/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 From \(x-y=0\) we have \(y=x .\)... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 9

From \(x-y=0\) we have \(y=x .\) Substituting into \(3 x^{2}-4 y=0\) we obtain \(3 x^{2}-4 x=x(3 x-4)=0 .\) It follows that (0,0) and \((4 / 3,4 / 3)\) are the critical points of the system.

Short Answer

Expert verified
The critical points are \((0, 0)\) and \(\left( \frac{4}{3}, \frac{4}{3} \right)\)."

Step by step solution

01

Understand the Problem

We are given two equations: \(x - y = 0\) and \(3x^2 - 4y = 0\). The task is to find the critical points by solving these equations simultaneously.
02

Express y in terms of x

From the equation \(x - y = 0\), we can express \(y\) in terms of \(x\):\[ y = x \].
03

Substitute y in the second equation

Substitute \(y = x\) into the second equation \(3x^2 - 4y = 0\):\[ 3x^2 - 4x = 0 \].
04

Factor the equation

Factor the equation \(3x^2 - 4x = 0\) by taking \(x\) common:\[ x(3x - 4) = 0 \].
05

Solve the factored equation

Set each factor of the equation \(x(3x - 4) = 0\) to zero:1. \(x = 0\)2. \(3x - 4 = 0\)Solving these gives \(x = 0\) and \(x = \frac{4}{3}\).
06

Determine y values

Using \(y = x\), find corresponding \(y\) values:- For \(x = 0\), \(y = 0\).- For \(x = \frac{4}{3}\), \(y = \frac{4}{3}\).
07

Identify Critical Points

The solutions \((x, y)\) give the critical points of the system:- \((0, 0)\)- \(\left( \frac{4}{3}, \frac{4}{3} \right)\).

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Key Concepts

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

Simultaneous Equations
Simultaneous equations are equations that are satisfied by the same values of variables. When solving these equations, we find a common solution that works for all equations in the system. This is essential in problems where two or more conditions must be true at the same time. In our exercise, the equations given are \(x - y = 0\) and \(3x^2 - 4y = 0\), and we aim to find the values of \(x\) and \(y\) that satisfy both equations simultaneously.
To tackle simultaneous equations, a typical approach is to either express one variable in terms of another or use both equations to eliminate one of the variables. This gives us a simpler equation to solve.
  • Expressing a variable: Convert one equation to express a variable in terms of another, like transforming \(x - y = 0\) to \(y = x\).
  • Substituting the expression into the other equation, simplifying the problem to a single equation, which completes the process of solving simultaneously.
Factoring Equations
Factoring equations involves breaking down complex expressions into simpler, multiplied factors. This technique helps us solve polynomial equations by setting each factor equal to zero.
In our exercise, after substituting \(y = x\) into \(3x^2 - 4y = 0\), we simplify it to \(3x^2 - 4x = 0\). The next step is factoring, where we take \(x\) as a common factor:
  • We factor out \(x\) to get \(x(3x - 4) = 0\).
  • This results in two separate simple equations: \(x = 0\) and \(3x - 4 = 0\).
  • Factoring simplifies the solving process as we set each factor to zero, allowing us to find the solutions for \(x\).
System of Equations
A system of equations is a collection of two or more equations with the same set of variables. Solving a system of equations means finding a set of values for the variables that satisfies all the equations simultaneously.
In our problem, the system includes two equations: \(x - y = 0\) and \(3x^2 - 4y = 0\). The goal is to identify the points \((x, y)\) where these equations intersect.
  • First, determine the relationship between the variables using one equation (e.g., \(y = x\) from \(x - y = 0\)).
  • Next, apply this relationship to another equation, converting the system from two equations in two variables to one equation in a single variable.
This approach allows solving the system efficiently to find the critical points, which are shared solutions to the entire system.
Algebraic Substitution
Algebraic substitution is a method used to simplify and solve systems of equations. It involves solving one of the equations for one variable and then substituting that expression into another equation.
In our example, we start by solving the simplest equation, \(x - y = 0\), to get \(y = x\). We then substitute \(y = x\) into the second equation, \(3x^2 - 4y = 0\), to transform it into a single variable equation, \(3x^2 - 4x = 0\).
  • The substitution method is particularly useful when one equation is simple and can be rearranged easily.
  • By reducing the system from two variables to one, it simplifies the algebra required to solve the system.
Ultimately, substitution is a very powerful tool for solving systems of equations, making it easier to find critical points.

