Differential Equation - Term Paper Example

Ordinary Differential Equations Name: Course: Instructor: Institution: Date of Submission: 1.0. Summary Evidence checklist Summary of evidence required by student Evidence presented LO4.1 The evidence needed is a differentia equation that presents the Newton’s Law of cooling. The law states that heat loss of any object/ body is proportional to the temperature differences, but only when the temperature is small and the radiating surface nature is also equal. The differential equation presents the involvement of the derivatives and ordinary functions. One function variable has the capacity of satisfying the differential equation. Thus, the evidence in task LO4.1 that was needed, was the differential equation where the temperature is shown in the hot object as proportional to the differential proportional temperature of T0, which the equation presents decreases at a proportion of T – T0. The evidence presented shows a differential equation of the Newton’s Law of Cooling. It is presented presents the function and constant function dT/dt satisfies the differential equation. The solution to the Newton’s Law of Cooling is given through LO4.2 The student was supposed to give the ordinary differential equation of the RL circuit given while applying the Kirchhoff’s Voltage Law. Question 2,(b), the student is anticipated to give the Maclaurin series that includes the x4 term to the ordinary differential question given when one applies the Kirchhoff’s Voltage Law, through the initial condition given as I (0) = 0 Question 2 (c), is supposed to solve the differential equation where t = 0, I = 0. Where the integrating factor is given as Considering that V is constant. Question 2 (d) is supposed to show the complementary function that was identified in 2b, including how it satisfies the homogenous part of the differential equation given in 2a. Question 2 (e) the student is supposed to provide evidence on how the integral part of the section b satisfies the differential equation of section 2 (a) through the integral context of the RL circuit given. Provide the solution of an initial value using Euler's improved method while modelling Newton's cooling. a). The ordinary differential equation of the circuit was the solution given: b). The Maclaurin series in the condition of I (0) = 0 of the Kirchhoff’s Law is including the term x4 is given as: c). The process of solving the equation is given, and the solution is presented as d). The complementary function is given and identified in b as While to satisfy the homogenous section identified in 2a, it is given through: e). The solution is given through Taking the particular integral as: , since g(t) = V/L Putting  in the differential equation, Euler’s solution is given through the five minutes of cooling using the temperature LO4.3 The student is expected to provide second order differential equations modelling oscillations. The first second order differential equation of modelling oscillations was used to describe the motion (oscillation of mass) using Hooke’s law and he second law of motion. It was given m, k, y and t representing, mass, constant, length and time respectively. The second formulated differential equation included c as the constant where c > 0, It was used to modify the mass differential equation. The third differential equation of oscillations using f(t) as the denoting force function while t is time, and the damping force equation is modified using this functions as given in the evidence. Motion of mass differential equations Differential equation with the damping mass. Differential equation with the driving force LO4.4 The student is supposed to present second order homogeneous and non-homogenous differential equations To satisfy the homogenous part of the equation Where the complementary function is given as The homogenous solution is given as . LO4.5 One is supposed to provide the Newton’s cooling equations. 2.0. Discussion 2.1. Newton’s Law of Cooling The simple differential equation of Newton’s law of cooling is given through dy / dt = ry r is considered constant, where r = k it can also be either r = -k, which is also presented in the curves below. Positive and negative solutions Curves of the cooling effects. The Newton’s Law of Cooling through the differential equation presents the function/ derivatives requirements for the initial problem presented. For instance, in the equation presented in L04.1, the function that satisfied – K (T – T0) = 0, which is a solution to the first problem presented. The solutions to the differential equations presented show that the constant solutions is a function, either a simple or constant function. In the first differential equation provided, K = 0 is a constant function. dT / dt = is equally proportional to (T – T0) According to the law of cooling of Newton, the process of cooling objects is comparative to the temperature difference including the ambient temperature. Thus, Newton’s Law of cooling is expressed in the differential equation as presented where the object (particle) temperature is given over time. Temperature of the object is expressed as time in (t). The rate of cooling presents the changes in the temperatures, where cooling is the negative derivative. The rate of cooling presents the proportional rate of temperature for an object is given through α, which is the first quantity that is proportional to the α, which is independent to the time. When the object was stated to have cooled from 100oC to 40oC in 20mintes when the temperature was a 20oC. After 3 minutes of cooling, the temperature was given in the following differential equation. As stated, in the first order, the solution is given through dy/dt where the equation presents , Where y is substituted through: K is given through K is given as  0.07 Thus, after 3 minutes, the temperature is given as Based on this equations, heat transfer through Newton’s law necessitates a constant coefficient of heat transfer, while showing that heat loss of an article is equally relative to the temperature difference, as well as the surrounding temperature of the environment. The equations present Newton’s law of cooling through the differential equations. In this equations, the law of Newton presents how temperature changes occur instantaneously. The object cooled for 100oC to 40oC which is presented through the derivative given in dT/ dt is negative given in , Which is negative. However, the constant K is a positive function, which is the evidence of the differential equation of Newton’s Law of Cooling (Frank, et al., 2007). The t in the equation is an independent variable that represents time, where we have to find the temperature and other constants given in the equation. The cooling of the object after 3 minutes is given as Thus, the temperature prediction changes through time t will be as Figure presenting the cooling solution of the object. Therefore, the Newton’s Law of cooling and solution is given as The Improved Euler Method is also identified as the Heun Method. It is computed in three steps as presented below. 1. m1 = f (xj, yj) 2. m2 = f (xj + 1, yj + hm1) 3. yj + 1 = yj + h (m1 + m2) / 2 In question LO4.1 (c), step one is used to estimate the temperature after 5 minutes of cooling including the percentage error. K = 1/10 In 2, is used in the equation. x = 0.1 x = 0.2 steps size = h = 0.1 Thus, dy / dx = x + y where y (0) = 1 Therefore, x0 = 0; x1 = 0.1, x2 = 0.2 Using Euler’s improved method, the solution is given through the initial problem. The Newton’s law of cooling is given as du / dt = c (Usur – U), t > 0 U (0) = U0 The equation presents the differential equation, where the initial value is u(0). The Euler’s differential equation method to be used for the Newton cooling model is given through the equation: dy / dt = c (Y sur – y), t > 0 y (0) = yo In the equation, the given step sizes are used for each set such as tn = 0 + nh Thus, using the finite difference equation, the solution is given by yn + 1 – yn / h = c (ysur – yn) , y0 = y0 yn is the given approximation, where the y (tn) = y (nh) Thus, in using K = ½ In 2 Using the step size of h = 0.1 and the estimated resolution interval of 0,1, the following is attained. Time (in minutes) Temperature (in C) Error (in percentage %) 0 100 0 1 94.647 0.0046 2 89.652 0.0091 3 84.991 0.0134 4 80.642 0.0175 It can also be presented in the graph below. Current in a Closed RL Circuit R = ohms () (Resistance) L = henries (H) (Inductance) Voltage across the resistor is given through R and the voltage through L given through the equation L (dl/dt) Inductance presents the magnetic flux that a given circuit produces given in Φ=Li The first order equations using the first order circuits such as the RL circuit above require one circuit with one inductor as presented in the figure above. Other parts of the circuits include the resistor, current and the voltage. Thus, the circuit as shown above is a two terminal resistive sub-circuit, where it can be reduced to the resistor. The RL differential equation is presented through the Kirchhoff’s voltage as shown below. Ri + L di / dt = V, Where the solution is given as i = V / R (1 – e –(R/L)t Which is also presented in the graph below. V/R is in the graph is given as the steady state of the circuit. According to the law of Kirchhoff’s Voltage, the voltage increases and decreases the loop circuit, which was equal to the 0. The law presents the RL series gives the differential equation as shown below. The equation present t < 0 while the L inductor transports the current Is while the resistor R carries no current. Thus, the equation above presents the ordinary differential equation through the initial condition. T > 0, where the circuit