Problem 35 - Advanced Engineering Mathematics

Question

The differential equation is \(d y / d x=y^{\prime} / x^{\prime}=\left(x^{3}-x\right) / y\) and so \(y^{2} / 2=x^{4} / 4-x^{2} / 2+c\) or \(y^{2}=x^{4} / 2-x^{2}+c_{1}\) since \(x(0)=0\) and \(y(0)=x^{\prime}(0)=v_{0},\) it follows that \(c_{1}=v_{0}^{2}\) and so $$y^{2}=\frac{1}{2} x^{4}-x^{2}+v_{0}^{2}=\frac{\left(x^{2}-1\right)^{2}+2 v_{0}^{2}-1}{2}$$The \(x\) -intercepts on this graph satisfy $$x^{2}=1 \pm \sqrt{1-2 v_{0}^{2}}$$ and so we must require that \(\left.1-2 v_{0}^{2} \geq 0 \text { (or }\left|v_{0}\right| \leq \frac{1}{2} \sqrt{2}\right)\) for real solutions to exist. If \(x_{0}^{2}=1-\sqrt{1-2 v_{0}^{2}}\) and \(-x_{0} < x < x_{0},\) then \(\left(x^{2}-1\right)^{2}+2 v_{0}^{2}-1 > 0\) and so there are two corresponding values of \(y .\) Therefore \(\mathbf{X}(t)\) with \(\mathbf{X}(0)=\left(0, v_{0}\right)\) is periodic provided that \(\left|v_{0}\right| \leq \frac{1}{2} \sqrt{2}\)

Step-by-Step Solution

Answer

The solution is periodic if \( |v_0| \leq \frac{1}{2} \sqrt{2} \).

Step 1: Analyze the Differential Equation

The given differential equation is \( \frac{dy}{dx} = \frac{x^3 - x}{y} \). The problem begins by presenting a solution to this differential equation in terms of \( y^2 \) and \( x \).

Step 2: Derive the Solution in Terms of Constants

It states that \( y^2/2 = x^4/4 - x^2/2 + c \). Which simplifies to \( y^2 = x^4/2 - x^2 + c_1 \), where \( c_1 \) is an integration constant.

Step 3: Consider Initial Conditions

Given \( x(0) = 0 \) and \( y(0) = x'(0) = v_0 \), it follows that \( c_1 = v_0^2 \).

Step 4: Rewrite the Equation with Initial Conditions

Substitute \( c_1 \) back into the solution to get \( y^2 = \frac{1}{2}(x^2 - 1)^2 + 2v_0^2 - 1 \).

Step 5: Determine Condition for Real-Valued Solutions

The \( x \)-intercepts solve \( x^2 = 1 \pm \sqrt{1 - 2v_0^2} \). For this to be real, \( 1 - 2v_0^2 \geq 0 \), implying \( |v_0| \leq \frac{1}{2} \sqrt{2} \).

Step 6: Analyze the Range of \( x \) for Periodicity

Set \( x_0^2 = 1 - \sqrt{1 - 2v_0^2} \). For \( -x_0 < x < x_0 \), the expression \( (x^2 - 1)^2 + 2v_0^2 - 1 > 0 \) holds, implying there are two values of \( y \).

Step 7: Conclude on Periodicity

Since the condition for \( |v_0| \leq \frac{1}{2} \sqrt{2} \) allows for two values of \( y \) within the range \( -x_0 < x < x_0 \), the solution \( \mathbf{X}(t) \) with initial point \( \mathbf{X}(0)=(0, v_0) \) is periodic.

Key Concepts

Initial Conditions

Initial conditions play a fundamental role in solving differential equations. In this exercise, the initial conditions are given as \( x(0) = 0 \) and \( y(0) = x'(0) = v_0 \). These conditions provide a starting point for determining particular solutions from a set of solutions defined by the differential equation.

When we solve \( y^2 = x^4/2 - x^2 + c_1 \), the constant \( c_1 \) represents the integration constant that can vary depending on the specific scenario. Given the initial conditions, we can solve for this constant as \( c_1 = v_0^2 \).

This solution means that at time \( t = 0 \), the behavior of the system is specified, allowing us to determine how solutions evolve over time. The initial conditions essentially anchor the solutions to specific values, making them unique for each problem scenario.

Periodicity

Periodicity in differential equations refers to solutions that repeat at regular intervals over time. In this exercise, it is found that the function \( \mathbf{X}(t) \) is periodic under particular conditions.

For periodic solutions, it's crucial to ensure the conditions are met. In this scenario, \( |v_0| \leq \frac{1}{2} \sqrt{2} \) ensures the solution is periodic. This constraint results from analyzing the equation's real-valued solutions and determining the interval where two values of \( y \) exist.

When \( -x_0 < x < x_0 \) within this condition, the expression \( (x^2 - 1)^2 + 2v_0^2 - 1 > 0 \) holds, indicating that the curve loops back, repeating at regular intervals. Understanding these loops helps in identifying and proving the periodic nature of solutions.

Integration Constant

An integration constant is a constant of integration appearing in the solutions of differential equations. This constant emerges because integration, the reverse operation of differentiation, does not provide details about certain lost information, which often represents initial conditions or a specific value in the context.

In this exercise, during the integration step, we observe an integration constant \( c_1 \). To find its specific value, initial conditions are implemented. For \( y^2 = x^4/2 - x^2 + c_1 \), knowing \( x(0) = 0 \) and \( y(0) = v_0 \), the constant translates to \( c_1 = v_0^2 \).

This approach underlines the importance of using initial conditions to solve differential equations comprehensively. By applying the constant, we ensure the derived solution satisfies the characteristics initially specified, thereby addressing specific needs or constraints for every different scenario.