# Second Order Differential Equation Problems And Solutions Pdf

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Published: 11.05.2021  For each of the equation we can write the so-called characteristic auxiliary equation :. The general solution of the homogeneous differential equation depends on the roots of the characteristic quadratic equation. There are the following options:. First we write the corresponding characteristic equation for the given differential equation:.

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We saw in the chapter introduction that second-order linear differential equations are used to model many situations in physics and engineering. In this section, we look at how this works for systems of an object with mass attached to a vertical spring and an electric circuit containing a resistor, an inductor, and a capacitor connected in series.

Models such as these can be used to approximate other more complicated situations; for example, bonds between atoms or molecules are often modeled as springs that vibrate, as described by these same differential equations. Consider a mass suspended from a spring attached to a rigid support. This is commonly called a spring-mass system. Gravity is pulling the mass downward and the restoring force of the spring is pulling the mass upward. If the mass is displaced from equilibrium, it oscillates up and down.

This behavior can be modeled by a second-order constant-coefficient differential equation. Let x t x t denote the displacement of the mass from equilibrium. Note that for spring-mass systems of this type, it is customary to adopt the convention that down is positive. Thus, a positive displacement indicates the mass is below the equilibrium point, whereas a negative displacement indicates the mass is above equilibrium. Displacement is usually given in feet in the English system or meters in the metric system.

Consider the forces acting on the mass. The force of gravity is given by mg. In the English system, mass is in slugs and the acceleration resulting from gravity is in feet per second squared.

The motion of the mass is called simple harmonic motion. Assume an object weighing 2 lb stretches a spring 6 in. What is the period of the motion? We also know that weight W equals the product of mass m and the acceleration due to gravity g. A g mass stretches a spring 5 cm. Find the equation of motion of the mass if it is released from rest from a position 10 cm below the equilibrium position.

What is the frequency of this motion? It is easy to see the link between the differential equation and the solution, and the period and frequency of motion are evident. This form of the function tells us very little about the amplitude of the motion, however. In some situations, we may prefer to write the solution in the form. Although the link to the differential equation is not as explicit in this case, the period and frequency of motion are still evident.

Furthermore, the amplitude of the motion, A , is obvious in this form of the function. What is the frequency of motion? The amplitude? With the model just described, the motion of the mass continues indefinitely. In the real world, there is almost always some friction in the system, which causes the oscillations to die off slowly—an effect called damping. Because damping is primarily a friction force, we assume it is proportional to the velocity of the mass and acts in the opposite direction.

Just as in Second-Order Linear Equations we consider three cases, based on whether the characteristic equation has distinct real roots, a repeated real root, or complex conjugate roots. The general solution has the form. Because the exponents are negative, the displacement decays to zero over time, usually quite quickly.

Overdamped systems do not oscillate no more than one change of direction , but simply move back toward the equilibrium position.

A lb mass is attached to a ft spring. When the mass comes to rest in the equilibrium position, the spring measures 15 ft 4 in. The system is immersed in a medium that imparts a damping force equal to times the instantaneous velocity of the mass. What is the position of the mass after 10 sec? Its velocity? The mass stretches the spring 5 ft 4 in. The system is then immersed in a medium imparting a damping force equal to 16 times the instantaneous velocity of the mass.

Find the equation of motion if it is released from rest at a point 40 cm below equilibrium. The motion of a critically damped system is very similar to that of an overdamped system.

It does not oscillate. However, with a critically damped system, if the damping is reduced even a little, oscillatory behavior results. From a practical perspective, physical systems are almost always either overdamped or underdamped case 3, which we consider next. A 1-kg mass stretches a spring 20 cm. The system is attached to a dashpot that imparts a damping force equal to 14 times the instantaneous velocity of the mass. A 1-lb weight stretches a spring 6 in. Find the equation of motion if the mass is released from rest at a point 6 in.

Underdamped systems do oscillate because of the sine and cosine terms in the solution. However, the exponential term dominates eventually, so the amplitude of the oscillations decreases over time. The system always approaches the equilibrium position over time.

A lb weight stretches a spring 3. Assume the damping force on the system is equal to the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point 9 in. A 1-kg mass stretches a spring 49 cm. The system is immersed in a medium that imparts a damping force equal to four times the instantaneous velocity of the mass.

Find the equation of motion if the mass is released from rest at a point 24 cm above equilibrium. For motocross riders, the suspension systems on their motorcycles are very important.

The off-road courses on which they ride often include jumps, and losing control of the motorcycle when they land could cost them the race. This suspension system can be modeled as a damped spring-mass system.