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Most popular questions from this chapter

since \(\mathbf{A}^{3}=\mathbf{0}, \mathbf{A}\) is nilpotent. since $$e^{\mathbf{A} t}=\mathbf{I}+\mathbf{A} t+\mathbf{A}^{2} \frac{t^{2}}{2 !}+\cdots+\mathbf{A}^{k} \frac{t^{k}}{k !}+\cdots.$$ if \(\mathbf{A}\) is nilpotent and \(\mathbf{A}^{m}=\mathbf{0},\) then \(\mathbf{A}^{k}=\mathbf{0}\) for \(k \geq m\) and $$e^{\mathbf{A} t}=\mathbf{I}+\mathbf{A} t+\mathbf{A}^{2} \frac{t^{2}}{2 !}+\cdots+\mathbf{A}^{m-1} \frac{t^{m-1}}{(m-1) !}.$$ In this problem \(\mathbf{A}^{3}=\mathbf{0},\) so $$\begin{aligned} e^{\mathbf{A} t}=\mathbf{I}+\mathbf{A} t+\mathbf{A}^{2} \frac{t^{2}}{2} &=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)+\left(\begin{array}{rrr} -1 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right) t+\left(\begin{array}{rrr} -1 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{array}\right) \frac{t^{2}}{2} \\ &=\left(\begin{array}{ccc} 1-t-t^{2} / 2 & t & t+t^{2} / 2 \\ -t & 1 & t \\ -t-t^{2} / 2 & t & 1+t+t^{2} / 2 \end{array}\right) \end{aligned}$$ and the solution of \(\mathbf{X}^{\prime}=\mathbf{A X}\) is $$\mathbf{X}(t)=e^{\mathbf{A} t} \mathbf{C}=e^{\mathbf{A} t}\left(\begin{array}{l} c_{1} \\ c_{2} \\ c_{3} \end{array}\right)=\left(\begin{array}{c} c_{1}\left(1-t-t^{2} / 2\right)+c_{2} t+c_{3}\left(t+t^{2} / 2\right) \\ -c_{1} t+c_{2}+c_{3} t \\ c_{1}\left(-t-t^{2} / 2\right)+c_{2} t+c_{3}\left(1+t+t^{2} / 2\right) \end{array}\right).$$

\(\tau=2 \alpha, \Delta=\alpha^{2}+\beta^{2}>0,\) and \(\tau^{2}-4 \Delta=-4 \beta<0 .\) If \(\alpha<0,(0,0)\) is a stable spiral point. If \(\alpha>0,(0,0)\) is an unstable spiral point. Therefore (0,0) cannot be a node or saddle point.

We have $$\mathbf{X}(0)=c_{1}\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right)+c_{2}\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)+c_{3}\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right)=\left(\begin{array}{l} c_{1} \\ c_{2} \\ c_{3} \end{array}\right)=\left(\begin{array}{r} 1 \\ -4 \\ 6 \end{array}\right).$$ Thus, the solution of the initial-value problem is $$\mathbf{X}=\left(\begin{array}{c} t+1 \\ t \\ -2 t \end{array}\right)-4\left(\begin{array}{c} t \\ t+1 \\ -2 t \end{array}\right)+6\left(\begin{array}{c} t \\ t \\ -2 t+1 \end{array}\right).$$

The corresponding plane autonomous system is \\[ x^{\prime}=y, \quad y^{\prime}=-x+\epsilon x^{3} \\] If \((x, y)\) is a critical point, \(y=0\) and \(-x+\epsilon x^{3}=0 .\) Hence \(x\left(-1+\epsilon x^{2}\right)=0\) and so \(x=0, \sqrt{1 / \epsilon},-\sqrt{1 / \epsilon} .\) The critical points are \((0,0),(\sqrt{1 / \epsilon}, 0)\) and \((-\sqrt{1 / \epsilon}, 0)\)

If we let \(d x / d t=y,\) then \(d y / d t=-x^{3}-x .\) From this we obtain the first-order differential equation $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=-\frac{x^{3}+x}{y}$$ Separating variables and integrating we obtain $$\int y d y=-\int\left(x^{3}+x\right) d x$$ and $$\frac{1}{2} y^{2}=-\frac{1}{4} x^{4}-\frac{1}{2} x^{2}+c_{1}$$ Completing the square we can write the solution as \(y^{2}=-\frac{1}{2}\left(x^{2}+1\right)^{2}+c_{2} .\) If \(\mathbf{X}(0)=\left(x_{0}, 0\right),\) then \(c_{2}=\frac{1}{2}\left(x_{0}^{2}+1\right)^{2}\) and so $$\begin{aligned} y^{2} &=-\frac{1}{2}\left(x^{2}+1\right)^{2}+\frac{1}{2}\left(x_{0}^{2}+1\right)^{2}=\frac{x_{0}^{4}+2 x_{0}^{2}+1-x^{4}-2 x^{2}-1}{2} \\ &=\frac{\left(x_{0}^{2}+x^{2}\right)\left(x_{0}^{2}-x^{2}\right)+2\left(x_{0}^{2}-x^{2}\right)}{2}=\frac{\left(x_{0}^{2}+x^{2}+2\right)\left(x_{0}^{2}-x^{2}\right)}{2} \end{aligned}$$ Note that \(y=0\) when \(x=-x_{0} .\) In addition, the right-hand side is positive for \(-x_{0}
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