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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.

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 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.

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.