Problem 4
Question
In Problems, write the given nonlinear second-order differential equation as a plane autonomous system. Find all critical points of the resulting system. $$ x^{\prime \prime}+4 \frac{x}{1+x^{2}}+2 x^{\prime}=0 $$
Step-by-Step Solution
Verified Answer
The critical point is (0, 0).
1Step 1: Recognize the Differential Equation Format
Start by identifying the form of the given second-order differential equation: \(x'' + 4\frac{x}{1+x^2} + 2x' = 0\). Notice it's a second-order with respect to \(x\).
2Step 2: Transform into Two First-order Equations
Introduce a substitution to reduce the second-order equation into two first-order equations. Set \(y = x'\), leading to the system:\[\begin{align*}x' &= y, \y' &= -4\frac{x}{1+x^2} - 2y.\end{align*}\]
3Step 3: Write the System in Standard Planar Form
The system of first-order equations can be written in the form:\[\begin{align*}\frac{dx}{dt} &= y, \\frac{dy}{dt} &= -4\frac{x}{1+x^2} - 2y.\end{align*}\]This is now in the standard plane autonomous system format.
4Step 4: Find Critical Points
Set both derivatives equal to zero to find the critical points. Solving \(y = 0\) and \(-4\frac{x}{1+x^2} - 2y = 0\), we find:- From \(y = 0\), substitute into \(-4\frac{x}{1+x^2} = 0\), this simplifies to \(x = 0\).- Thus, the only critical point is \((x, y) = (0, 0)\).
Key Concepts
Second-order Differential EquationsAutonomous SystemsCritical PointsPlane Systems
Second-order Differential Equations
A second-order differential equation involves the second derivative of a function. These equations are key in describing systems where the rate of change of momentum or velocity varies, such as in harmonic oscillators. Our exercise presents the equation: \[ x'' + 4\frac{x}{1+x^2} + 2x' = 0 \] Here, \(x''\) signifies the second derivative of \(x\) concerning time \(t\). The goal is to understand how such an equation can model dynamic systems. We first convert it into an equivalent system of first-order differential equations. This helps in analyzing and solving complex systems by simplifying them into more manageable chunks. By breaking it down, we can more easily study system behavior and stability. Remember, second-order linear equations have constant coefficients and solutions exhibit predictable behaviors when compared to nonlinear forms as in our case.
Autonomous Systems
An autonomous system in differential equations is one where the equations do not explicitly depend on the independent variable, usually time \(t\). Instead, it relies on the functions themselves. Our exercise involves transforming a second-order differential equation into a system of first-order equations:
- \(x' = y\)
- \(y' = -4\frac{x}{1+x^2} - 2y\)
Critical Points
Critical points, also known as equilibrium points, are where the state variables of a system do not change over time. In our context, they are the solutions to the system of equations when both derivatives are zero:
- \(y = 0\)
- \(-4\frac{x}{1+x^2} = 0\)
Plane Systems
Plane systems refer to the representation of differential equations in a two-dimensional space, generally in terms of \(x\) and \(y\). In such a setup, an autonomous system can be visualized as vector fields on the \(xy\)-plane:
- \(\frac{dx}{dt} = y\)
- \(\frac{dy}{dt} = -4\frac{x}{1+x^2} - 2y\)
Other exercises in this chapter
Problem 4
The general solution of the linear system \(\mathbf{X}^{\prime}=\mathbf{A} \mathbf{X}\) is given. (a) In each case discuss the nature of the solution in a neigh
View solution Problem 4
In Problems \(3-10\), without solving explicitly, classify the critical points of the given first-order autonomous differential equation as either asymptoticall
View solution Problem 4
Show that the given plane autonomous system (or second-order differential equation) has no periodic solutions. $$ \begin{aligned} &x^{\prime}=x y^{2}-x^{2} y \\
View solution Problem 4
Without solving explicitly, classify the critical points of the given first- order autonomous differential equation as either asymptotically stable or unstable.
View solution