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Stars indicate the approximate level of difficulty, from easiest ( ) to most difficult (). Warning: Even when the motion is nonchaotic, it can be very sensitive to tiny errors. In several of the computer problems you may need to increase your working precision to get satisfactory results. Consider the nonlinear first-order equation x˙=2x1. (a) By separating variables, find a solution x1(t). (b) Your solution should contain one constant of integration k, so you might reasonably expect it to be the general solution. Show, however, that there is another solution, x2(t)=1, that is not of the form of x1(t) whatever the value of k. (c) Show that although x1(t) and x2(t) are solutions, neither Ax1(t), nor Bx2(t), nor x1(t)+x2(t) are solutions. (That is, the superposition principle does not apply to this equation.)

Short Answer

Expert verified
(a) x1(t)=(t+C2)2+1; (b) x2(t)=1; (c) Superposition does not apply.

Step by step solution

01

Separate Variables

Separate variables in the differential equation x˙=2x1. This gives the form dxx1=2dt.
02

Integrate Both Sides

Integrate both sides: dxx1=2dt. The left side integrates to 2x1 and the right side to 2t+C, where C is a constant of integration. Thus, 2x1=2t+C.
03

Solve for x

To express x in terms of t, first divide both sides by 2: x1=t+C2. Squaring both sides results in x1=(t+C2)2. Add 1 to both sides to solve for x: x1(t)=(t+C2)2+1.
04

Find Alternative Solution

Verify if x2(t)=1 is a solution to the differential equation. Substitute x2(t)=1 into x˙=2x1: x˙=0 and 211=0. Since both sides are equal, x2(t)=1 is indeed a solution.
05

Verify Non-linear Superposition

For non-linear equations, solutions Ax1(t), Bx2(t), or x1(t)+x2(t) generally do not hold. Verify each case.
06

Test x1 + x2 for Superposition

Compute x1(t)+x2(t)=(t+C2)2+2 and substitute into x˙. The equation x˙=2(t+C2)2+21 yields a different form, confirming that x1(t)+x2(t) is not a solution.
07

Confirm Non-Applicability of Superposition

Repeat similar calculations for Ax1(t) or Bx2(t) by substituting them into the differential equation. In all traditional linear combinations, they should fail to satisfy the original differential equation x˙=2x1. This confirms the non-linearity and invalidity of superposition for this problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Superposition Principle
In the context of differential equations, the superposition principle tells us that if we have two or more solutions to a linear differential equation, then any linear combination of them is also a solution. This principle is fundamental when dealing with linear systems, allowing solutions to be constructed from simpler components.

However, in the case of nonlinear differential equations, such as the one presented in this problem, the superposition principle does not apply. That is because the combination of solutions can lead to terms that do not satisfy the original nonlinear equation. This is why in this exercise, even though x1(t) and x2(t) are solutions, neither Ax1(t) nor Bx2(t) nor x1(t)+x2(t) satisfy the differential equation x˙=2x1. Nonlinear equations are picky; each solution needs to be verified independently.
Separation of Variables
Separation of variables is a powerful technique used to solve differential equations. It involves rearranging an equation so that each variable occurs on only one side of the equation. This method works particularly well for differential equations that can be expressed in a form where the function of x is separated from the function of t.

In this exercise, the differential equation x˙=2x1 was separated into dxx1=2dt, enabling us to integrate each side independently. By performing the integrals, we solved for the function of x in terms of t, thus arriving at the potential solution x1(t). An essential aspect of separation of variables is ensuring that all solutions fit within the functional constraints posed by the problem.
Constant of Integration
The constant of integration appears when we integrate a differential equation. It represents the family of solutions that stem from the addition of an arbitrary constant after an indefinite integral. This constant is crucial because it accounts for all possible solution shifts along the t-axis.

In the solution presented, after integrating both sides of the equation dxx1=2dt, we acquired an expression with the constant of integration C. Solving for x, we found x1(t)=(t+C2)2+1. This constant C is the bridge to different particular solutions which explain initial conditions or specific scenarios in the context of the problem.
First-Order Differential Equation
A first-order differential equation is an equation involving the first derivative of an unknown function. It provides valuable insights into the behavior of a system by defining the rate of change of a quantity over time.

The provided exercise dealt with the first-order differential equation x˙=2x1. Here, x˙ signifies the first derivative of x with respect to time t. Solving such equations involves finding the function x(t) that satisfies this relationship. By solving the equation using separation of variables, the students derived potential solutions. Understanding the nature and solutions of first-order differential equations is fundamental in many areas of engineering, physics, and beyond, as it helps describe systems' dynamics effectively.

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Most popular questions from this chapter

Here is an iterated map that is easily studied with the help of your calculator: Let xt+1=f(xt) where f(x)=cos(x). If you choose any value for x0, you can find x1,x2,x3, by simply pressing the cosine button on your calculator over and over again. (Be sure the calculator is in radians mode.) (a) Try this for several different choices of x0, finding the first 30 or so values of xt. Describe what happens. (b) You should have found that there seems to be a single fixed attractor. What is it? Explain it, by examining (graphically, for instance) the equation for a fixed point f(x)=x and applying our test for stability [namely, that a fixed point x is stable if |f(x)|<1].

Here is a different example of the disagreeable things that can happen with nonlinear equations. Consider the nonlinear equation x˙=2x. since this is first-order, one would expect that specification of x(0) would determine a unique solution. Show that for this equation there are two different solutions, both satisfying the initial condition x(0)=0. [Hint: Find one solution x1(t) by separating variables, but note that x2(t)=0 is another. Fortunately none of the equations normally encountered in classical mechanics suffer from this disagreeable ambiguity.]

Consider an undamped, undriven simple harmonic oscillator - a mass m on the end of a spring whose force constant is k. (a) Write down the general solution x(t) for the position as a function of time t. Use this to sketch the state-space orbit, showing the motion of the point [x(t),x˙(t)] in the two dimensional state space with coordinates (x,x˙). Explain the direction in which the orbit is traced as time advances. (b) Write down the total energy of the system and use conservation of energy to prove that the state-space orbit is an ellipse.

Consider an inhomogeneous second-order linear equation of the form p(t)x¨(t)+q(t)x˙(t)+r(t)x(t)=f(t) Let xp(t) denote a solution (a "particular" solution) of this equation and prove that any solution x(t) can be written as x(t)=xp(t)+a1x1(t)+a2x2(t) where x1(t) and x2(t) are two independent solutions of the corresponding homogeneous equation that is, (12.59) with f(t) deleted. [Hint: Write down the equations for x(t) and xp(t) and subtract.] This result shows that to find all solutions of (12.59), we have only to find one particular solution and two independent solutions of the corresponding homogeneous equation. (b) Explain clearly why the result you proved in part (a) is not, in general, true for a nonlinear equation such as p(t)x¨(t)+q(t)x˙(t)+r(t)x(t)=f(t)

Consider the iterated map xt+1=f(xt) where f(x)=x2. (a) Show that it has exactly two fixed points of which just one is stable. What are they? (b) Show that xt approaches the stable fixed point if and only if \(-11\). (Thus we could say the map has a second stable fixed point at x= and the basin of attraction for this fixed point is the set |x0|>1. ) For chaotic systems, the basins of attraction can be much more complicated than these examples and are often fractals.

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