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Chapter 17: Problem 5

Argue that \(y=\tan t \quad\) is the only solution to \(\quad y(0)=0, \quad y^{\prime}(t)=1+y^{2}(t)\) Use the integral formula, \(\int \frac{y^{\prime}(\tau)}{1+y^{2}(\tau)} d \tau=\arctan y(\tau)+C\).

Short Answer

Expert verified
The only solution is \(y(t) = \tan(t)\).

Step by step solution

01

Recognize the Integral Formula

The problem gives us an integral formula, \( \int \frac{y'(\tau)}{1+y^2(\tau)} d \tau = \arctan y(\tau) + C \). This will be fundamental in finding the solution for \( y(t) \). Before proceeding, note that \( y'(t) = 1 + y^2(t) \) is the derivative given in the problem.
02

Identifying Relevant Terms

Given \( y'(t) = 1 + y^2(t) \), recognize that the expression inside the integral's denominator, \( 1 + y^2(\tau) \), matches the derivative \( y'(t) \). This is crucial because it suggests a natural alignment with the integral's structure.
03

Applying the Integral Formula

Use the integral formula to integrate both sides of \( y'(\tau) = 1 + y^2(\tau) \). Taking the left side, we have \( \int_{0}^{t} \frac{y'(\tau)}{1 + y^2(\tau)} d \tau \). According to the given formula, this equals \( \arctan y(t) - \arctan y(0) \) assuming C is represented via the initial condition.
04

Using Initial Condition

With the initial condition \( y(0) = 0 \), we have \( \arctan y(0) = \arctan 0 = 0 \). Therefore, \( \int_{0}^{t} \frac{y'(\tau)}{1 + y^2(\tau)} d \tau = \arctan y(t) \).
05

Solving for \( y(t) \)

Knowing the integral of \( 1 \) over \([0,t]\) is simply \( t \), we equate this with \( \arctan y(t) \). Thus, \( t = \arctan y(t) \). Solving for \( y(t) \) gives us \( y(t) = \tan t \). This confirms \( y(t) = \tan t \) is a solution.
06

Uniqueness of Solution

The differential equation \( y'(t) = 1 + y^2(t) \) prescribed in the problem and its corresponding initial condition that produced \( y(t) = \tan(t) \), combined with the integral solution, suggests uniqueness due to function properties and alignment.

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Key Concepts

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

Initial Conditions
Understanding initial conditions is crucial for solving differential equations. These conditions help specify a particular solution from a family of possible solutions. In our exercise, the initial condition is given as \( y(0) = 0 \).

Initial conditions are used to determine the constant of integration when solving differential equations. Without these conditions, we would face an infinite number of solutions. They relate to where exactly we start on a graph or what specific path the function should take.

In this particular equation, the initial condition \( y(0) = 0 \) indicates how the curve should behave at \( t = 0 \). It tells us that at this point, the solution should pass through the origin (on a graph), aligning precisely with the point where the tangent (\( \tan \)) of 0 is indeed 0.
Integral Calculus
Integral calculus is a powerful tool in solving differential equations. It allows us to find antiderivatives, which helps us understand how a function accumulates over an interval.

In this exercise, the provided integral formula \( \int \frac{y'(\tau)}{1+y^2(\tau)} d \tau = \arctan y(\tau) + C \) is central to solving the equation. This formula implies that integrating the given derivative over a specific range will provide us an arc tangent relationship with \( y \), hence connecting differential and integral calculus.

The process involves identifying the correct setup to apply the integral formula by carefully matching integral parts with the corresponding differential equation terms. Thus, recognizing how the derivative \( y'(t) \) fits into the integral structure is crucial. Our solution steps involve using this integration to connect \( t \) with \( \arctan y(t) \), eventually leading us to the solution \( y(t) = \tan t \).
  • Using integral calculus, we're able to bridge the rate of change (given by the derivative) and the actual function (\( y(t) \)).
Uniqueness of Solutions
The uniqueness of a solution in a differential equation ensures that, under given conditions, only one function satisfies both the equation and initial conditions.

In this exercise, the differential equation \( y'(t) = 1 + y^2(t) \) paired with the initial condition \( y(0) = 0 \) strongly suggests a unique solution through the interaction of these components.

Typically, uniqueness is guaranteed by theorems such as the Picard-Lindel枚f theorem, which states that if a function and its partial derivative are continuous over a region including an initial condition, the solution will be unique in that region. With \( y = \tan t \), the behavior of the \( \tan \) function and its association with the derivative \( y' = 1 + y^2 \) confirm this uniqueness because the properties of \( \tan \) naturally fulfill these conditions within a specific interval.
  • Uniqueness is vital in ensuring the solution is definitive and specific to the problem setup.
  • This is especially important in physical applications, where different solutions could imply different real-world behaviors.

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Most popular questions from this chapter

The special case of \(y^{\prime}=f(t, y)\) in which \(f(t, y)=F(t)(f\) is independent of \(y)\) has a familiar solution from the Fundamental Theorem of Calculus I. Check by substitution that $$y(t)=y_{a}+\int_{a}^{t} F(x) d x \quad \text { solves } \quad y(a)=y_{a} \quad \text { and } \quad y^{\prime}(t)=F(t)$$ The differential equation $$y(a)=y_{a} \quad y^{\prime}(t)=F(t)$$ has therefore been completely solved. Henceforth we will consider that \(f\) is dependent on \(y\) and possibly also on \(t\).

a. Show that Ricker's equation, $$p^{\prime}(t)=\alpha p e^{-p / \beta}-\gamma p$$ is equivalent to $$v^{\prime}(\tau)=v e^{-v}-\gamma_{0} v$$ with the substitutions, \(u(t)=p(t) / \beta, \tau=\alpha t,\) and \(\gamma_{0}=\gamma / \alpha\). b. Show that the Beverton-Holt equation, $$p^{\prime}(t)=\frac{r \times p}{1+p / \beta}$$ is equivalent to $$v^{\prime}(\tau)=\frac{v}{1+v}$$ with proper substitutions. c. Show that the Gompertz equation, $$p^{\prime}(t)=-r \ln \frac{p}{\beta}$$ with proper substitutions, is equivalent to an equation with no parameters.

