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Plot the direction field for the differential equation by hand. Do this by drawing short lines of the appropriate slope centered at each of the integer valued coordinates \((t, y)\), where \(-2 \leq t \leq 2\) and \(-1 \leq y \leq 1\). $$ y^{\prime}=\left(t^{2} y\right) /\left(1+y^{2}\right) $$

Short Answer

Expert verified
Sketch short line segments at each integer coordinate based on calculated slopes.

Step by step solution

01

Understand the Function

We are given the differential equation \( y' = \frac{t^2 y}{1 + y^2} \). This equation defines the slope of the tangent lines at each point \((t, y)\) on the graph. Our goal is to sketch these slopes at integer coordinates within the given bounds.
02

Define the Grid

Set up a grid with integer coordinates \((t, y)\), where \(-2 \leq t \leq 2\) and \(-1 \leq y \leq 1\). This results in a grid with the points: \((-2, -1), (-2, 0), (-2, 1), ..., (2, 1)\).
03

Calculate Slopes

For each point \((t, y)\), substitute the values into the equation \( y' = \frac{t^2 y}{1 + y^2} \) to find the slope. For example, at \( (0, 0) \), the slope is \( y' = \frac{0^2 \cdot 0}{1 + 0^2} = 0 \). Repeat this calculation for all points.
04

Draw Direction Field

At each grid point, draw a short line with the calculated slope centered on the point. For instance, if the slope at a point is 0, draw a horizontal line; if it's positive or negative, draw it inclined accordingly.
05

Analyze the Pattern

After drawing the lines, analyze the overall pattern of the direction field. This can help understand the behavior of the solutions to the differential equation at various initial conditions.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that involve functions and their derivatives. They are used to describe various real-world phenomena where change is a constant factor, such as in physics, engineering, and biology. In our exercise, the differential equation is given as \( y' = \frac{t^2 y}{1 + y^2} \). Here, it represents the rate of change (or the slope of the function) at various points \((t, y)\) on the coordinate grid. Understanding these equations is essential because they provide insights into how a system evolves over time. By analyzing the differential equation provided, we can predict the behavior of the function owing to the changing variables involved, in this case, \(t\) and \(y\). The goal is to visualize this change using a direction field, which is a graphical representation of the slope of the solution curves at given points.
Slope Calculation
Slope calculation in the context of a direction field involves determining the steepness or incline of the tangent lines at specific points. For our differential equation, \( y' = \frac{t^2 y}{1 + y^2} \), the slope \( y' \) is found by substituting the \(t\) and \(y\) values of each grid point into the formula. For instance:
  • At \((t, y) = (0, 0)\), the slope is calculated as \( y' = \frac{0^2 \, \cdot 0}{1 + 0^2} = 0 \).
  • At \((t, y) = (1, 1)\), substitute to get \( y' = \frac{1^2 \, \times 1}{1 + 1^2} = \frac{1}{2} \).
  • Repeat these calculations for all integer coordinate points within the range.
These slope values provide the inclination angle of the short lines you draw at each grid location, representing how the solution curve behaves at that specific point.
Grid Setup
For creating a direction field manually, setting up a proper grid is crucial. The grid serves as a visual framework for plotting each slope. In this exercise, the grid is defined by integer values within the bounds \(-2 \leq t \leq 2\) and \(-1 \leq y \leq 1\). To construct the grid:
  • List all integer points \((t, y)\) that fall within these limits. This results in: \((-2, -1), (-2, 0), (-2, 1), \ldots, (2, 1)\).
  • Each point on this grid represents a location where a short line (with a slope calculated from the differential equation) will be drawn.
Establishing this structured grid allows for a systematic approach to sketching and observing the direction field, ensuring each slope is accurately represented at its correct position.
Tangent Lines
Tangent lines are essential in sketching direction fields as they visually indicate the direction and rate of change of the solution curves to the differential equation. For each grid point \((t, y)\), a tangent line is drawn with a slope calculated earlier.Here's how to draw them:
  • At each coordinate on the grid where you know the slope, draw a short line centered at the point.
  • If the slope is 0, the line is horizontal. For positive slopes, the lines incline upwards. For negative slopes, they incline downwards.
  • These lines are short, representing the instantaneous rate of change without extending too far, thus not suggesting an entire solution curve.
When all tangent lines are plotted, they form a complete direction field. This gives you a vivid conception of how solutions to the differential equation behave at and between the plotted points.

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Most popular questions from this chapter

Uniqueness is not just an abstraction designed to please theoretical mathematicians. For example, consider a cylindrical drum filled with water. A circular drain is opened at the bottom of the drum and the water is allowed to pour out. Imagine that you come upon the scene and witness an empty drum. You have no idea how long the drum has been empty. Is it possible for you to determine when the drum was full? (a) Using physical intuition only, sketch several possible graphs of the height of the water in the drum versus time. Be sure to mark the time that you appeared on the scene on your graph. (b) It is reasonable to expect that the speed at which the water leaves through the drain depends upon the height of the water in the drum. Indeed, Torricelli's law predicts that this speed is related to the height by the formula \(v^{2}=2 g h\), where \(g\) is the acceleration due to gravity near the surface of the earth. Let \(A\) and \(a\) represent the area of a cross section of the drum and drain, respectively. Argue that \(A \Delta=a v \Delta t\), and in the limit, \(A d h / d t=a v\). Show that \(d h / d t=-(a / A) \sqrt{2 g h}\). (c) By introducing the dimensionless variables \(\omega=\alpha h\) and \(s=\beta t\) and then choosing parameters $$ \alpha=\frac{1}{h_{0}} \quad \text { and } \quad \beta=\left(\frac{a}{A}\right) \sqrt{\frac{2 g}{h_{0}}}, $$ where \(h_{0}\) represents the height of a full tank, show that the equation \(d h / d t=-(a / A) \sqrt{2 g h}\) becomes \(d w / d s=-\sqrt{w}\). Note that when \(w=0\), the tank is empty, and when \(w=1\), the tank is full. (d) You come along at time \(s=s_{0}\) and note that the tank is empty. Show that the initial value problem, \(d w / d s=-\sqrt{w}\), where \(w\left(s_{0}\right)=0\), has an infinite number of solutions. Why doesn't this fact contradict the uniqueness theorem? Hint: The equation is separable and the graphs you drew in part (a) should provide the necessary hint on how to proceed.

Use a numerical solver to sketch the solution of the given initial value problem. (i) Where does your solver experience difficulty? Why? Use the image of your solution to estimate the interval of existence. (ii) For 11-14 only, find an explicit solution; then use your formula to determine the interval of existence. How does it compare with the approximation found in part (i)? $$ \frac{d y}{d t}=\frac{t-2}{y+1}, \quad y(-1)=1 $$

The equations are not exact. However, if you multiply by the given integrating factor, then you can solve the resulting exact equation. $$ \left(x^{2}+y^{2}-x\right) d x-y d y=0, \quad \mu(x, y)=\frac{1}{x^{2}+y^{2}} $$

Is it possible to find a function \(f(t, x)\) that is continuous and has continuous partial derivatives such that the functions \(x_{1}(t)=\cos t\) and \(x_{2}(t)=1-\sin t\) are both solutions to \(x^{\prime}=f(t, x)\) near \(t=\pi / 2 ?\)

Suppose that \(y d x+\left(x^{2} y-x\right) d y=0\) has an integrating factor that is a function of \(x\) alone [i.e., \(\mu=\mu(x)]\). Find the integrating factor and use it to solve the differential equation.

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