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Chapter 11: Problem 14

From \(\sin y=0\) we have \(y=\pm n \pi .\) From \(e^{x-y}=1,\) we can conclude that \(x-y=0\) or \(x=y .\) The critical points of the system are therefore \((\pm n \pi, \pm n \pi)\) for \(n=0,1,2, \ldots\)

Short Answer

Expert verified
The critical points are \((\pm n \pi, \pm n \pi)\) for \(n = 0, 1, 2, \ldots\).

Step by step solution

01

Understanding the First Equation

The given equation is \( \sin y = 0 \). The sine function is zero at integer multiples of \( \pi \), which is why the solutions for \( y \) are \( y = \pm n \pi \), where \( n \) is an integer (\( n = 0, 1, 2, \ldots \)). This gives us possible values for \( y \).
02

Analyzing the Second Equation

The next equation is \( e^{x-y} = 1 \). We know that \( e^0 = 1 \), thus \( x-y = 0 \). From this, we conclude that \( x = y \). This suggests that whatever value \( y \) takes, \( x \) must take the same value.
03

Finding the Critical Points

Given that \( y = \pm n \pi \) from the first equation and \( x = y \) from the second equation, the critical points are formed by pairs where \( x = y = \pm n \pi \). Hence, the critical points of this system are \( (\pm n \pi, \pm n \pi) \) for \( n = 0, 1, 2, \ldots \).

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Key Concepts

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

Trigonometric Functions
Trigonometric functions play a crucial role in mathematics, especially in solving problems involving angles and periodic phenomena. One of the most fundamental trigonometric functions is the sine function, denoted as \( \sin(y) \).

Understanding when the sine function equals zero is key to solving the original exercise. The sine of an angle, \( y \), becomes zero at integral multiples of \( \pi \), such as \( 0, \pi, -\pi, 2\pi, \), and so forth. This means for \( \sin(y) = 0 \), the solutions are \( y = \pm n \pi \), where \( n \) is any non-negative integer.

This periodic nature is due to the wave shape of trigonometric functions, which repeat values in a regular cycle. Recognizing these cycles helps in pinpointing specific solutions and simplifying trigonometric computations.
Exponential Functions
Exponential functions, such as \( e^{x} \), are a backbone of continuous growth and decay models in natural sciences and finance. In our problem, we encounter the equation \( e^{x-y} = 1 \).

Understanding this requires knowing that any base raised to the power of zero equals one, so \( e^{0} = 1 \). Thus, for \( e^{x-y} = 1 \), it must be that \( x - y = 0 \). This mean \( x = y \).

Exponential functions have the property of rapid change, but they also affect systems of equations by implementing constraints like in the equation above. This reveals more about the relationship between \( x \) and \( y \): they must be equal for the equation to hold true.

Incorporating exponential functions into problems helps in understanding growth rates and interactions between variables in mathematical systems.
Mathematical Solutions
Solving a system of equations involves finding all possible solutions for the variables that satisfy every equation simultaneously. In the given exercise, we combined trigonometric and exponential functions to find critical points.

First, the trigonometric solution \( y = \pm n \pi \) provides discrete values for \( y \). Then, using this result in the exponential equation \( x - y = 0 \) implies \( x = y \), which effectively couples the solutions of \( x \) with those of \( y \).

Combining these, we find that the critical points are those where \( x \) and \( y \) are equal, specifically \( (\pm n \pi, \pm n \pi) \) for any non-negative integer \( n \). These solutions suggest a structured pairing of values that satisfy both equations in the system.

Understanding how to combine solutions from different types of equations is essential in mathematics. It helps create a cohesive strategy to tackle complex problems through a logical sequence of steps.

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Most popular questions from this chapter

Critical points are (1,0) and \((-1,0),\) and $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{rr} 2 x & -2 y \\ 0 & 2 \end{array}\right)$$ At \(\mathbf{X}=(1,0), \tau=4, \Delta=4,\) and so \(\tau^{2}-4 \Delta=0 .\) We can conclude that (1,0) is unstable but we are unable to classify this critical point any further. At \(\mathbf{X}=(-1,0), \Delta=-4<0\) and so (-1,0) is a saddle point.