We define our frame of reference with respect to the frame of the motorcycle. Assume the end of the shock absorber attached to the motorcycle frame is fixed. We measure the position of the wheel with respect to the motorcycle frame. This may seem counterintuitive, since, in many cases, it is actually the motorcycle frame that moves, but this frame of reference preserves the development of the differential equation that was done earlier.

As with earlier development, we define the downward direction to be positive. When the motorcycle is lifted by its frame, the wheel hangs freely and the spring is uncompressed. This system can be modeled using the same differential equation we used before:.

A motocross motorcycle weighs lb, and we assume a rider weight of lb. When the rider mounts the motorcycle, the suspension compresses 4 in. The suspension system provides damping equal to times the instantaneous vertical velocity of the motorcycle and rider. Therefore, the differential equation that models the behavior of the motorcycle suspension is.

Now, to determine our initial conditions, we consider the position and velocity of the motorcycle wheel when the wheel first contacts the ground. Since the motorcycle was in the air prior to contacting the ground, the wheel was hanging freely and the spring was uncompressed.

Therefore the wheel is 4 in. NASA is planning a mission to Mars. To save money, engineers have decided to adapt one of the moon landing vehicles for the new mission. However, they are concerned about how the different gravitational forces will affect the suspension system that cushions the craft when it touches down. The acceleration resulting from gravity on the moon is 1.

The suspension system on the craft can be modeled as a damped spring-mass system. We retain the convention that down is positive. Despite the new orientation, an examination of the forces affecting the lander shows that the same differential equation can be used to model the position of the landing craft relative to equilibrium:.

The last case we consider is when an external force acts on the system. In the case of the motorcycle suspension system, for example, the bumps in the road act as an external force acting on the system.

Another example is a spring hanging from a support; if the support is set in motion, that motion would be considered an external force on the system. We model these forced systems with the nonhomogeneous differential equation. As we saw in Nonhomogenous Linear Equations , differential equations such as this have solutions of the form.

A mass of 1 slug stretches a spring 2 ft and comes to rest at equilibrium. The system is attached to a dashpot that imparts a damping force equal to eight times the instantaneous velocity of the mass. What is the transient solution? What is the steady-state solution? Find the equation of motion if there is no damping.

Find the particular solution before applying the initial conditions. Consider an undamped system exhibiting simple harmonic motion. In the real world, we never truly have an undamped system; —some damping always occurs.

For theoretical purposes, however, we could imagine a spring-mass system contained in a vacuum chamber. ## Differential Equations

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System Simulation and Analysis. Plant Modeling for Control Design. High Performance Computing.

In mathematics, a differential equation is an equation that relates one or more functions and their derivatives. Such relations are common; therefore, differential equations play a prominent role in many disciplines including engineering , physics , economics , and biology. Mainly the study of differential equations consists of the study of their solutions the set of functions that satisfy each equation , and of the properties of their solutions. Only the simplest differential equations are solvable by explicit formulas; however, many properties of solutions of a given differential equation may be determined without computing them exactly.

### Second-Order Differential Equations

Documentation Help Center Documentation. Solve a differential equation analytically by using the dsolve function, with or without initial conditions. To solve a system of differential equations, see Solve a System of Differential Equations. Solve Differential Equation with Condition.

We have fully investigated solving second order linear differential equations with constant coefficients. Now we will explore how to find solutions to second order linear differential equations whose coefficients are not necessarily constant. Below, we will investigate only ordinary points. Instead, we use the fact that the second order linear differential equation must have a unique solution. We can express this unique solution as a power series. Fortunately, we can easily take derivatives. Over the last hundred years, many techniques have been developed for the solution of ordinary differential equations and partial differential equations. While quite a major portion of the techniques is only useful for academic purposes, there are some which are important in the solution of real problems arising from science and engineering. In this chapter, only very limited techniques for solving ordinary differential and partial differential equations are discussed, as it is impossible to cover all the available techniques even in a book form. The readers are then suggested to pursue further studies on this issue if necessary. After that, the readers are introduced to two major numerical methods commonly used by the engineers for the solution of real engineering problems. Dynamical Systems - Analytical and Computational Techniques.

Two basic facts enable us to solve homogeneous linear equations. The first of these says that if we know two solutions and of such an equation, then the linear​. We presented particular solutions to the considered problem. Finally, a few illustrative examples are shown. The second-order differential equations provide an important mathematical tool for modelling the phenomena occurring in dynamical systems.

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Let y 1 and y 2 be two solutions of the homogeneous linear equation 2 on the interval I.

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