Problem 31 - Calculus

Question

(a) Find the general solution to the differential equation modeling how a person learns: $$\frac{d y}{d t}=100-y$$ (b) Plot the slope field of this differential equation and sketch solutions with \(y(0)=25\) and \(y(0)=110\) (c) For each of the initial conditions in part (b), find the particular solution and add to your sketch. (d) Which of these two particular solutions could represent how a person learns?

Step-by-Step Solution

Answer

The general solution is \( y(t) = 100 - Ce^{-t} \), and the solution \( y(t) = 100 - 75e^{-t} \) fits a realistic learning model.

Step 1: Identify the Type of Differential Equation

The given differential equation \( \frac{dy}{dt} = 100 - y \) is a first-order linear differential equation. It can be solved using the separation of variables method.

Step 2: Separate Variables

Rearrange the equation to separate variables by moving terms involving \( y \) to one side and \( t \) to the other: \( \frac{dy}{100-y} = dt \).

Step 3: Integrate Both Sides

Integrate both sides to find the general solution. The left side integrates to \( -\ln|100-y| \), and the right side integrates to \( t + C \), where \( C \) is the constant of integration. This gives: \( -\ln|100-y| = t + C \).

Step 4: Solve for y

Solve for \( y \) to express it explicitly: \( |100 - y| = e^{-t-C} \). Let \( e^{-C} = C_1 \), then \( y = 100 - C_1 e^{-t} \). Thus, the general solution is \( y(t) = 100 - Ce^{-t} \) where \( C \) is a constant.

Step 5: Plot the Slope Field

The slope field of the differential equation can be plotted using graphical tools or software. The field will have lines indicating the direction of the solution curves based on the slope \( 100 - y \).

Step 6: Sketch Solutions for Given Initial Conditions

For \( y(0) = 25 \), substitute into the general solution: \( 25 = 100 - C \) gives \( C = 75 \). The specific solution is \( y(t) = 100 - 75e^{-t} \). For \( y(0) = 110 \), \( 110 = 100 - C \) gives \( C = -10 \). The specific solution is \( y(t) = 100 + 10e^{-t} \). Plot these on the slope field.

Step 7: Determine Realistic Learning Curve

The solution \( y(t) = 100 - 75e^{-t} \) starts below the learning limit (100) and increases towards it, which is consistent with a learning process where performance improves over time. The other solution starts above the limit (110) and decreases towards 100, which may not accurately represent a typical learning process if 100 is the performance limit. Thus, \( y(t) = 100 - 75e^{-t} \) is more realistic.

Key Concepts

Separation of Variables

Separation of variables is a powerful technique often used to solve first-order differential equations like \( \frac{dy}{dt} = 100 - y \). This method involves rearranging the equation so that all terms involving one variable are on one side and all terms involving the other variable are on the other side. Here, by moving all the \( y \)-related terms to one side and all \( t \)-related terms to the other side, we get \( \frac{dy}{100-y} = dt \). This separation allows us to integrate each side independently.

By integrating the left-hand side with respect to \( y \) and the right-hand side with respect to \( t \), we find a general solution that relates \( y \) and \( t \).

• The integration of \( \frac{dy}{100-y} \) results in \( -\ln|100-y| \).
• The integration of \( dt \) gives \( t + C \), where \( C \) is the constant of integration.

These steps lead us to the implicit solution \( -\ln|100-y| = t + C \), which can be further solved to express \( y \) explicitly as a function of \( t \): \( y = 100 - Ce^{-t} \). This form clearly shows how \( y \) changes with respect to time \( t \), given the initial conditions.

Slope Field

A slope field, or direction field, is a visual representation of solutions to a differential equation. It is a tiled grid of arrows that represent the slope \( 100-y \) at various points \( (t, y) \) in the plane. Each arrow indicates the direction that a solution curve would take through that point.

Creating a slope field for the equation \( \frac{dy}{dt} = 100-y \) involves computing the slope at a variety of points in a \( t \)-\( y \) plot, then drawing short line segments at each point with that slope.

• When \( y = 0 \), the slope is positive and strong, as the formula results in a slope of 100.
• When \( y = 100 \), the slope is zero, indicating a horizontal tangent, hence equilibrium.

By understanding slope fields, you can grasp the behavior of the solution over time just by looking at the direction of the arrows. It helps visualize how solutions behave for different initial conditions.

Initial Conditions

Initial conditions provide specific values for the solution of a differential equation, usually at \( t=0 \). These conditions are essential for finding a unique particular solution from a general solution. For the exercise, we have two initial conditions: \( y(0) = 25 \) and \( y(0) = 110 \).

Substituting these initial conditions into the general solution \( y = 100 - Ce^{-t} \) helps determine the specific constant \( C \) for each case:

• For \( y(0) = 25 \), substituting into the equation gives \( 25 = 100 - C \), resulting in \( C = 75 \).
• For \( y(0) = 110 \), the equation \( 110 = 100 - C \) leads to \( C = -10 \).

Initial conditions thus tailor the general behavior described by the differential equation to a situation that is unique and specific, reflecting different starting points in the learning model.

Particular Solution

A particular solution of a differential equation is a solution that satisfies the initial conditions. It is an expression that represents a specific scenario modeled by the differential equation.

Using the solutions derived from the initial conditions, let's draw particular solutions from \( y(t) = 100 - Ce^{-t} \).

• With \( y(0) = 25 \), the particular solution is \( y(t) = 100 - 75e^{-t} \). This solution shows how learning progresses by increasing towards the limit of 100.
• With \( y(0) = 110 \), the particular solution is \( y(t) = 100 + 10e^{-t} \). This scenario shows a decrease from above 100, unrealistic if 100 is assumed as a maximal learning level.

In practical terms, the solution for \( y(0) = 25 \) is ideal as it adheres to a typical learning curve, where performance starts below the capacity and grows towards it over time.