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draw a direction field and plot (or sketch) several solutions of the given differential equation. Describe how solutions appear to behave as \(t\) increases, and how their behavior depends on the initial value \(y_{0}\) when \(t=0\). $$ y^{\prime}=-y(3-t y) $$

Short Answer

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Question: Based on the given differential equation \(y' = -y(3 - ty)\), describe the behavior of the solutions as \(t\) increases and discuss the dependence of the solutions' behavior on the initial value \(y_0\) when \(t=0\). Answer: To analyze the behavior of the solutions, we should first sketch the direction field by drawing short line segments with slopes equal to the given equation evaluated at points in the \((t, y)\) plane. Then, we will observe the direction field to determine the trends of the solutions as \(t\) increases, such as increasing, decreasing, or oscillatory behaviors. The dependence of the solutions' behavior on the initial value \(y_0\) can be examined by analyzing the direction field at \(t=0\) and observing how the line segments change as you move along the \(y\)-axis. If the direction field varies significantly as \(y\) changes, it indicates that the solutions are sensitive to the initial values and may exhibit large differences in long-term behavior depending on \(y_0\).

Step by step solution

01

Set up the Direction Field

A direction field is a graphical representation of a given ordinary differential equation. It consists of many short line segments, which indicate the direction of the tangent of the function at the corresponding points in the \((t, y)\) plane. To create a direction field, you need to draw short line segments at regular intervals across the \((t, y)\) plane, with slopes equal to the given differential equation evaluated at each point. In this case, we have the equation \(y' = -y(3 - ty)\). We will draw short line segments at regular points across the plane, with slopes equal to the value of the right-hand side.
02

Sketching the Direction Field

To sketch the direction field, you can either do it manually or use software like Mathematica, MATLAB, or online tools like Desmos. Here's how you can sketch the direction field by hand: 1. Choose a regular grid of points across the (t, y) plane, such as every multiple of 0.5 or 1. 2. Evaluate the right-hand side of the differential equation \(-y(3 - ty)\) at each grid point, and find the slope of the tangent line at this point. 3. Draw the tangent line segments at each point, reflecting the calculated slope. This should give you an overall impression of the behavior of the solutions, and you can visually see where they will increase or decrease, as well as detecting any stationary points and equilibrium solutions.
03

Analyzing the Behavior of the Solutions and the Effect of Initial Values

After sketching the direction field, you can analyze the behavior of the solutions by looking at the line segments. 1. As \(t\) increases, the line segments of the direction field will show the trend of the solutions. If the line segments are mostly positive, the solutions will generally increase, while if the line segments are mostly negative, the solutions will decrease. Depending on where the equilibrium points are, you can also identify oscillatory behaviors or the presence of limit cycles. 2. The dependence of the solutions' behavior on initial values \(y_0\) at \(t=0\) can be observed by looking at the direction field at \(t=0\) and seeing how the line segments change as you move along the \(y\)-axis. If the direction field varies significantly as \(y\) changes, it means the solutions are sensitive to the initial values, and small changes in \(y_0\) may lead to large differences in the long-term behavior of the solutions. By following these steps, you can analyze the behavior of the solutions of the differential equation and determine the effect of the initial values on their behavior.

