Problem 57 - Calculus for Biology and Medicine

Question

For Problems \(57-66\) 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). \(\frac{d y}{d t}=3 y-2\) (a) \(y(0)=2\), (b) \(y(0)=0\).

Step-by-Step Solution

Answer

Sketch the vector field and apply initial conditions to find the solution paths for \( y(t) \) on the plot.

Step 1: Understanding the Differential Equation

We start with the differential equation \( \frac{dy}{dt} = 3y - 2 \). This equation describes how \( y \) changes with respect to \( t \). Notice that this is a linear first-order differential equation.

Step 2: Sketching the Vector Field

To draw the vector field, observe that the slope \( \frac{dy}{dt} \) depends linearly on \( y \). For different values of \( y \), calculate the slope. For example, when \( y = 0 \), \( \frac{dy}{dt} = -2 \); when \( y = 1 \), \( \frac{dy}{dt} = 1 \); and so on. Use these slopes at different values of \( y \) to sketch the vectors on the plane, with \( t \) on the horizontal axis.

Step 3: Initial Conditions for Sketching Solutions

We are given the initial conditions \( y(0) = 2 \) and \( y(0) = 0 \). Using these points, we will sketch the trajectories in the direction of the vector field from these initial points.

Step 4: Solving the Differential Equation

Solve \( \frac{dy}{dt} = 3y - 2 \) by separating variables and integrating. Rewrite the equation as \( \frac{dy}{3y-2} = dt \). Integrate both sides to find \( y(t) \). The integral on the left is \( \int \frac{1}{3y-2} \, dy = \frac{1}{3}\ln|3y-2| \) and on the right is just \( t + C \), leading to \( \frac{1}{3}\ln|3y-2| = t + C \). Simplify to get the explicit form of \( y(t) \).

Step 5: Applying the Initial Condition (a)

For \( y(0) = 2 \), substitute \( t = 0 \) and \( y = 2 \) into the equation to find \( C \). You will find that for this initial condition, \( y(t) \) evolves accordingly. Solve for \( C \) and sketch the solution over the vector field, following the trajectory determined by this initial condition.

Step 6: Applying the Initial Condition (b)

For \( y(0) = 0 \), similarly substitute \( t = 0 \) and \( y = 0 \) into the equation to find another \( C \) specific to this initial condition. Again, solve for \( C \) and draw this different trajectory on the vector field. This will show how different initial conditions affect the solution paths.

Key Concepts

Vector Field

Visualizing a vector field can be immensely useful when dealing with differential equations. In simple terms, a vector field is a collection of vectors that represent the direction and speed of a moving particle in space. In the context of differential equations like \( \frac{d y}{d t}=3 y-2 \), each vector is a small arrow indicating the slope or direction of \( y \)'s movement at any given point.

The vector field of this equation can be plotted by picking various points for \( y \) and calculating \( \frac{d y}{d t} \) for these points. This gives us:

• For \( y = 0 \), the slope is \( -2 \) (a downward direction).
• For \( y = 1 \), the slope is \( 1 \) (an upward direction).
• For \( y = 2 \), the slope becomes \( 4 \) (a steeper upward direction).

Plotting these vectors with \( t \) as the horizontal axis will result in a graphical representation of how \( y \) changes over time. This requires consistently plotting arrows or lines that represent these tiny changes. Through this map, we gain a visual insight into the system's behavior at any given point.

Initial Conditions

Initial conditions serve as the starting points for solving differential equations. They are crucial, as they provide specific values that help define a unique solution trajectory. For the equation \( \frac{d y}{d t} = 3y - 2 \), two initial conditions are offered:

• \( y(0) = 2 \)
• \( y(0) = 0 \)

These conditions specify that at time \( t = 0 \), \( y \) is either 2 or 0 respectively. They significantly influence the path or trajectory \( y(t) \) will take over time.

To obtain a full solution, you need to use these initial values while integrating the differential equation. This process determines the constant of integration, \( C \), which ensures the solution accurately fits the initial state of the system. As each initial condition will yield a different \( C \), the end result will be a unique path representing the solution.

Solution Trajectories

Solution trajectories illustrate how the dependent variable \( y \) evolves over time based on the initial conditions and the vector field. When we solve the differential equation \( \frac{d y}{d t} = 3y - 2 \) while incorporating initial conditions, we obtain specific algebraic solutions, \( y(t) \).

For instance:

• With \( y(0) = 2 \), after substituting into the solved equation and determining \( C \), we plot \( y(t) \) on the graph. This trajectory is a curve that follows the vector field pattern, showing \( y \) increases rapidly due to the initial upward slope.
• For \( y(0) = 0 \), when substituted, this results in a different \( C \). The trajectory will be distinct, illustrating the slower growth of \( y \) compared to the previous condition.

These solution paths allow us to comprehend how even a slight change in initial conditions can lead to different outcomes. They provide a dynamic picture of the system's future behavior and potential equilibrium points. Examining these trajectories can help better predict and understand complex systems in various fields like physics, engineering, and biology.