Chapter 8: Problem 66
Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d y}{d x}=-y-y^{3}\) (a) \(y(0)=0\), (b) \(y(0)=1\), (c) \(y(0)=2\), (d) \(y(0)=3\).
Short Answer
Step by step solution
Analyze the Differential Equation
Identify Critical Points
Draw the Vector Field
Sketch the Solution Curves
Reflect on the Behavior of Solutions
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equation
\[ \frac{d y}{d x} = -y - y^3 \]
This is a first-order differential equation, which means it only includes the first derivative of the function, in this case, \( \frac{d y}{d x} \). It models the rate at which \( y \) changes with respect to \( x \). The equation is also nonlinear because of the \( y^3 \) term, meaning its behavior can become complex and is not just a simple line or curve.
- This equation suggests that the rate of change of \(y\) is influenced by both its current value \(y\) and its cubic value \(y^3\).
- The negative signs indicate that the function tends to decrease as \(y\) becomes positive, implying a pulling effect back toward zero.
Initial Conditions
- \(y(0) = 0\)
- \(y(0) = 1\)
- \(y(0) = 2\)
- \(y(0) = 3\)
For example, if \(y(0) = 0 \), the curve will start at zero for \(x = 0\) and stay flat since the initial slope is zero. On the other hand, if \(y(0) = 1 \), the curve will start at one and will decrease because of the negative slope directed by the differential equation.
Solution Curves
- For \(y(0) = 0 \), the solution is a constant curve at zero since the slope is zero at that point.
- For \(y(0) = 1, 2, \) or \(3\), the solution curves start at these initial values and slope downwards toward zero.
Critical Points
By setting the differential equation
\[ \frac{d y}{d x} = 0, \]
we determine the roots of
\[ -y - y^3 = 0, \]
which results in the critical point being \( y = 0 \). At this point, the slope is zero, indicating neither growth nor decay, which characterizes a stable state.
- If disturbed, small changes in \(y\) around zero result in trajectories that revert back to zero, evidencing a local stability.
- As shown by the solution curves, all trajectories eventually head towards this critical point as \( x \) increases.