use the phase-plane method to show that the solution to the nonlinear second-order differential equation $x + 6x - x^2 = 0$ that satisfies $x(0) = 1$ and $x(0) = 0$ is periodic.

let $\frac{dx}{dt} = y$. then the differential equation $\frac{dy}{dx} =$ can be solved by separating variables. it follows that the general solution is , and since $\mathbf{x}(0) = (x(0), x(0)) = (1, 0)$ then the particular solution is . but for each $x$ such that $4 - 2\sqrt{6}

Question

use the phase-plane method to show that the solution to the nonlinear second-order differential equation $x + 6x - x^2 = 0$ that satisfies $x(0) = 1$ and $x(0) = 0$ is periodic.

let $\frac{dx}{dt} = y$. then the differential equation $\frac{dy}{dx} =$   can be solved by separating variables. it follows that the general solution is  , and since $\mathbf{x}(0) = (x(0), x(0)) = (1, 0)$ then the particular solution is  . but for each $x$ such that $4 - 2\sqrt{6} < x < 1$, the particular solution has   corresponding value(s) of $y$. therefore $\mathbf{x}(t)$ is a periodic solution.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Phase Portrait Analysis", "Nonlinear Systems of ODEs" ], "new_concepts": [ "Phase-Plane Method", "Periodic Solutions of ODEs" ], "current_concepts": [ "Phase Portrait Analysis", "Nonlinear Systems of ODEs", "Phase-Plane Method", "Periodic Solutions of ODEs" ] } ## Convert to a first-order system Using the **Nonlinear Systems of ODEs** knowledge point $$ \begin{aligned} &\frac{dx}{dt} = y\ &\frac{dy}{dt} = x'' = x^2 - 6x \end{aligned} $$ ## Formulate the phase-plane equation Using the **Phase Portrait Analysis** knowledge point $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{x^2 - 6x}{y} $$ ## Solve by separation of variables Using the **Phase Portrait Analysis** knowledge point $$ \begin{aligned} &y \, dy = (x^2 - 6x) \, dx\ &\frac{1}{2}y^2 = \frac{1}{3}x^3 - 3x^2 + C\ &y^2 - \frac{2}{3}x^3 + 6x^2 = C_1 \end{aligned} $$ ## Apply the initial conditions Using the **Phase Portrait Analysis** knowledge point $$ \begin{aligned} &x(0) = 1, \quad y(0) = 0\ &0^2 - \frac{2}{3}(1)^3 + 6(1)^2 = \frac{16}{3}\ &y^2 - \frac{2}{3}x^3 + 6x^2 = \frac{16}{3} \end{aligned} $$ ## Analyze the periodic behavior Using the **Phase Portrait Analysis** knowledge point $$ \begin{aligned} &y^2 = \frac{2}{3}x^3 - 6x^2 + \frac{16}{3}\ &y = \pm \sqrt{\frac{2}{3}x^3 - 6x^2 + \frac{16}{3}} \end{aligned} $$ For each $x$ in the interval $4 - 2\sqrt{6} < x < 1$, the term inside the square root is strictly positive, yielding two real, symmetric values of $y$. Use the phase-plane method to show that the solution to the nonlinear second-order differential equation $x'' + 6x - x^2 = 0$ that satisfies $x(0) = 1$ and $x'(0) = 0$ is periodic. Let $\frac{dx}{dt} = y$. Then the differential equation $\frac{dy}{dx} =$ $\frac{x^2-6x}{y}$ can be solved by separating variables. It follows that the general solution is $y^2 - \frac{2}{3}x^3 + 6x^2 = C$, and since $\mathbf{X}(0) = (x(0), x'(0)) = (1, 0)$ then the particular solution is $y^2 - \frac{2}{3}x^3 + 6x^2 = \frac{16}{3}$. But for each $x$ such that $4 - 2\sqrt{6} < x < 1$, the particular solution has two real corresponding value(s) of $y$. Therefore $\mathbf{X}(t)$ is a periodic solution. { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Nonlinear Systems of ODEs" ] }

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QUESTION IMAGE

Question

Explanation:

Convert to a first-order system

Formulate the phase-plane equation

Solve by separation of variables

Apply the initial conditions

Analyze the periodic behavior

Answer: