Question

Graph several functions that satisfy the following differential equations. Then find and graph the particular function that satisfies the given initial condition. $$f^{\prime}(t)=1 / t ; f(1)=4$$

Short answer

Question: Find the particular function that satisfies the differential equation \(f^{\prime}(t)=\frac{1}{t}\) with the initial condition \(f(1)=4\), and describe its graph. Answer: The particular function that satisfies the given differential equation and initial condition is \(f(t)=\ln|t|+4\). Its graph is a logarithmic curve, which passes through the point (1, 4), with the curve moving upward as \(t\) increases.

Step-by-step solution

Solve the differential equation

To find the general solution to the given differential equation, integrate both sides of the equation with respect to \(t\): $$\int f^{\prime}(t) dt = \int \frac{1}{t} dt$$ The integral of \(\frac{1}{t}\) with respect to \(t\) is: $$\ln|t|+C$$ Where \(C\) is the integration constant. Now we obtain the general solution: $$f(t)=\ln|t|+C$$

Apply the initial condition

To find the particular function that satisfies the given initial condition, substitute the initial condition into the general solution: $$f(1)=\ln|1|+C = 4$$ Since \(\ln(1) = 0\), the integration constant, \(C\), can be determined as: $$C=4$$

Find the particular function

Substitute the value of \(C\) back into the general solution to obtain the particular function: $$f(t)=\ln|t|+4$$

Graph the general solution and the particular function

Graph several functions that satisfy the general solution by varying the value of \(C\) (e.g., \(C=-2,0,2,4\)). Then, graph the particular function \(f(t)=\ln|t|+4\) and observe that it passes through the initial condition \((1, 4)\). You can use graphing software like Desmos or GeoGebra to create the graphs.

Key concepts

Solving Differential Equations

Differential equations describe how one variable changes in relation to another. They are the backbone of many natural and physical phenomena. To solve these equations, the integration process is often employed. Integration, which can be thought of as the inverse operation of differentiation, helps us to find a function when we know its derivative. In the case of the given differential equation, \( f'(t) = 1/t \), the first step is to integrate both sides with respect to the variable \( t \). Integration turns the derivative \( f'(t) \) back into the function, \( f(t) \). While doing this, an integration constant \( C \) is added because when we differentiate a constant, it becomes zero, hence the original constant is unknown and could be any real number. This constant must be determined by applying the initial conditions provided.

Steps for solving typically include integrating both sides of the differential equation, finding the general solution that includes the integration constant, and then applying any given initial conditions to find the specific solution to the problem at hand. This process yields a particular solution that uniquely satisfies the conditions laid out in the problem.

Integration Constant

The integration constant \( C \) is a crucial part of solving differential equations. Because the derivative of a constant is zero, we acknowledge that the antiderivative is not unique; an infinite number of potential functions could serve as the antiderivative, differing only by a constant. To generalize this solution, we express it as \( f(t) = \text{antiderivative} + C \).

In the exercise provided, the integration process led to \( f(t) = \ln|t| + C \). Here, \( C \) represents an indefinite number of solutions, with each unique value of \( C \) corresponding to a different function. These different functions are all part of the general solution to the differential equation, but only one of them will satisfy the particular initial condition given. The integration constant is determined by substituting known values from an initial condition into the general solution.

Initial Condition

The initial condition in a differential equation problem provides a specific value for the function and its derivatives at a particular point. These conditions are essential for pinpointing the exact solution within the family of functions that satisfy the general solution represented by the integration constant. Initial conditions represent a starting point for the function, and with that fixed point, the otherwise indefinite integration constant can be uniquely determined.

In this specific exercise, the initial condition \( f(1) = 4 \) serves as our guide to find the value of \( C \). By substituting \( t = 1 \) into \( f(t) = \ln|t| + C \) and knowing \( f(1) = 4 \) from the initial condition, we easily deduce that \( C = 4 \). Thus, the particular solution becomes \( f(t) = \ln|t| + 4 \) which is the function passing through the point \( (1, 4) \) in the Cartesian plane.

Function Graphing

Graphing functions is an integral part of understanding differential equations, as it provides a visual representation of the solutions. Each function represented by a different value of the integration constant will have a slightly different shape and position on the graph, but all will follow the same general trends depicted by the differential equation. When dealing with initial value problems, graphing can confirm whether the particular solution accurately reflects the initial condition provided.

In our case, after determining the particular solution to the initial value problem, graphing several functions with diverse values of \( C \) can display the family of solutions. However, the graph of the particular solution \( f(t) = \ln|t| + 4 \) should uniquely pass through the point \( (1, 4) \) on the graph, confirming it as the correct solution. Various tools such as Desmos or GeoGebra can be utilized for plotting these graphs, showcasing how they are distinct yet related to the underpinning equation. This visualization is immensely helpful for students, as it reinforces the connection between the algebraic manipulation of equations and their graphical consequences on a coordinate plane.