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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 τI 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 mg12ρv2 A (ω φ)cD v 1 , associated with the fall, which varies with both Iω 12ρv2 A (ω φ)βcD 12ρrω 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 LWωΔ 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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Homogeneous Differential Equation (Maths)

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