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Consider the initial value problem $$ y^{\prime}=y^{2}-t^{2}, \quad y(0)=\alpha $$ where \(\alpha\) is a given number. (a) Draw a direction field for the differential equation. Observe that there is a critical value of \(\alpha\) in the interval \(0 \leq \alpha \leq 1\) that separates converging solutions from diverging ones. Call this critical value \(\alpha_{\theta}\). (b) Use Euler's method with \(h=0.01\) to estimate \(\alpha_{0} .\) Do this by restricting \(\alpha_{0}\) to an interval \([a, b],\) where \(b-a=0.01 .\)

Short Answer

Expert verified
Answer: The critical value, \(\alpha_{\theta}\), is a specific initial condition for the given differential equation that separates converging and diverging solutions. It can be roughly observed from the direction field plot. To estimate this value numerically, implement Euler's method with a step size of \(h=0.01\), iterating through possible values of \(\alpha\). As \(\alpha\) approaches \(\alpha_{\theta}\), the numerical solution will show a significant change in behavior, indicating the presence of the critical value.

Step by step solution

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01

Draw the direction field for the given differential equation

To visualize the behavior of the solutions of the ODE, we will draw the direction field, or slope field. A direction field is a graphical representation that shows the slope of the tangent line along with the solution curve. In the given equation \(y^{\prime}=y^2-t^2\), plug in different values of \(t\) and \(y\) to calculate the corresponding slopes. Then, draw small line segments at each grid point, matching the slope of the solution passing through that point.
02

Observe the critical value separating converging and diverging solutions

From the plot of the direction field, we can visualize the behavior of the solutions. Observe the critical value \(\alpha_{\theta}\) that separates converging solutions from diverging ones. The slopes in the initial condition indicate that there exists such a critical value \(\alpha_{\theta}\) that separates these two types of solutions.
03

Euler's method for solving the initial value problem

The next step is to use Euler's method to estimate the critical value, \(\alpha_{\theta}\). Recall that Euler's method is a numerical method used to approximate solutions of initial value problems. The formula for the Euler method is given by: $$ y_{i+1}=y_i + hf(t_i,y_i) $$ Where \(y_{i+1}\) and \(y_i\) are the numerical approximations for \(y(t)\) at points \(t_{i+1}\) and \(t_i\), respectively, and \(h\) is the step size.
04

Implement Euler's method with \(h=0.01\) to estimate \(\alpha_{\theta}\)

Using Euler's method with the given step size \(h=0.01\), we can approximate \(\alpha_{\theta}\). First, restrict the value of \(\alpha_{\theta}\) to an interval \([a, b]\) such that \(b-a=0.01\). Then, loop through the interval \([a, b]\) using the Euler's method formula to compute the approximate solution. As the value of \(\alpha\) approaches \(\alpha_{\theta}\), the numerical solution should show a significant change in its behavior, indicating the presence of the critical value separating the two types of solutions.

Key Concepts

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

Euler's Method
When dealing with the complex landscapes of differential equations, often the exact solution is difficult to pinpoint. That's where Euler's method steps in as a numerical lifeline. This basic yet powerful algorithm approximates the solutions of initial value problems, performing like a mathematical compass, guiding us through the unknown territory one small step at a time.

To understand how this method works, imagine plotting a course on a map. At each step, you use your current location and direction to move a tiny bit forward. The direction and distance covered are determined by the slope of the differential equation, much like the local terrain's steepness would dictate your walking path. Using Euler's method, you start with an initial condition given by the problem, and then you move forward step by step. Each step size is represented by the variable 'h.' As the steps accumulate, a path—a numerical solution to the equation—emerges.

Applying this method involves the calculation: \[y_{i+1} = y_i + h \times f(t_i,y_i)\]. This equation means that the new estimated point \(y_{i+1}\) is the sum of the previous point \(y_i\) and the product of your step size 'h' and the slope at that point. It's like making a guess of where you'll be after taking a stride based on your current direction.
Direction Field
The direction field, or slope field, is the visual storyteller of differential equations. It shows a bird's-eye view of how solutions to these equations behave without solving them analytically. Each function solution is a path on this field, and the arrows on the field represent the slope or derivative at each point. It's akin to seeing numerous tiny compass needles on a map, each pointing to the local 'north' dictating the slope of the tangent to the solution curve.

Creating a direction field involves calculating the slopes at numerous points in the region of interest. For the given exercise, you calculate the slope (the value of the derivative \(y'\)) at various coordinate points \((t, y)\) according to the differential equation \(y' = y^2 - t^2\). Each line segment or arrow represents the slope at that point and gives a glimpse into the path the solution curve would take.