We have \(d y / d x=y^{\prime} / x^{\prime}=-f(x) / y\) and so, using separation of variables, $$\frac{y^{2}}{2}=-\int_{0}^{x} f(\mu) d \mu+c \quad \text { or } \quad y^{2}+2 F(x)=c$$ . We can conclude that for a given value of \(x\) there are at most two corresponding values of \(y .\) If (0,0) were a stable spiral point there would exist an \(x\) with more than two corresponding values of \(y .\) Note that the condition \(f(0)=0\) is required for (0,0) to be a critical point of the corresponding plane autonomous system \(x^{\prime}=y, y^{\prime}=-f(x)\).

The corresponding plane autonomous system is $$x^{\prime}=y, \quad y^{\prime}=-x+\left(\frac{1}{2}-3 y^{2}\right) y-x^{2}$$ If \((x, y)\) is a critical point, \(y=0\) and \(s o-x-x^{2}=-x(1+x)=0 .\) Therefore (0,0) and (-1,0) are the only two critical points. We next use the Jacobian matrix $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} 0 & 1 \\ -1-2 x & \frac{1}{2}-9 y^{2} \end{array}\right)$$ to classify these critical points. For \(\mathbf{X}=(0,0), \tau=1 / 2, \Delta=1,\) and \(\tau^{2}-4 \Delta<0 .\) Therefore (0,0) is an unstable spiral point. For \(\mathbf{X}=(-1,0), \tau=1 / 2, \Delta=-1\) and so (-1,0) is a saddle point.

The equation $$x^{\prime}=\alpha \frac{y}{1+y} x-x=x\left(\frac{\alpha y}{1+y}-1\right)=0$$ implies that \(x=0\) or \(y=1 /(\alpha-1) .\) When \(\alpha>0, \hat{y}=1 /(\alpha-1)>0 .\) If \(x=0,\) then from the differential equation for \(y^{\prime}, y=\beta .\) On the other hand, if \(\hat{y}=1 /(\alpha-1), \hat{y} /(1+\hat{y})=1 / \alpha\) and so \(\hat{x} / \alpha-1 /(\alpha-1)+\beta=0\) It follows that $$\hat{x}=\alpha\left(\beta-\frac{1}{\alpha-1}\right)=\frac{\alpha}{\alpha-1}[(\alpha-1) \beta-1]$$ and if \(\beta(\alpha-1)>1, \hat{x}>0 .\) Therefore \((\hat{x}, \hat{y})\) is the unique critical point in the first quadrant. The Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} \alpha \frac{y}{y+1}-1 & \frac{\alpha x}{(1+y)^{2}} \\ -\frac{y}{1+y} & \frac{-x}{(1+y)^{2}}-1 \end{array}\right)$$ and for \(\mathbf{X}=(\hat{x}, \hat{y}),\) the Jacobian can be written in the form $$\mathbf{g}^{\prime}((\hat{x}, \hat{y}))=\left(\begin{array}{cc} 0 & \frac{(\alpha-1)^{2}}{\alpha} \hat{x} \\ -\frac{1}{\alpha} & -\frac{(\alpha-1)^{2}}{\alpha^{2}}-1 \end{array}\right)$$ It follows that $$\tau=-\left[\frac{(\alpha-1)^{2}}{\alpha^{2}} \hat{x}+1\right]<0, \quad \Delta=\frac{(\alpha-1)^{2}}{\alpha^{2}} \hat{x}$$ and so \(\tau=-(\Delta+1) .\) Therefore \(\tau^{2}-4 \Delta=(\Delta+1)^{2}-4 \Delta=(\Delta-1)^{2}>0 .\) Therefore \((\hat{x}, \hat{y})\) is a stable node.

From \(x\left(1-x^{2}-3 y^{2}\right)=0\) we have \(x=0\) or \(x^{2}+3 y^{2}=1 .\) If \(x=0,\) then substituting into \(y\left(3-x^{2}-3 y^{2}\right)\) gives \(y\left(3-3 y^{2}\right)=0 .\) Therefore \(y=0,1,-1 .\) Likewise \(x^{2}=1-3 y^{2}\) yields \(2 y=0\) so that \(y=0\) and \(x^{2}=1-3(0)^{2}=1\) The critical points of the system are therefore \((0,0),(0,1),(0,-1),(1,0),\) and (-1,0)
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