Problem 65 - 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). \(\frac{d y}{d x}=(y-1)(y-2)(y-5)\) (a) \(y(0)=0\), (b) \(y(0)=4\) (c) \(y(0)=3 / 2\), (d) \(y(0)=6\).

Step-by-Step Solution

Answer

The solutions based on initial conditions go as follows: (a) moves down, (b) moves towards 2, (c) moves towards 2 and, (d) moves up indefinitely.

Step 1: Understanding the Differential Equation

We start by analyzing the given differential equation \( \frac{d y}{d x} = (y-1)(y-2)(y-5) \). This is an autonomous equation because the right-hand side is a function of \( y \) only. The roots of this polynomial, \( y = 1, 2, \) and \( 5 \), are the equilibrium points, where \( \frac{d y}{d x} = 0 \).

Step 2: Analyzing the Behavior Around Equilibria

For \( y < 1 \), \( \frac{d y}{d x} < 0 \) indicating \( y \) is decreasing. For \( 1 < y < 2 \), \( \frac{d y}{d x} > 0 \), so \( y \) is increasing. Similarly, for \( 2 < y < 5 \), \( \frac{d y}{d x} < 0 \), and for \( y > 5 \), \( \frac{d y}{d x} > 0 \). This shows \( y = 1 \) and \( y = 5 \) are stable equilibria, while \( y = 2 \) is unstable.

Step 3: Drawing the Vector Field

Sketch the vector field based on the sign of \( \frac{d y}{d x} \) identified in Step 2. Draw arrows for increasing functions (positive slopes) and reversed arrows for decreasing functions (negative slopes), showing how \( y \) moves with \( x \).

Step 4: Solving the Differential Equation with Initial Condition (a): \( y(0) = 0 \)

The initial condition \( y(0) = 0 \) shows that \( y < 1 \). From Step 2, we see \( y \) will decrease. Sketch the solution curve initially at \( y = 0 \) going downward over time.

Step 5: Solving the Differential Equation with Initial Condition (b): \( y(0) = 4 \)

The initial condition \( y(0) = 4 \) means \( y \) starts between 2 and 5. According to Step 2, \( y \) will decrease until it approaches the stable equilibrium \( y = 2 \). Sketch the solution curve starting from \( y = 4 \) heading downward toward \( y = 2 \).

Step 6: Solving the Differential Equation with Initial Condition (c): \( y(0) = 3/2 \)

With \( y(0) = 3/2 \), \( y \) starts between 1 and 2. As seen in Step 2, \( y \) will increase towards the equilibrium at \( y = 2 \). Draw the solution starting from \( y = 3/2 \) moving upwards towards \( y = 2 \).

Step 7: Solving the Differential Equation with Initial Condition (d): \( y(0) = 6 \)

Starting with \( y(0) = 6 \), means \( y > 5 \). From Step 2, we know \( y \) increases, confirming it will move further away from 5. Sketch this solution curve by showing \( y \) starting from 6 and moving upwards.

Step 8: Sketching All Solutions on One Graph

Combine the individual solutions onto one graph. Place the \( x \)-axis as the independent variable axis, and \( y \)-axis for the dependent variable. Mark the equilibria points \( y = 1, 2, 5 \) clearly and plot all solutions together, representing each initial condition.

Key Concepts

Vector Field Plot

A vector field plot is a visual representation of the differential equation's behavior. It shows how the dependent variable, \( y \), changes with respect to the independent variable, often denoted as \( x \). To create this plot for the differential equation \( \frac{d y}{d x} = (y-1)(y-2)(y-5) \), we analyze the equation's structure. Here, we assess the sign of \( \frac{d y}{d x} \) across different ranges of \( y \).

By examining \( y < 1 \), \( 1 < y < 2 \), \( 2 < y < 5 \), and \( y > 5 \), we observe whether \( y \) increases or decreases.

• For \( y < 1 \), \( \frac{d y}{d x} < 0 \), showing a downward trend for \( y \).
• At \( 1 < y < 2 \), \( \frac{d y}{d x} > 0 \), illustrating an upward motion.
• When \( 2 < y < 5 \), \( \frac{d y}{d x} < 0 \), suggesting that \( y \) decreases again.
• For \( y > 5 \), \( \frac{d y}{d x} > 0 \), meaning \( y \) starts to increase.

In the field plot, arrows display the direction and behavior of \( y \) depending on where it lies relative to these points.

Equilibrium Points

Equilibrium points are values of \( y \) where the rate of change, \( \frac{d y}{d x} \), equals zero. Essentially, these are points at which the system is in balance and there is no change in the value of \( y \). For our differential equation, the equilibrium points occur at the roots of \( (y-1)(y-2)(y-5) = 0 \), which are \( y = 1, 2, \) and \( 5 \).

Knowing whether these points are stable or unstable is important.

• Stable equilibrium means small perturbations will return to the equilibrium.
• Unstable equilibrium means that disturbances grow larger as time progresses.

In our case:

• \( y = 1 \) and \( y = 5 \) are stable equilibria because small changes result in \( y \) returning to these points.
• \( y = 2 \) is unstable because disturbances result in \( y \) moving away.

These properties help determine the expected behavior of solutions as they approach or diverge from these points.

Autonomous Equation

An autonomous equation is a type of differential equation where the derivative \( \frac{d y}{d x} \) depends solely on the dependent variable \( y \), and not explicitly on the independent variable \( x \). The equation \( \frac{d y}{d x} = (y-1)(y-2)(y-5) \) qualifies as autonomous because \( (y-1)(y-2)(y-5) \) is a function only of \( y \).

The significance of autonomous equations is that their solutions can often be analyzed through qualitative means, focusing on behavior over time rather than exact values.

• You can determine long-term behavior by looking at equilibrium points and how \( y \) behaves in their vicinity.
• Solutions are typically visualized through sketches like vector fields or phase lines rather than explicit functions of \( x \).

Overall, autonomous equations simplify the process of understanding a system's dynamics by centering the analysis around \( y \)-values.

Initial Conditions

Initial conditions are specific values set at the start of analyzing a differential equation, typically given in the form of \( y(x_0) = y_0 \). They serve as starting points for finding particular solutions to the differential equation. For our example, we have initial conditions like \( y(0) = 0 \), \( y(0) = 4 \), \( y(0) = \frac{3}{2} \), and \( y(0) = 6 \).

To understand how these initial conditions affect the solution:

• If \( y(0) = 0 \), since \( 0 < 1 \), \( y \) will move downward due to the negative derivative.
• For \( y(0) = 4 \), as \( 2 < 4 < 5 \), \( y \) decreases towards \( y = 2 \).
• With \( y(0) = \frac{3}{2} \), \( y \) starts below 2 and is drawn upward to this equilibrium.
• When \( y(0) = 6 \), above 5, \( y \) is propelled higher due to the positive derivative.

Graphically, these conditions anchor solution curves to specific trajectories within the vector field, highlighting the dynamic paths \( y \) takes over time.