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Chapter 8: Problem 20

Find the equilibrium solution of the following equations, make a sketch of the direction field, for \(t \geq 0,\) and determine whether the equilibrium solution is stable. The direction field needs to indicate only whether solutions are increasing or decreasing on either side of the equilibrium solution. $$y^{\prime}(t)-\frac{y}{4}-1=0$$

Short Answer

Expert verified
Based on the given differential equation \(y'(t) - \frac{y}{4} - 1 = 0\), we found that the equilibrium solution is \(y = -4\). The solution is stable because when \(y < -4\) the solutions are decreasing, and when \(y > -4\) the solutions are increasing. To sketch the direction field, draw a horizontal line at \(y = -4\) and draw small arrows pointing downward when \(y < -4\) and upward when \(y > -4\).

Step by step solution

01

Find the equilibrium solution

First, let's set the derivative of the given equation equal to zero: $$y'(t)-\frac{y}{4}-1=0$$ Rearrange this equation to solve for the derivative of the function, \(y'(t)\): $$y'(t)=\frac{y}{4}+1$$ We can then set \(y'(t) = 0\) and solve for \(y\): $$0=\frac{y}{4}+1$$ $$-\frac{y}{4}=1$$ $$y=-4$$ So, the equilibrium solution is \(y = -4\).
02

Determine the stability of the equilibrium solution

Now we need to analyze the stability of the equilibrium solution by examining the sign of \(y'(t)\) on either side of \(y=-4\). Let's first consider when \(y < -4\), say \(y = -5\): $$y'(t) = \frac{-5}{4} + 1 = -\frac{1}{4}$$ Since \(y'(t) < 0\), the solutions are decreasing when \(y < -4\). Now consider when \(y > -4\), say \(y = -3\): $$y'(t) = \frac{-3}{4} + 1 = \frac{1}{4}$$ Since \(y'(t) > 0\), the solutions are increasing when \(y > -4\). Since solutions are decreasing when \(y < -4\) and increasing when \(y > -4\), we can conclude that the equilibrium solution \(y=-4\) is stable.
03

Sketch the direction field

To sketch the direction field, we need to consider the behavior of solutions on either side of the equilibrium solution \(y = -4\). As determined in the previous step, solutions are decreasing when \(y < -4\) and increasing when \(y > -4\). Keeping this in mind, we can develop the following sketch: - Draw the horizontal line \(y = -4\). - When \(y < -4\), draw small arrows pointing downward along the vertical lines (since solutions decrease). - When \(y > -4\), draw small arrows pointing upward along the vertical lines (since solutions increase).

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Most popular questions from this chapter

An object in free fall may be modeled by assuming that the only forces at work are the gravitational force and air resistance. By Newton's Second Law of Motion (mass \(\times\) acceleration \(=\) the sum of the external forces), the velocity of the object satisfies the differential equation $$\underbrace {m}_{\text {mass}}\quad \cdot \underbrace{v^{\prime}(t)}_{\text {acceleration }}=\underbrace {m g+f(v)}_{\text {external forces}}$$ where \(f\) is a function that models the air resistance (assuming the positive direction is downward). One common assumption (often used for motion in air) is that \(f(v)=-k v^{2},\) where \(k>0\) is a drag coefficient. a. Show that the equation can be written in the form \(v^{\prime}(t)=g-a v^{2},\) where \(a=k / m\) b. For what (positive) value of \(v\) is \(v^{\prime}(t)=0 ?\) (This equilibrium solution is called the terminal velocity.) c. Find the solution of this separable equation assuming \(v(0)=0\) and \(0 < v^{2} < g / a\) d. Graph the solution found in part (c) with \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) \(m=1,\) and \(k=0.1,\) and verify that the terminal velocity agrees with the value found in part (b).

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Make a sketch of the population function (as a function of time) that results from the following growth rate functions. Assume the population at time \(t=0\) begins at some positive value.

Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(t)=\frac{e^{t}}{2 y}, y(\ln 2)=1$$

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