Question

For the following problems, use a software program or your calculator to generate the directional fields. Solve explicitly and draw solution curves for several initial conditions. Are there some critical initial conditions that change the behavior of the solution?[T] \(y^{\prime}=y^{3} e^{x}\)

Short answer

The critical initial condition is \( y(0) = 0 \), where \( y = 0 \) remains constant. Other initial conditions change the solution's divergence due to \( y^3 \).

Step-by-step solution

Directional Field Interpretation

To generate the directional field for the differential equation \(y' = y^3e^x\), use a software tool like Desmos or a graphing calculator that supports differential equations. This field represents the slope of the tangent line at each point \((x,y)\). Examine the slopes to understand how the solutions behave across different regions in the plane.

Find Critical Points

Since \( y' = 0 \) when \( y = 0 \), we identify \( y = 0 \) as an equilibrium solution or critical point. This means if \( y(0) = 0 \), the solution remains at zero for all \( x \). Critical initial conditions often arise here and can indicate different behavior patterns in solutions.

Solve the Differential Equation Explicitly

To solve \( y' = y^3 e^x \), we separate variables: \( \frac{dy}{y^3} = e^x dx \). Integrate both sides to obtain \( -\frac{1}{2y^2} = e^x + C \), where \( C \) is a constant. Solving for \( y \), we get \( y(x) = \pm \sqrt{-\frac{1}{2(e^x + C)}} \).

Analyze Solution Behavior

The solution behavior varies significantly based on initial conditions. For small \( y \), \( y^3 \) is approximately zero and solutions tend to resemble exponential functions. For larger values of \( y \), the solutions become steeper. Analyze how solutions diverge from or approach critical points by testing various initial conditions.

Graph Solution Curves

Plot several solution curves representing different initial conditions on the directional field. This visualization, which can be created manually or with graphing software, helps observe how solutions vary and offers insight into stability around critical points.

Key concepts

Directional Fields

Directional fields provide a tool to visually grasp the behavior of differential equations like \( y' = y^3 e^x \). Each tiny line or arrow in this field represents the slope of the solution at a specific point \((x, y)\). By observing these slopes, you can see how the solution curves generally behave without even solving the equation.

Here's how you can create a directional field:

• Use graphing calculators or software that supports differential equations, such as Desmos or GeoGebra.
• Plot the slope of the equation at various points in the plane.

The directional field helps identify where solutions might stabilize, rise, or fall rapidly. It’s like having a roadmap of possible behaviors of the solution across the plane.

Equilibrium Solutions

Equilibrium solutions are critical points where the differential equation's rate of change is zero, meaning the system remains constant. For the equation \( y' = y^3 e^x \), an equilibrium solution is found at \( y = 0 \), since substituting it in yields \( y' = 0 \).

Key characteristics of equilibrium solutions include:

• They show where solutions can stabilize.
• They act as possible starting points for behavior changes in solutions.

In this exercise, starting with \( y(0) = 0 \) means the solution stays at zero for all \( x \). Understanding equilibrium solutions shows how and why solutions diverge or converge in the larger context of the differential equation.

Separation of Variables

Separation of variables is a classic technique used to solve differential equations. The idea is to manipulate the equation so each variable and its differential are on opposite sides of the equation.

For the example \( y' = y^3 e^x \), we achieve this by separating it to \( \frac{dy}{y^3} = e^x dx \).

Next, integrate both sides:

• Integrate \( \frac{dy}{y^3} \) to get \( -\frac{1}{2y^2} \).
• Integrate \( e^x dx \) to get \( e^x \).

This results in the implicit solution: \[ -\frac{1}{2y^2} = e^x + C \]. Solving for \( y \) gives: \[ y(x) = \pm \sqrt{-\frac{1}{2(e^x + C)}} \]. This reveals the general solution pattern and helps predict solution behavior across different initial conditions.

Initial Conditions

Initial conditions specify the starting point of a solution curve on the graph of a differential equation. They are essential for determining the specific behavior of a solution among a family of curves.

Initial conditions are given as \( y(x_0) = y_0 \), where \( x_0 \) is the starting point on the x-axis and \( y_0 \) is the solution value at that point.

• Using \( y(0) = 0 \) leads to the equilibrium solution, where the curve remains flat along the \( x \)-axis.
• Different initial conditions will give rise to different solution curves.

Testing various initial conditions reveals patterns of stability and points where solution behavior changes, helping you understand the scope of potential solutions.