is reduced through the IC inductor energy through i(0+) = i(0-) = I0 = Is Therefore, using the Kirchhoff’s law of voltage, the ordinary differential equation is given through Where L is given as the independent of the currents i and t(Bali, 2011). It is solved through the equation. Thus, Given that V = 0 We solve the equation through L d/dt i (t) + Ri (t)= 0 The initial condition is given as: IC: i (0+) = I0 = Is L (di) + Ri (dt) = 0 di / i = -R / L dt In i (t) – In i (0) = In i (t) / I0 = - R / L (t) i (t) = I0e –(t/г) ґ = L / R, which presents the time constant in the circuit operation. Possibly, the loop current i (t), which drops to the e-1, given in the initial value through the I0, within the time given in (ґ). It is given through the value of 0 as the proportional constant, the differential equation can be modified to Which also when the damp force is included, the equation attained is: Damped force presents that the mass on any spring cannot oscillate at the same amplitude, though the oscillations end due to the poor strength of the springs in the process of storing and releasing energy. As friction occurs, heat is dissipated, this leads to the damped harmonic motion, where the differential solution of the damped motion is given through the equation: F = m (d2 X / dt2) = - kx – b (dx / dt) Fricition force = - C dy / dt where c > 0 According to the law of newton, it can be attained through restoring the frictional force using the equation presented above. Th differential equation of the damped oscillation is given as m (d2y /dt2) + C (dy / dt) + ky = 0 The term – b (dx/ dt) is considered as a damping term that is added to a differential equation. The damping term role is given in the opposite direction of the proportional velocity and experienced motion. The damping force forces the particles with high velocities to quickly slow down. The forced or driven oscillations give solutions to a given initial condition of the damped motion through the initial behavior evaluation leading to the damping process as the steady state is attained. The steady state of a forced or driven oscillations presents the long-term periodic behavior of the system. In the solution given in question 3 (d), the solution is The transient and steady state part are identified. The transient term in the solution is presented as t increased while the steady state solution part is given through the forcing function. It is a damped harmonic force, where the driving force is oscillatory leading to the amplitude force function. That is; the transient term is the initial condition function while the steady state is the solution linked to the damping /driven forcing term. 3.0. Conclusion Newton’s law of cool states that when any force is applied to a particle, it is equal in the mass and the acceleration of the article given in F= m (d2y / dt2). The second order ordinary differential equations through the Newton’s law of cooling relates force, acceleration and mass of the objects given as t to represent time and y (t) to represent the position of the object. When Hooke’s law is applied to the motions perceived in springs as presented, the mass is applied at the end of the spring, the initial condition of the through either damped, or driven force is used. Hooke’s law depicts that if the y (in the newton’s law) cannot deform the spring, the restoring force is given through: Fspring = -ky. When Hooke’s and Newton second laws are applied together to solve a differential equation, Hooke’s law is used a function that satisfies the newton’s developed equation as depicted above. In RL circuits, the differential equations is given through the Kirchhoff’s law. Simple harmonic motion is given through the equation yn + w2y = 0. The solutions have c as constants mainly, which depicts the oscillatory solution through the frequency given. The t (time) presents the time an oscillation takes to complete a single cycle. The cycles may include the vibrations as well. Damped harmonic motion presents the oscillation that is resistance to the motion where the friction or force applied is proportional to the velocity. Given in F fricition = -C dy / dt, where c is constant and since the force is opposite to the motion, it is considered to be C > 0. Homogenous equations are perceived when f(t) = 0 while when the equation is f(t) = 0 it can be presented as the non-homogenous equation. 4.0. References Bali, P. N., 2011. A Textbook of Engineering Mathematics (PTU, Jalandhar) Sem-II. New York: Laxmi Publications. Frank, I., Theodore, L. B., David, D. & Adrienne, S. L., 2007. Fundamentals of Heat and Mass Transfer. Print ed. New York: John Wiley & Sons. HWU, 2015. Mathematics for Engineers and Scientist 3. F18XC1. White Paper. Heriot Watt University. Teaching Math, pp. 1-82. Taylor, R. J., 2005. Classical Mechanics. 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