Problem 1

Question

Draw the direction field for $y^{\prime}(t)=\sqrt{y(t)}$ and decide whether the equilibrium solution $y(t)=0$ is stable.

Step-by-Step Solution

Verified

Answer

The equilibrium solution $y(t) = 0$ is not stable.

1Step 1: Understand the Differential Equation

The given differential equation is $ y'(t) = \sqrt{y(t)} $. Here, $ y'(t) $ is the derivative of $ y(t) $ with respect to $ t $, and $ \sqrt{y(t)} $ suggests that this equation describes the rate of change of $ y(t) $ based on its current value.

2Step 2: Identify Equilibrium Solutions

An equilibrium solution is a constant solution where the slope or derivative is zero. In $ y'(t) = \sqrt{y(t)} $, setting $ y'(t) = 0 $ gives $ \sqrt{y(t)} = 0 $, implying $ y(t) = 0 $. Thus, $ y(t) = 0 $ is an equilibrium solution.

3Step 3: Draw the Direction Field

A direction field visualizes the slopes of the solution curves. For $ y'(t) = \sqrt{y(t)} $, plot several points $(t, y(t))$ and draw small lines with a slope that equals $ \sqrt{y(t)} $. Notice that when $ y(t) > 0 $, the slopes $ \sqrt{y(t)} $ are positive, and when $ y(t) = 0 $, the slope is zero.

4Step 4: Analyze Stability of Equilibrium Solution

An equilibrium solution is stable if small perturbations lead to solutions returning to the equilibrium. For $ y'(t) = \sqrt{y(t)} $, slightly increasing $ y(t) = 0 $ leads to $ y'(t) > 0 $, meaning $ y(t) $ increases, moving away from 0. Thus, $ y(t) = 0 $ is not stable since perturbed solutions do not return to $ y = 0 $.

Key Concepts

Understanding Differential EquationsEquilibrium SolutionsStability Analysis of Equilibrium Solutions

Understanding Differential Equations

A differential equation like $ y'(t) = \sqrt{y(t)} $ describes how a function $ y(t) $ changes over time. In this context, $ y'(t) $ is the derivative of $ y(t) $, showing its rate of change concerning the variable $ t $. The expression $ \sqrt{y(t)} $ implies that this rate depends on the current value of $ y(t) $.
Here are some important ideas about differential equations:

The primary goal is often to find all possible functions $ y(t) $ that satisfy the equation.
The behavior and solutions of the equation can give insights into real-world phenomena it models.
The rate of change $ \sqrt{y(t)} $ highlights an intrinsic relationship between $ y(t) $ and its derivative.

In our specific equation, if $ y(t) $ is zero or positive, the expression $ \sqrt{y(t)} $ is valid, guiding the pattern of solutions. Understanding these concepts allows us to anticipate how solutions might behave over time.

Equilibrium Solutions

Equilibrium solutions are critical points where the system remains constant over time, i.e., the derivative $ y'(t) $ is zero. For the differential equation $ y'(t) = \sqrt{y(t)} $, setting $ y'(t) = 0 $ leads to the equation $ \sqrt{y(t)} = 0 $. This implies that $ y(t) = 0 $ is a point where the system does not change.
Equilibrium solutions have unique features:

They represent steady states where the function $ y(t) $ ceases to evolve.
Identifying these points is essential for understanding the long-term behavior of solutions.

Knowing that $ y(t) = 0 $ is an equilibrium solution helps in analyzing whether other solutions might converge to or diverge from this point. This forms the basis for further investigation, such as stability analysis, which determines how solutions behave near these equilibrium states.

Stability Analysis of Equilibrium Solutions

Stability analysis evaluates whether small perturbations around an equilibrium solution will decay over time, bringing the system back to equilibrium, or grow, moving the system away.
In our exercise with the equation $ y'(t) = \sqrt{y(t)} $, the equilibrium solution is $ y(t) = 0 $. By slightly increasing $ y(t) $ from zero, we find that $ y'(t) > 0 $. This means $ y(t) $ begins to increase further away from zero, indicating that $ y(t) = 0 $ is unstable.
Key aspects of stability analysis include:

Stable equilibria: Small changes result in forces that bring the system back to equilibrium.
Unstable equilibria: Small changes lead to responses that further displace the system.
Identifying stability also helps predict the long-term behavior of dynamic systems.

Through this analysis, you can predict the behavior of solutions around equilibrium, essential for applications ranging from physics to economics.

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