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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 fundamental in expressing how things change. They relate a function associated with a variable, often time, to the rates at which it changes. In our case, the differential equation is given as \( y' = -y(3 - ty) \). This tells us how the rate of change of \( y \) varies depending on both \( y \) and \( t \).
In a direction field, these equations are visualized through line segments. Each segment represents the slope or direction of the change at a given point in the \((t, y)\) plane.
This particular equation is nonlinear because of the \( ty \) term. Nonlinear differential equations can produce complex and interesting dynamics compared to linear ones, potentially leading to behaviors such as cycles or equilibrium solutions.
Initial Value Problem
An initial value problem (IVP) consists of a differential equation coupled with an initial condition. This condition specifies the value of the function at a particular time, typically \( t = 0 \). In this scenario, considering an initial value problem means understanding how different starting values \( y_0 \) affect the behavior of solutions to the differential equation as \( t \) increases.
For our equation, setting \( y_0 \) means choosing a point \((0, y_0)\) on the \( y \)-axis. The direction field then helps visualize how the solution evolves from this point in the \((t, y)\) plane.
The influence of \( y_0 \) is profound. Small variations in initial values can lead to drastically different long-term behaviors. In nonlinear systems, specific choices of \( y_0 \) might stabilize onto an equilibrium or diverge significantly. Therefore, solving an initial value problem involves tracing a path from the initial point dictated by the direction field.
Equilibrium Solutions
Equilibrium solutions occur where the rate of change is zero, meaning the solution does not evolve over time. For the differential equation \( y' = -y(3 - ty) \), equilibrium occurs when the expression \(-y(3 - ty)\) equals zero.
Solving for equilibrium points involves setting the derivative \( y' \) to zero and solving the resulting equation. In this context, equilibrium can occur when \( y = 0 \) or \( 3 = ty \). The point \( y = 0 \) is known as a trivial solution, as it means no change at all, while other values of \( y \) depend on \( t \) for non-trivial solutions.
Understanding equilibrium solutions is crucial because they indicate stable points that solutions might approach over time. In settings like direction fields, these are seen as spots where the arrows, or slopes, flatten out. These points help anticipate how the system might behave for prolonged periods, depending on the initial conditions.

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Most popular questions from this chapter

Let \(v(t)\) and \(w(t)\) be the horizontal and vertical components of the velocity of a batted (or thrown) bascball. In the absence of air resistance, \(v\) and \(w\) satisfy the equations $$ d v / d t=0, \quad d w / d t=-g $$ $$ \text { (a) Show that } $$ $$ \mathbf{v}=u \cos A, \quad w=-g t+u \sin A $$ $$ \begin{array}{l}{\text { where } u \text { is the initial speed of the ball and } A \text { is its initial angle of elevation, }} \\ {\text { (b) Let } x(t) \text { and } y(t), \text { respectively, be the horizontal and vertical coordinates of the ball at }} \\ {\text { time } t \text { . If } x(0)=0 \text { and } y(0)=h, \text { find } x(t) \text { and } y(t) \text { at any time } t} \\ {\text { (c) Let } g=32 \text { flsec', } u=125 \mathrm{ft} / \mathrm{sec}, \text { and } h=3 \mathrm{ft} \text { . Plot the trajectory of the ball for }} \\ {\text { several values of the angle } A, \text { that is, plot } x(t) \text { and } y(t) \text { parametrically. }} \\ {\text { (d) Suppose the outfield wall is at a distance } L \text { and has height } H \text { . Find a relation between }}\end{array} $$ $$ \begin{array}{l}{u \text { and } A \text { that must be satisfied if the ball is to clear the wall. }} \\ {\text { (e) Suppose that } L=350 \mathrm{ft} \text { and } H=10 \mathrm{ft} \text { . Using the relation in part (d), find (or estimate }} \\ {\text { from a plot) the range of values of } A \text { that correspond to an initial velocity of } u=110 \mathrm{ft} \text { sec. }} \\\ {\text { (f) For } L=350 \text { and } H=10 \text { find the minimum initial velocity } u \text { and the corresponding }} \\ {\text { optimal angle } A \text { for which the ball will clear the wall. }}\end{array} $$

Find the escape velocity for a body projected upward with an initial velocity \(v_{0}\) from a point \(x_{0}=\xi R\) above the surface of the earth, where \(R\) is the radius of the earth and \(\xi\) is a constant. Neglect air resistance. Find the initial altitude from which the body must be launched in order to reduce the escape velocity to \(85 \%\) of its value at the earth's surface.

Involve equations of the form \(d y / d t=f(y)\). In each problem sketch the graph of \(f(y)\) versus \(y,\) determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. $$ d y / d t=-2(\arctan y) /\left(1+y^{2}\right), \quad-\infty

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

Solve the given initial value problem and determine at least approximately where the solution is valid. $$ (2 x-y) d x+(2 y-x) d y=0, \quad y(1)=3 $$

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