As this field takes shape, it segments into zones, each with its distinct flow determined by the equation. Notably, an initial value problem's direction field allows students to visually predict how different initial conditions will influence the behavior of solutions.
Differential Equations
Differential equations are the language of change. They are mathematical equations that tie a function to its derivatives, describing how phenomena evolve over time or space. Understanding differential equations means you're deciphering the undercurrents of motion, growth, and dynamics at play in countless scientific and engineering contexts.

The differential equation in this exercise, \(y' = y^2 - t^2\), is a canvas where \(y'\) (the first derivative) sketches out how the function \(y\) changes with respect to time \(t\). This particular equation is nonlinear, indicated by the square of \(y\) - meaning its solutions offer a rich tapestry of behaviors. Solving this equation analytically can be challenging – if not impossible – but numerical methods like Euler's method provide a pragmatic approach to deciphering what the function \(y\) looks like as time marches on.
Critical Value
A critical value, in the context of differential equations, is that special number that acts as a watershed between different solution behaviors. For an orderly queue of parameters leading to solutions, this is the one that either makes or breaks a pattern.

Imagine you are hiking up a hill and the path forks. One way gently slopes down, leading you back safely, while the other path steepens, propelling you to unknown heights. This junction is the critical value - in this case, \(\alpha_{\theta}\) - that separates converging solutions (the descending path) from diverging ones (the ascending path).

In the given exercise, identifying \(\alpha_{\theta}\) requires keen observation of how solutions behave given different initial values \(\alpha\). With Euler's method, and by closely examining the nature of the solutions produced as \(\alpha\) varies, we focus on locating this critical juncture. In the big picture of functions' exploratory journey, estimating the critical value becomes a landmark that dictates the rates at which solutions either fan out wildly or merge serenely.

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Most popular questions from this chapter

(a) Solve the Gompertz equation $$ d y / d t=r y \ln (K / y) $$ subject to the initial condition \(y(0)=y_{0}\) (b) For the data given in Example 1 in the text \([ \leftr=0.71 \text { per year, } K=80.5 \times 10^{6} \mathrm{kg}\), \right. \(\left.y_{0} / K=0.25\right]\), use the Gompertz model to find the predicted value of \(y(2) .\) (c) For the same data as in part (b), use the Gompertz model to find the time \(\tau\) at which \(y(\tau)=0.75 K .\) Hint: You may wish to let \(u=\ln (y / K)\).

Consider the initial value problem \(y^{\prime}=y^{1 / 3}, y(0)=0\) from Example 3 in the text. (a) Is there a solution that passes through the point \((1,1) ?\) If so, find it. (b) Is there a solution that passes through the point \((2,1)\) ? If so, find it. (c) Consider all possible solutions of the given initial value problem. Determine the set of values that these solutions have at \(t=2\)

Show that the equations are not exact, but become exact when multiplied by the given integrating factor. Then solve the equations. $$ x^{2} y^{3}+x\left(1+y^{2}\right) y^{\prime}=0, \quad \mu(x, y)=1 / x y^{3} $$

Draw a direction field for the given differential equation and state whether you think that the solutions are converging or diverging. $$ y^{\prime}=5-3 \sqrt{y} $$

Convergence of Euler's Method. It can be shown that, under suitable conditions on \(f\) the numerical approximation generated by the Euler method for the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=y_{0}\) converges to the exact solution as the step size \(h\) decreases. This is illustrated by the following example. Consider the initial value problem $$ y^{\prime}=1-t+y, \quad y\left(t_{0}\right)=y_{0} $$ (a) Show that the exact solution is \(y=\phi(t)=\left(y_{0}-t_{0}\right) e^{t-t_{0}}+t\) (b) Using the Euler formula, show that $$ y_{k}=(1+h) y_{k-1}+h-h t_{k-1}, \quad k=1,2, \ldots $$ (c) Noting that \(y_{1}=(1+h)\left(y_{0}-t_{0}\right)+t_{1},\) show by induction that $$ y_{n}=(1+h)^{x}\left(y_{0}-t_{0}\right)+t_{n} $$ for each positive integer \(n .\) (d) Consider a fixed point \(t>t_{0}\) and for a given \(n\) choose \(h=\left(t-t_{0}\right) / n .\) Then \(t_{n}=t\) for every \(n .\) Note also that \(h \rightarrow 0\) as \(n \rightarrow \infty .\) By substituting for \(h\) in \(\mathrm{Eq}\). (i) and letting \(n \rightarrow \infty,\) show that \(y_{n} \rightarrow \phi(t)\) as \(n \rightarrow \infty\). Hint: \(\lim _{n \rightarrow \infty}(1+a / n)^{n}=e^{a}\).

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