Problem 24 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For each differential equation below, sketch the slope eld and nd the general solution. (a) \(\frac{d y}{d t}=-y\) (b) \(\frac{d y}{d t}=-t\) (c) \(\frac{d y}{d t}=e^{-t}\)

Step-by-Step Solution

Answer

The general solutions to the differential equations are \(y(t) = C e^{-t}\), \(y(t) = -t^2/2 + C\), and \(y(t) = -e^{-t} + C\), respectively. Plugins values of 't' in each equation should give the correct slope, allowing for correct plotting of the slope fields.

Step 1 - Solve equation and sketch for (a)

Integrating both sides of the equation \(d y / d t = - y\) gives \(y(t) = C e^{-t}\) where C is the constant of integration. For the slope field, at each point (t, y) on the plane, a short line segment is drawn with slope given by the value of the derivative, in this case, -y.

Step 2 - Solve equation and sketch for (b)

For the equation \(d y / d t = - t\), integration gives \(y(t) = -t^2/2 + C\), where C is the constant of integration. For the slope field, as in the previous case, we draw short line segments at each point (t, y), but in this case with slope -t

Step 3 - Solve equation and sketch for (c)

The integral of \(d y / d t = e^{-t}\) gives \(y(t) = -e^{-t} + C\). For the slope field, the line segment at each point (t, y) has slope \(e^{-t}\)

Key Concepts

Slope Fields

Visualizing differential equations through slope fields is a helpful tool for understanding their behavior. A slope field, also known as a direction field, consists of small line segments drawn at various points on the plane to represent the direction of the solutions.
\[\frac{d y}{d t} = f(t, y)\]
For this equation, the slope field involves drawing line segments such that the slope at each point \(t, y\) is determined by the function \(f(t, y):\)

• They offer a graphical representation of differential equations without finding exact solutions.
• Slope fields help visualize how solutions to differential equations behave over time.
• They are particularly useful for identifying patterns and tendencies in nonlinear equations.
• In our exercises, we calculated slopes for each scenario, such as \(y= -t\), \(y= -y\), and \(y = e^{-t}\).

Sketching a slope field informs us about the general direction in which solutions travel, guiding us intuitively even before solving the equations analytically.

General Solution

The general solution of a differential equation encompasses a family of functions, encompassing all possible specific solutions. When we solve a differential equation, we integrate to find this general expression.
Consider the form:
\[ \text{\int} f(t) \, dt = F(t) + C \] where \(C\) stands for the constant of integration.

• The general solution often contains one or more arbitrary constants, reflecting the infinite number of curves satisfying the equation.
• These constants allow the adjustment of the general solution to meet initial conditions.
• In exercise (a), the general solution is \(y(t) = Ce^{-t}\).
• For (b), it is \(y(t) = -\frac{t^2}{2} + C\).
• Finally, for (c), the expression is \(y(t) = -e^{-t} + C\).

Understanding the general solution is key to analyzing how systems evolve, sometimes needing further conditions to yield unique solutions.

Integration

Integration is the crucial process used to solve differential equations, often leading us to the general solution. Integrating a function reverses differentiation, essentially finding the original function given its derivative.
The integration process is represented by:
\[ \text{\int} f'(x) \, dx = f(x) + C \]

• To solve \(\frac{d y}{d t} = -y\), we integrate using separation of variables, finding \(y(t) = Ce^{-t}\).
• For the equation \(\frac{d y}{d t} = -t\), integrating directly gives \(y(t) = -\frac{t^2}{2} + C\).
• The step \(\int e^{-t} dt = -e^{-t} + C\) solves equation (c).
• Each integral solution includes \(C\), the constant of integration, because indefinite integrals account for all possible shifts in functions.

Mastering integration is essential for effectively tackling differential equations, unlocking their broad set of solutions.

Initial Value Problems

Initial Value Problems (IVPs) specify a unique solution within the family of solutions described by a general solution. By providing an initial condition, usually of the form \(y(t_0) = y_0\), we can determine the value of the constant \(C\).
This condition allows for solving the particular solution:

• An IVP looks for a specific answer based on given starting values.
• For example, if a condition \(y(0) = 1\) were given alongside the general solution \(y(t) = Ce^{-t}\), you’d substitute to solve for \(C\).
• Plugging into the equation, we have \(1 = Ce^{0} = C\), showing that \(C = 1\).
• This yields the specific solution \(y(t) = e^{-t}\).

By applying initial conditions, IVPs help in finding the precise path a solution will take, importantly informing real-world predictions based on the system's early state.