/*! 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 40 (a) From the phase portrait we s... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 40

(a) From the phase portrait we see that critical points are \(\alpha\) and \(\beta .\) Let \(X(0)=X_{0} .\) If \(X_{0} < \alpha\) we see that \(X \rightarrow \alpha\) as \(t \rightarrow \infty .\) If \(\alpha< X_{0} < \beta,\) we see that \(X \rightarrow \alpha\) as \(t \rightarrow \infty .\) If \(X_{0} > \beta,\) we see that \(X(t)\) increases in an unbounded manner, but more specific behavior of \(X(t)\) as \(t \rightarrow \infty\) is not known. (b) When \(\alpha=\beta\) the phase portrait is as shown. If \(X_{0}< \alpha,\) then \(X(t) \rightarrow \alpha\) as \(t \rightarrow \infty .\) If \(X_{0}> \alpha\) then \(X(t)\) increases in an unbounded manner. This could happen in a finite amount of time. That is, the phase portrait does not indicate that \(X\) becomes unbounded as \(t \rightarrow \infty\). (c) When \(k=1\) and \(\alpha=\beta\) the differential equation is \(d X / d t=(\alpha-X)^{2} .\) For \(X(t)=\alpha-1 /(t+c)\) we have \(d X / d t=1 /(t+c)^{2}\) and $$(\alpha-X)^{2}=\left[\alpha-\left(\alpha-\frac{1}{t+c}\right)\right]^{2}=\frac{1}{(t+c)^{2}}=\frac{d X}{d t}$$ For \(X(0)=\alpha / 2\) we obtain $$X(t)=\alpha-\frac{1}{t+2 / \alpha}$$ For \(X(0)=2 \alpha\) we obtain $$X(t)=\alpha-\frac{1}{t-1 / \alpha}$$ For \(X_{0}> \alpha, X(t)\) increases without bound up to \(t=1 / \alpha .\) For \(t> 1 / \alpha, X(t)\) increases but \(X \rightarrow \alpha\) as \(t \rightarrow \infty\).

Short Answer

Expert verified
For \( X_0 < \alpha \), \( X \to \alpha \). For \( X_0 > \beta \), \( X(t) \to \infty \). For \( \alpha = \beta \), \( X(t) \to \infty \) quickly.

Step by step solution

01

Identify Critical Points

From the problem, the critical points of the system given by the phase portrait are \( \alpha \) and \( \beta \). These are the points where the system's behavior changes.
02

Initial Condition Analysis for \( X_0 < \alpha \)

Given \( X(0) = X_0 \), if \( X_0 < \alpha \), the phase portrait shows that \( X \to \alpha \) as \( t \to \infty \). This means that the solution trajectory approaches \( \alpha \) over time, indicating stability at \( \alpha \).
03

Initial Condition Analysis for \( \alpha < X_0 < \beta \)

For \( X(0) = X_0 \) where \( \alpha < X_0 < \beta \), the system also approaches \( \alpha \) as \( t \to \infty \). This indicates that even when starting closer to \( \beta \), the system is attracted back to \( \alpha \).
04

Initial Condition Analysis for \( X_0 > \beta \)

When \( X_0 > \beta \), \( X(t) \) increases without bound, meaning it diverges to infinity as \( t \to \infty \). The specific behavior is unbknown from the phase portrait.
05

Analyze Scenario \( \alpha = \beta \)

When \( \alpha = \beta \), the critical point is \( \alpha \). If \( X_0 < \alpha \), \( X(t) \to \alpha \) as \( t \to \infty \). If \( X_0 > \alpha \), \( X(t) \) increases without bound potentially in finite time.
06

Solve Differential Equation for \( k = 1, \alpha = \beta \)

Given the differential equation \( \frac{dX}{dt} = (\alpha - X)^2 \) and using the solution \( X(t) = \alpha - \frac{1}{t+c} \), it follows that \( \frac{dX}{dt} = \frac{1}{(t+c)^2} \) which matches \( (\alpha - X)^2 \).
07

Calculate Specific Solutions for Given Initial Conditions

For \( X(0) = \frac{\alpha}{2} \), the solution is \( X(t) = \alpha - \frac{1}{t + \frac{2}{\alpha}} \). For \( X(0) = 2\alpha \), the solution is \( X(t) = \alpha - \frac{1}{t - \frac{1}{\alpha}} \). In either case, for \( X_0 > \alpha \), \( X(t) \) increases unboundedly until \( t = \frac{1}{\alpha} \), then \( X \approx \alpha \) as \( t \to \infty \).

