ωv 
ω  v v v  2ω     
I    I      I     angular velocities, which
we have determined
experimentally
2
β ωω
tK
mg                    .
4915
vt =                        γ mg
K v + K vωωt

βmg
ω
t
γ ω
K
33
Our experimental determination of β and γ
provide our final numerical solutions

1.2
Li near Veloci ty
•   vt = 0.60 m/s
0.96
– vt (old) = 0.62 m/s
Velocity (m/s)

0.72
0.48                                                                              – ωt (old) = 21.49 rad/s

0.24
•   As expected, the terminal
0                                                                           linear velocity is predicted
exactly
0        0.4          0.8          1.2         1.6         2
Ti me (s)
Ang ular Vel ocity
•   Terminal angular velocity is
30

24                                                                     higher than the previous
prediction

– No reliable data
18

12

6

0
0         0.4          0.8          1.2         1.6       2
34
Ti me (s)
Futures projects would be expected to obtain
better experimental data

•   Other considerations include:
– Accounting for the processional motion (wobble) about the z-
axis as the helicopter falls

– Incorporate a variable wing flex angle φ, which will provide a
more accurate solution

– Eliminate physical distortion by using a rigid model

– Construct a better enclosure (Circular?)

– Obtain better values for coefficient of drag and the scaling
factors (β, γ)

35
Conclusion/Summary

Application of differential equations to model
the motion of a paper helicopter

•   Analytical solutions made significant improvements over
previous models

•   Data collection was only successful for linear motion, but
further measures must be taken for determining the
angular motion
– Sonic motion sensor may not be a possible data collection
source

•   Several major assumptions were made for the sake of
simplification, and would help us understand the motion
much better

•   Future projects may wish to test different helicopter
dimensions, to ensure the model is accurate
36

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