Question

Consider the following autonomous first order differential equations: $$ y^{\prime}=-y^{2}, \quad y^{\prime}=y^{3}, \quad y^{\prime}=y(4-y) . $$ Match each of these equations with one of the solution graphs shown. Note that each solution satisfies the initial condition \(y(0)=1\). Can you match them without solving the differential equations?

Short answer

Question: Match the given autonomous first-order differential equations with their solution graphs without solving them. Differential Equations: 1. \(y' = -y^2\) 2. \(y' = y^3\) 3. \(y' = y(4-y)\) Answer: 1. The first equation corresponds to the graph with a decreasing curve, starting from the point (0,1) and always having a negative slope. 2. The second equation corresponds to the graph with an increasing curve that does not change its direction, always having a positive slope. 3. The third equation corresponds to the graph with an increasing curve at first which then starts decreasing at some point, indicating a change in the slope's sign.

Step-by-step solution

Understand the behavior of each equation

First, let's analyze each of the given differential equations: 1. \(y' = -y^2\): The negative sign in front of \(y^2\) means that the slope will always be negative regardless of the value of y. Since y(0) = 1, the slope at the initial point will be negative. 2. \(y' = y^3\): The slope will always be positive regardless of the value of y as cube of any real number is always positive. Since y(0) = 1, the slope at the initial point will be positive. 3. \(y' = y(4-y)\): The slope is dependent on the value of y. If \(0 < y < 4\), the slope will be positive, and if \(y > 4\), the slope will be negative. At y(0) = 1, the slope will be positive.

Match the equations with their graphs

Based on the analysis in Step 1, we can match the given differential equations with their corresponding graphs without solving them: 1. For the first equation, \(y' = -y^2\), we need a graph with a negative slope at the initial point. This can only be the graph which has a decreasing curve, starting from the point (0, 1). So, the first equation corresponds to this graph. 2. For the second equation, \(y' = y^3\), we need a graph with a positive slope at the initial point, and the curve must keep increasing since the slope is always positive. This can only be the graph which has an increasing curve that does not change its direction. So, the second equation corresponds to this graph. 3. For the third equation, \(y' = y(4-y)\), we need a graph with a positive slope at the initial point, and the slope must change its sign at some point (when \(y > 4\)). This can only be the graph which has an increasing curve at first and then starts decreasing at some point. So, the third equation corresponds to this graph. In conclusion, you can match the given autonomous first-order differential equations with their solution graphs without solving them by carefully analyzing their behavior in accordance with the initial condition and the slopes.

Key concepts

Differential Equations Behavior

Understanding the behavior of differential equations is essential for predicting the nature of their solutions without explicitly solving the equations. Autonomous first-order differential equations, such as those in our example \(y' = -y^2\), \(y' = y^3\), and \(y' = y(4-y)\), rely solely on the variable \(y\) and are not directly dependent on the independent variable, typically time (\(t\)).

Slope Behavior and Its Significance

For \(y' = -y^2\), the slope of the solution curve is always negative, indicating that as time progresses, the value of \(y\) decreases. Similar logic applies to \(y' = y^3\) where the slope is always positive, leading to an increase in \(y\) over time. The last equation \(y' = y(4-y)\) has a more complex behavior, with slope changing based on the value of \(y\), which leads to different phases in the solution's behavior.

Equilibrium Points

Equilibrium points, where \(y' = 0\), play a crucial role in understanding the dynamic behavior of solutions. For instance, in \(y' = y(4-y)\), \(y = 0\) and \(y = 4\) are equilibrium points. Between these values, the slope is positive, leading to growth, while beyond \(y = 4\), the slope becomes negative, leading to decline in \(y\) values.

By analyzing the behavior of these equations, we can match them with the correct solution graphs without needing to solve them explicitly. This understanding is particularly useful in fields such as biology and economics, where such models are frequently applied.

Initial Value Problem

An initial value problem (IVP) in differential equations is a scenario where a function is subject to particular starting conditions. In the given exercise, each of the autonomous first-order differential equations comes with an initial condition \(y(0) = 1\).

Significance of Initial Conditions

An IVP's pivotal point is the initial condition which impacts how the solution curve will behave and evolve over time. For example, a positive initial condition with a positive derivative indicates that the curve will be ascending initially.

Uniqueness of Solutions

The existence and uniqueness theorem in differential equations ensures that, under certain conditions, there exists a unique function that satisfies both the differential equation and the initial condition. This is crucial for ensuring the predictability of solutions derived from mathematical models.

With the given equations, the initial condition \(y(0) = 1\) allows us to analyze potential solutions for their immediate direction of change. This guides us in visually matching these equations to their solution curves simply by examining the provided graphs and assessing which comply with the defined initial behavior.

Slope Analysis

Slope analysis in the context of differential equations involves assessing the rate of change of the dependent variable, which is typically the slope of the tangent to the curve at any given point. By investigating the slope, we can predict the increasing or decreasing nature of the solution functions over intervals.

Positive vs. Negative Slopes

Slope analysis tells us that a positive derivative, or slope, signifies growth, while a negative derivative indicates decay in the value of \(y\). For instance, any positive initial condition combined with a positive slope will produce an upward trend in the solution's graph.

Deciphering Behavior without Solutions

Using slope analysis, one can determine the general behavior of the solution curve without the actual solutions. For example, in our autonomous equations, the first suggests a decreasing function over time due to the negative slope, while the second suggests an increasing trend. The third equation's slope changes sign, indicating that the solution will initially increase and then decrease past the turning point.

The role of slope analysis is indispensable in matching the differential equations to their graphs, primarily because it provides information about the instantaneous rate of change at each point along the function. This tool is incredibly powerful in the study of dynamic systems where explicit solutions are not readily available.