Problem 21

Question

For Problems (a) Determine all equilibrium solutions. (b) Determine the regions in the $x y$ -plane where the solutions are increasing, and determine those regions where they are decreasing. (c) Determine the regions in the $x y$ -plane where the solution curves are concave up, and determine those regions where they are concave down. (d) Sketch representative solution curves in each region of the $x y$ -plane identified in (c) $$y^{\prime}=y(y-1)(y+1)$$

Step-by-Step Solution

Verified

Answer

In summary, the equilibrium solutions for the given differential equation $y'=y(y-1)(y+1)$ are $y=-1$, $y=0$, and $y=1$. The solutions are increasing in regions $-\infty < y < -1$ and $1 < y < \infty$, and decreasing in regions $-1 < y < 0$ and $0 < y < 1$. Moreover, the solutions are concave up in the region $-\frac{1}{\sqrt{3}} < y < \frac{1}{\sqrt{3}}$, and concave down in regions $-\infty < y <-\frac{1}{\sqrt{3}}$ and $\frac{1}{\sqrt{3}} < y < \infty$. By sketching the solution curves, we can observe the different behaviors of the solutions based on their initial conditions in the xy-plane.

1Step 1: Set the derivative equal to zero.

Set the given equation equal to zero to find the equilibrium solutions: $y'(x) = y(y - 1)(y + 1) = 0$.

2Step 2: Solve for y.

From the equation above, we have three values for y where the derivative equals zero: $y = 0$, $y = 1$, and $y = -1$. These are the equilibrium solutions. #b. Determine regions where solutions are increasing and decreasing.#

3Step 3: Evaluate y' in different y-intervals.

We have three equilibrium solutions, split the real line into four intervals: $-∞ < y < -1$, $-1 < y < 0$, $0 < y < 1$, and $1 < y < ∞$. Analyze the sign of $y'(x)$ in each of these intervals: 1. When $-∞ < y < -1$, all terms in the DE are positive ($y < 0$, $y-1 < 0$, and $y+1 < 0$). Therefore, $y'(x) > 0$. 2. When $-1 < y < 0$, only the term $(y+1)$ is positive. Therefore, $y'(x) < 0$. 3. When $0 < y < 1$, only the term $(y-1)$ is negative. Therefore, $y'(x) < 0$. 4. When $1 < y < ∞$, all terms in the DE are positive ($y > 0$, $y-1 > 0$, and $y+1 > 0$). Therefore, $y'(x) > 0$. Thus, solutions are decreasing in regions $-1 < y < 0$ and $0 < y < 1$, while solutions are increasing in regions $-∞ < y < -1$ and $1 < y < ∞$. #c. Determine regions where solutions are concave up and concave down.#

4Step 4: Compute the second derivative.

Differentiate $y'$ with respect to x to find the second derivative $y''(x)$: $y''(x) = \frac{d}{dx}(y'(x)) = \frac{d}{dx}(y(y-1)(y+1))$ Using the product rule and chain rule, we get: $y''(x) = (1)(y-1)(y+1) + y(1)(y+1) + y(y-1)(1) = 3y^2-1$

5Step 5: Analyze the sign of y'' in different intervals.

We have a quadratic function $3y^2-1$. Determine its roots and analyze the sign of y'' in different intervals: 1. The roots of the quadratic function are $y = \frac{1}{\sqrt{3}}$ and $y = -\frac{1}{\sqrt{3}}$. 2. When $-∞ < y <\frac{1}{-\sqrt{3}}$, $y''(x) < 0$, making the solutions concave down. 3. When $-\frac{1}{\sqrt{3}} < y < \frac{1}{\sqrt{3}}$, $y''(x) > 0$, making the solutions concave up. 4. When $\frac{1}{\sqrt{3}} < y < ∞$, $y''(x) < 0$, making the solutions concave down. #d. Sketch representative solution curves.#

6Step 6: Plot the solution curves.

Based on our analysis, sketch the solution curves: 1. Plot the equilibrium solutions: horizontal lines $y = -1$, $y = 0$, and $y = 1$. 2. Solutions are increasing in regions $-∞ < y < -1$ and $1 < y < ∞$, and decreasing in regions $-1 < y < 0$ and $0 < y < 1$. 3. Solutions are concave up in the region $-\frac{1}{\sqrt{3}} < y < \frac{1}{\sqrt{3}}$ and concave down in regions $-∞ < y <\frac{1}{-\sqrt{3}}$ and $\frac{1}{\sqrt{3}} < y < ∞$. Draw the solution curves accordingly, showing different behaviors for different starting points in the xy-plane.

Key Concepts

Equilibrium SolutionsConcave Up and DownSolution Curves

Equilibrium Solutions

Equilibrium solutions occur when the derivative, $y'(x)$, is equal to zero. This happens at specific values of $y$ where the slope of the solution curve is flat, indicating no change over time. To find these points, we solve the equation $y(y-1)(y+1) = 0$. This equation is derived from the given differential equation $y' = y(y-1)(y+1)$.

Solving this, we attain the equilibrium solutions $y = 0$, $y = 1$, and $y = -1$. Each of these solutions represents a state where the system doesn't change as time progresses. In the context of the $xy$-plane, these are horizontal lines where the solution's slope, $y'(x)$, is zero.

$y = 0$: The function remains steady along the x-axis.
$y = 1$: The curve remains at this constant state along the line $y = 1$.
$y = -1$: Similarly, the curve is static at $y = -1$.

Understanding equilibrium solutions is crucial because it helps us identify long-term behaviors of a system. These are points of stability that a dynamic system may settle into over time.

Concave Up and Down

The concavity of solution curves is influenced by the second derivative, $y''(x)$. This tells us how the slope of $y'(x)$ itself changes, providing insight into the curvature of the solutions.

To find when a curve is concave up or down, we compute $y''(x)$ from $y' = y(y-1)(y+1)$. Differentiating, we obtain:\[y'' = 3y^2 - 1\]

This results in a quadratic function whose roots, $y = \frac{1}{\sqrt{3}}$ and $y = -\frac{1}{\sqrt{3}}$, help us identify the regions of concavity:

In the interval $-\infty < y < -\frac{1}{\sqrt{3}}$, $y'' < 0$. Here, solution curves are concave down, resembling a downward bowl shape.
In the region $-\frac{1}{\sqrt{3}} < y < \frac{1}{\sqrt{3}}$, $y'' > 0$. Curves are concave up like an upward bowl.
For $y > \frac{1}{\sqrt{3}}$, $y'' < 0$, and the curves are again concave down.

Understanding the concavity is useful for visualizing how solutions behave between critical points, allowing us to predict the progression of changes within these intervals.

Solution Curves

Solution curves depict how a system evolves over time and can be understood by examining the first derivative $y'(x)$. This equation indicates whether a function is increasing or decreasing depending on the sign of $y'(x)$.

We explore this by looking at various intervals set by the equilibrium solutions. For $y' = y(y-1)(y+1)$, the sign shows us curve behaviors:

$-\infty < y < -1$: All terms are positive, thus $y'(x) > 0$, indicating increasing behavior.
$-1 < y < 0$: Only $(y+1)$ is positive, so $y'(x) < 0$, indicating decreasing behavior.
$0 < y < 1$: Only $(y-1)$ is negative, so $y'(x) < 0$.
$1 < y < \infty$: All terms positive means $y'(x) > 0$.

The behavior of solution curves can be visualized as increasing outside $[-1, 1]$ and decreasing within. This division aids in sketching the solution curves with interactions at equilibrium points, demonstrating shifts in direction. Observing and understanding these patterns help in predicting system trends and behaviors.

Problem 21

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