Parameter reduced population models are shown below. In the Ricker equation, find a condition on \(\gamma_{0}\) that will insure that there is a value of \(v\) for which the population is growing the fastest. In the Beverton-Holt equation show that there is no value of \(v\) for which the population is growing the fastest. In the Gompertz equation, find the value of \(v\) for which the population is growing the fastest. $$\text { a. } \quad v^{\prime}=v e^{-v}-\gamma_{0} v \text { Ricker }$$$$\begin{array}{l} \text { b. } \quad v^{\prime}=\frac{v}{1+v} \\\\\text { c. } v^{\prime}=-v \ln (v)\end{array}$$ Beverton-Holt Gompertz The three previous problems have important implications for wildlife management, at least conceptually. Suppose you are managing a wildlife population, salmon, for example, as a renewable resource, and wish to annually harvest as many salmon as possible. If you harvest too severely, the next years spawn will be low, and four years later the harvest will be limited. Your optimum strategy is to maintain the population at the level where the growth is the greatest.

Show that each solution satisfies the initial condition and the differential equation. $$\begin{aligned} &\text { Solution } \quad \text { Initial Condition } \quad \text { Differential Equation }\\\ &\text { a. } \quad y(t)=e^{2 t}+e^{t} \quad y(0)=2 \quad y^{\prime}(t)-y(t)=e^{2 t} \end{aligned}$$ $$\text { b. } \quad y(t)=\frac{1}{3} e^{t}+\frac{2}{3} e^{-2 t}, \quad y(0)=1, \quad y^{\prime}(t)+2 y(t)=e^{t}$$ c. \(\quad y(t)=t e^{t} \quad y(0)=0 \quad y^{\prime}(t)-y(t)=e^{t}\) d. \(\quad y(t)=\frac{t^{2}}{3}+\frac{1}{t}, \quad y(1)=\frac{4}{3}, \quad t \times y^{\prime}(t)+y(t)=t^{2}\) e. \(\quad y(t)=\sqrt{t+1} \quad y(0)=1 \quad y(t) \times y^{\prime}(t)=\frac{1}{2}\) f. \(\quad y(t)=\sqrt{1+t^{2}}, \quad y(0)=1, \quad y(t) \times y^{\prime}(t)=t\) \(\begin{array}{lll}\text { g. } y(t)=\sqrt{4+t^{2}} & y(0)=2 & y(t) \times y^{\prime}(t)=t\end{array}\) \(\begin{array}{lll}\text { Solution } & \text { Initial } & \text { Differential Equation }\end{array}\) Condition h. \(y(t)=\frac{1}{t+1}, \quad y(0)=1, \quad y^{\prime}(t)+(y(t))^{2}=0\) i. \(\quad y(t)=0.5+0.5 e^{-0.2 \sin t} \quad y(0)=1, \quad y^{\prime}(t)+0.2(\cos t) y(t)=0.1 \cos t\) j. \(\quad y(t)=\tan t, \quad y(0)=0, \quad y^{\prime}(t)=1+(y(t))^{2}\) k. \(y(t)=3\) \(y(0)=3, \quad y^{\prime}(t)=(y(t)-1) \times(y(t)-3) \times(y(t)-5)\) l. \(y(t)=5\) \(y(0)=5, \quad y^{\prime}(t)=(y(t)-1) \times(y(t)-3) \times(y(t)-5)\)

Some population scientists have argued that population density can get so low that reproduction will be less than natural attrition and the total population will be lost. Named the Allee effect. after W. C. Allee who wrote extensivelv about it \(^{6}\), this mav be a basis for arguing. for example that marine fishing of a certain species (Atlantic cod on Georges Bank, for example \({ }^{7}\) ee Paul Greenberg, "Four Fish, the Future of the Last Wild Food, The Penguin Press, 2010 . The model is a lot more complicated than we present here.) should be suspended, despite the presence of a small residual population. How should we modify the logistic differential equation, \(u^{\prime}=u(1-u),\) to incorporate such a threshold? Assume a fixed area and uniform density throughout the area and a threshold number, \(\epsilon .\) If the population number is less than \(\epsilon\) the population will decline; if the population number is more than \(\epsilon\) the population will increase. a. Modify the direction field for \(u^{\prime}=u(1-u)\) to account for the Allee effect. That is, a \(u, t\) plane, draw the line \(u=1\) and a threshold line \(u=\epsilon\) where \(\epsilon=0.1,\) say. Arrows below \(u=\epsilon\) should point downward; arrows between \(u=\epsilon\) and \(u=1\) should point upwards. Draw enough direction field arrows to indicate the paths of solutions for a threshold model. b. Draw the the logistic phase plane graph and the phase plane graphs for the following three candidates of a threshold logistic differential equation where \(\epsilon=0.1\). \(u^{\prime}=f(u)=u \times(1-u) \quad\) Logistic \(u^{\prime}=f_{1}(u)=u^{\frac{2}{3}} \times(u-\epsilon)^{\frac{1}{3}} \times(1-u) \quad\) Candidate 1 \(u^{\prime}=f_{2}(u)=u \times \frac{u-\epsilon}{u+\epsilon} \times(1-u)\) Candidate 2 \(u^{\prime}=f_{3}(u)=u \times(u-\epsilon) \times(1-u)\) Candidate 3
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