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Key Concepts

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

Understanding Critical Points
In phase portrait analysis, **critical points** play a crucial role.These points, often referred to as equilibrium points, are where the derivative of the function is zero.This means that at these points, the system exhibits a balance, and there is no change in the state variable with respect to time.
For our exercise, the critical points are denoted by \( \alpha \) and \( \beta \).These are the locations where the behavior of the differential equation solutions significantly alters.In a practical sense, critical points act as attractors or repulsors, guiding the trajectory of solutions over time.
Identifying these points helps in determining the long-term behavior of systems modeled by differential equations.They are pivotal in analyzing whether solutions will converge, diverge, or reach a steady pattern.
Analyzing Differential Equation Solutions
Differential equations model how variables change over time, and the solution to these equations describes the trajectory of this change.In our specific example, the differential equation \( \frac{dX}{dt} = (\alpha - X)^2 \) provides insights into the system's dynamics.
Solving these equations involves finding a function \( X(t) \) that satisfies the relationship dictated by the equation.The solutions to the given differential equation take the form \( X(t) = \alpha - \frac{1}{t+c} \).These solutions let us predict how the value of \( X \) evolves over time, given initial conditions.
For instance, if \( X(0) = \frac{\alpha}{2} \), it dictates the constant \( c \) and helps detail the trajectory \( X(t) \) will follow.The solutions imply that, depending on initial conditions, \( X(t) \) could increase indefinitely or approach a particular value, usually one of the critical points, as \( t \to \infty \).
Performing Stability Analysis
**Stability analysis** assesses how the system behaves as time progresses, particularly focusing on the response of the system to small disturbances.When critical points are identified, understanding their stability is key to predicting the system's future behavior.
A critical point is considered stable if small deviations from this point result in trajectories that return to the point.For instance, if \( X_0 < \alpha \), the phase portrait suggests the system will converge back to \( \alpha \), showing stability.This indicates that \( \alpha \) is a stable point in our system.
Conversely, if \( X_0 > \beta \), \( X(t) \) increases without bounds鈥攑otentially indicating an unstable scenario or divergence.This means the system will not return to its original point, reflecting instability.Performing stability analysis is essential in various fields, like engineering and economics, where anticipating system behavior is crucial to design and decision-making processes.

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Most popular questions from this chapter

An integrating factor for \(y^{\prime}-2 x y=1\) is \(e^{-x^{2}} .\) Thus \\[ \begin{aligned} \frac{d}{d x}\left[e^{-x^{2}} y\right] &=e^{-x^{2}} \\ e^{-x^{2}} y &=\int_{0}^{x} e^{-t^{2}} d t=\frac{\sqrt{\pi}}{2} \operatorname{erf}(x)+c \\ y &=\frac{\sqrt{\pi}}{2} e^{x^{2}} \operatorname{erf}(x)+c e^{x^{2}} \end{aligned} \\] From \(y(1)=(\sqrt{\pi} / 2) e \operatorname{erf}(1)+c e=1\) we get \(c=e^{-1}-\frac{\sqrt{\pi}}{2} \operatorname{erf}(1) .\) The solution of the initial-value problem is \\[ y=\frac{\sqrt{\pi}}{2} e^{x^{2}} \operatorname{erf}(x)+\left(e^{-1}-\frac{\sqrt{\pi}}{2} \operatorname{erf}(1)\right) e^{x^{2}} \\] \\[ =e^{x^{2}-1}+\frac{\sqrt{\pi}}{2} e^{x^{2}}(\operatorname{erf}(x)-\operatorname{erf}(1)) \\]

