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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
    loading, stress, strain, and deflection are all related this
    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
                    •   Wing blade has negligible
                        thickness (100 μm)

                    •   Angular velocity is relatively
                        low (ωT ≈ 20 rad/s)
                        [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                    
                                                   


                        •   This correction adds
                            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
                                         – r = rotor radius
                                         – 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
          Velocity (rad/s)




                                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
                                                                                                             •   ωt = 27.56 rad/s
                                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
                 Velocity (rad/s)




                                                                                                                  – 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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Description: Homogeneous Differential Equation (Maths)