Problem 66 - Calculus for Biology and Medicine

Question

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) \(y(0)=3\).

Step-by-Step Solution

Answer

The vector field indicates stability at \(y=0\); solutions for all conditions converge to \(y=0\).

Step 1: Analyze the Differential Equation

The given differential equation is \( \frac{d y}{d x} = -y - y^3 \). It is a first-order, nonlinear ordinary differential equation. The rate of change of \(y\) with respect to \(x\) depends on both \(y\) and its cube, \(y^3\).

Step 2: Identify Critical Points

Set \( \frac{d y}{d x} = 0\) to find critical points, leading to \(-y - y^3 = 0\). Factoring gives \(y(1 + y^2) = 0\), so the critical point is \(y = 0\).

Step 3: Draw the Vector Field

In the vector field plot, draw small arrows at several points \((x,y)\) (here between \(-3\) and \(3\) on the y-axis), indicating the direction and magnitude of \( \frac{d y}{d x} = -y - y^3\). Arrows should point towards zero as \(y\) approaches zero and away from zero as \(y\) increases or decreases beyond zero.

Step 4: Sketch the Solution Curves

For each initial condition, sketch the solution curve (trajectory): (a) At \(y(0)=0\), the slope is zero, so the solution stays constant at zero. (b) At \(y(0)=1\), the solution will decrease towards zero due to \( \frac{d y}{d x} < 0\). (c) Similarly, at \(y(0)=2\) and (d) \(y(0)=3\), the solutions will initially decrease but even more so due to larger negative values from \(-y-y^3\). Solution curves for \(y > 0\) exhibit steep initial decline.

Step 5: Reflect on the Behavior of Solutions

The critical point at \(y=0\) is stable because small deviations lead to solutions approaching back to \(y=0\). For initial conditions greater than zero, the solutions demonstrate exponential decay approaching the stable critical point as \(x\) increases.

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates some function with its derivatives. In simpler terms, it describes how a certain quantity changes over time or space, taking into account its current value and rate of change. In this exercise, you are looking at the 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

Initial conditions are values that specify the state of a system at a specific point. They are crucial for solving differential equations because they help determine a unique solution from infinitely many possibilities. Here, initial conditions are given for \(y\) at \(x = 0\), such as:

• \(y(0) = 0\)
• \(y(0) = 1\)
• \(y(0) = 2\)
• \(y(0) = 3\)

Each of these points represents a different starting situation described by the equation. These conditions help in sketching the solution curve by indicating where the curve begins, which in turn affects its shape and trajectory in the \(xy\)-plane.
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

Solution curves, or trajectories, show how the dependent variable (\(y\)) changes as the independent variable (\(x\)) varies. Each curve corresponds to a particular initial condition and depicts a possible path for the system described by the differential equation.

• 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.

All these solutions are typically sketchable on a vector field plot, illustrating the motion or flow of the system over time. Their behavior exemplifies how differential equations can represent real-world systems that respond dynamically based on their initial state.

Critical Points

Critical points in a differential equation refer to points where the rate of change is zero. This means there is no change in the system in the vicinity of these points, which often indicates a point of equilibrium or stability.
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.

Understanding critical points allows us to predict steady states of a system and how it reacts to small perturbations, a valuable concept in physics, biology, and engineering.