We first note that \(s(t)+i(t)+r(t)=n .\) Now the rate of change of the number of susceptible persons, \(s(t)\) is proportional to the number of contacts between the number of people infected and the number who are susceptible; that is, \(d s / d t=-k_{1} s i .\) We use \(-k_{1}<0\) because \(s(t)\) is decreasing. Next, the rate of change of the number of persons who have recovered is proportional to the number infected; that is, \(d r / d t=k_{2} i\) where \(k_{2}>0\) since \(r\) is increasing. Finally, to obtain \(d i / d t\) we use \\[\frac{d}{d t}(s+i+r)=\frac{d}{d t} n=0\\] This gives \\[\frac{d i}{d t}=-\frac{d r}{d t}-\frac{d s}{d t}=-k_{2} i+k_{1} s i\\] The system of differential equations is then $$\begin{aligned}&\frac{d s}{d t}=-k_{1} s i\\\&\frac{d i}{d t}=-k_{2} i+k_{1} s i\\\&\frac{d r}{d t}=k_{2} i\end{aligned}$$ A reasonable set of initial conditions is \(i(0)=i_{0},\) the number of infected people at time \(0, s(0)=n-i_{0},\) and \(r(0)=0\).

(a) Initially the tank contains 300 gallons of solution. since brine is pumped in at a rate of 3 gal/min and the mixture is pumped out at a rate of 2 gal/min, the net change is an increase of 1 gal/min. Thus, in 100 minutes the tank will contain its capacity of 400 gallons. (b) The differential equation describing the amount of salt in the tank is \(A^{\prime}(t)=6-2 A /(300+t)\) with solution $$A(t)=600+2 t-\left(4.95 \times 10^{7}\right)(300+t)^{-2}, \quad 0 \leq t \leq 100$$ as noted in the discussion following Example 5 in the text. Thus, the amount of salt in the tank when it overflows is $$A(100)=800-\left(4.95 \times 10^{7}\right)(400)^{-2}=490.625 \mathrm{lbs}$$. (c) When the tank is overflowing the amount of salt in the tank is governed by the differential equation $$\begin{aligned} \frac{d A}{d t} &=(3 \mathrm{gal} / \mathrm{min})(2 \mathrm{lb} / \mathrm{gal})-\left(\frac{A}{400} \mathrm{lb} / \mathrm{gal}\right)(3 \mathrm{gal} / \mathrm{min}) \\ &=6-\frac{3 A}{400}, \quad A(100)=490.625 \end{aligned}$$ Solving the equation, we obtain \(A(t)=800+c e^{-3 t / 400} .\) The initial condition yields \(c=-654.947,\) so that $$A(t)=800-654.947 e^{-3 t / 400}$$. When \(t=150, A(150)=587.37\) lbs. (d) As \(t \rightarrow \infty,\) the amount of salt is 800 lbs, which is to be expected since \((400 \text { gal })(2 \mathrm{lb} / \mathrm{gal})=800 \mathrm{lbs}\) (e)

For \(y^{\prime}+\frac{1}{x+1} y=\frac{\ln x}{x+1}\) an integrating factor is \(e^{\int[1 /(x+1)] d x}=x+1\) so that \(\frac{d}{d x}[(x+1) y]=\ln x\) and \(y=\frac{x}{x+1} \ln x-\frac{x}{x+1}+\frac{c}{x+1}\) for \(0< x<\infty .\) If \(y(1)=10\) then \(c=21\) and \(y=\frac{x}{x+1} \ln x-\frac{x}{x+1}+\frac{21}{x+1}\).

Using separation of variables to solve \(d T / d t=k\left(T-T_{m}\right)\) we get \(T(t)=T_{m}+c e^{k t} .\) Using \(T(0)=70\) we find \(c=70-T_{m},\) so \(T(t)=T_{m}+\left(70-T_{m}\right) e^{k t} .\) Using the given observations, we obtain $$\begin{array}{c} T\left(\frac{1}{2}\right)=T_{m}+\left(70-T_{m}\right) e^{k / 2}=110 \\ T(1)=T_{m}+\left(70-T_{m}\right) e^{k}=145 \end{array}$$ Then, from the first equation, \(e^{k / 2}=\left(110-T_{m}\right) /\left(70-T_{m}\right)\) and $$\begin{aligned} e^{k}=\left(e^{k / 2}\right)^{2}=\left(\frac{110-T_{m}}{70-T_{m}}\right)^{2} &=\frac{145-T_{m}}{70-T_{m}} \\ \frac{\left(110-T_{m}\right)^{2}}{70-T_{m}} &=145-T_{m} \\ 12100-220 T_{m}+T_{m}^{2} &=10150-250 T_{m}+T_{m}^{2} \\ T_{m} &=390 \end{aligned}$$ The temperature in the oven is \(390^{\circ}\).
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