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Chapter 11: Problem 32

Solve $$ \frac{d^{2} x}{d t^{2}}=-4 x $$ with \(x(0)=0\) and \(x^{\prime}(0)=6\).

Short Answer

Expert verified
The solution is \( x(t) = 3\sin(2t) \).

Step by step solution

01

Identify the Type of Equation

This is a second-order linear homogeneous differential equation with constant coefficients: \( \frac{d^{2} x}{dt^{2}} = -4x \). The general approach to solve such equations involves characteristic equations.
02

Form the Characteristic Equation

Assume a solution of the form \( x(t) = e^{rt} \). Plugging this into the differential equation gives the characteristic equation: \( r^2 = -4 \).
03

Solve the Characteristic Equation

The characteristic equation \( r^2 = -4 \) has solutions \( r = \pm 2i \), indicating complex conjugate roots.
04

Write the General Solution

For complex roots \( \pm bi \), the solution takes the form \( x(t) = C_1 \cos(bt) + C_2 \sin(bt) \). Here, \( b = 2 \), so \( x(t) = C_1 \cos(2t) + C_2 \sin(2t) \).
05

Apply Initial Conditions to Find Constants

First, use the condition \( x(0) = 0 \). Substituting gives \( C_1 = 0 \). Next, apply \( x'(0) = 6 \). Calculating the derivative: \( x'(t) = 0 \cdot \cos(2t) + 2C_2 \cdot \cos(2t) = 2C_2 \cdot \cos(2t) \). Substituting \( x'(0) = 6 \) gives \( 2C_2 = 6 \), hence \( C_2 = 3 \).
06

Write the Specific Solution

Now that we have \( C_1 = 0 \) and \( C_2 = 3 \), the specific solution is \( x(t) = 3\sin(2t) \).

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Key Concepts

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

Characteristic Equation
In the realm of solving second-order linear homogeneous differential equations, the characteristic equation plays a pivotal role. These equations typically appear in the form:
  • \( \frac{d^2 x}{dt^2} + a \frac{dx}{dt} + bx = 0 \)
To solve these equations, we assume a solution of the form \( x(t) = e^{rt} \), where \( r \) is a constant. This assumption helps transform the differential equation into an algebraic equation known as the characteristic equation. This equation is vital because it allows us to find the correct form of the solution based on the roots of the equation.
  • For our problem, the characteristic equation resulting from \( \frac{d^{2} x}{d t^{2}}=-4 x \) is \( r^2 + 4 = 0 \).
  • Solving this gives us the possible values of \( r \), which inform the general solution form.
Initial Conditions
Initial conditions are essential as they transform our general solution into a specific one that satisfies the given conditions of a problem. These conditions typically specify the function and its derivatives at a particular point. They are usually given as:
  • \( x(0) = x_0 \)
  • \( x'(0) = v_0 \)
For our exercise:
  • The initial conditions are \( x(0)=0 \) and \( x'(0)=6 \).
  • By applying \( x(0)=0 \), we find \( C_1 = 0 \) in the general solution \( x(t) = C_1 \cos(2t) + C_2 \sin(2t) \).
  • Using \( x'(0)=6 \) helps us identify \( C_2 = 3 \). With these constants, we narrow down to a specific solution.
By applying these initial conditions correctly, we arrive at \( x(t) = 3\sin(2t) \), which accurately fulfills both conditions laid out in the problem.
Complex Roots
When solving a characteristic equation, the nature of the roots has a direct impact on the solution form:
  • Real and distinct roots yield solutions involving exponential functions.
  • Real repeated roots result in solutions with polynomials and exponentials.
  • Complex conjugate roots introduce trigonometric functions into the solution.
In our scenario, solving \( r^2 = -4 \) yields complex roots: \( r = \pm 2i \). Complex roots always come in conjugate pairs \( a \pm bi \), leading the general solution to take the form:
  • \( x(t) = C_1 \cos(bt) + C_2 \sin(bt) \)
This specific structure reflects the periodic nature of solutions when complex roots are involved, corresponding to the sinusoidal functions that emerge from the Euler's formula. Thus, when tackling complex roots, expect to see these trigonometric elements, which offer both a solution method and insights into the system's behavior.

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Most popular questions from this chapter

Assume that $$ \begin{array}{l} \frac{d N}{d t}=N-4 P N \\ \frac{d P}{d t}=2 P N-3 P \end{array} $$ (a) Show that this system has two equilibria: the trivial equilibrium \((0,0)\), and a nontrivial one in which both species have positive densities. (b) Use the eigenvalue approach to show that the trivial equilibrium is unstable. (c) Determine the eigenvalues corresponding to the nontrivial equilibrium. Does your analysis allow you to infer anything about the stability of this equilibrium? (d) Use a graphing calculator to sketch curves in the \(N-P\) plane. Also, sketch solution curves of the prey and the predator densities as functions of time.

Write each system of differential equations in matrix form. \(\frac{d x_{1}}{d t}=x_{3}-2 x_{1}\) \(\frac{d x_{2}}{d t}=-x_{1}+x_{3}\) \(\frac{d x_{3}}{d t}=x_{1}+x_{2}+x_{3}\)

Solve $$ \frac{d^{2} x}{d t^{2}}=-9 x $$ with \(x(0)=0\) and \(x^{\prime}(0)=12\).

An unrealistic feature of the Lotka-Volterra model is that the prey exhibits unlimited growth in the absence of the predator. The model described by the following system remedies this shortcoming (in the model, we assume that the prey evolves according to logistic growth in the absence of the predator; the other features of the model are retained): $$ \begin{array}{l} \frac{d N}{d t}=N\left(1-\frac{N}{K}\right)-4 P N \\ \frac{d P}{d t}=P N-5 P \end{array} $$ Here, \(K>0\) denotes the carrying capacity of the prey in the absence of the predator. In what follows, we will investigate how the carrying capacity affects the outcome of this predator-prey interaction. (a) Draw the zero isoclines of \((11.78)\) for (i) \(K=10\) and (ii) \(K=3\). (b) When \(K=10\), the zero isoclines intersect, indicating the existence of a nontrivial equilibrium. Analyze the stability of this nontrivial equilibrium. (c) Is there a minimum carrying capacity required in order to have a nontrivial equilibrium? If yes, find it and explain what happens when the carrying capacity is below this minimum and what happens when the carrying capacity is above this minimum.

Based on each system of equations modeling Romeo and Juliet's relationship, describe in words how Romeo and Juliet are both behaving (you do not need to solve any of the systems). \(\frac{d J}{d t}=J-0.2 R\) \(\frac{d R}{d t}=J-0.1 R\)
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