Question

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

Short answer

Question: Find the particular solution of the given differential equation and initial condition, and describe the graph of the solution. Differential equation: \(f^{\prime}(s) = 4\sec s \tan s\) Initial condition: \(f(\pi / 4) = 1\) Answer: The particular solution for the given differential equation and initial condition is \(f(s) = 4 \sec s + 1 - 4\sqrt{2}\). The graph of the solution shows the behavior of the function \(f(s)\) and confirms that it satisfies the initial condition \(f(\pi / 4) = 1\).

Step-by-step solution

Identify the differential equation

The given differential equation is: $$f^{\prime}(s) = 4 \sec s \tan s$$

Integrate the differential equation

To find the general solution for the given function, we need to integrate: $$\int 4 \sec s \tan s \, ds$$ Let's recall that the derivative of secant function is: $$\frac{d(\sec s)}{ds} = \sec s \tan s$$ So, the integral becomes: $$\int 4 \frac{d(\sec s)}{ds}\, ds = 4\int \frac{d(\sec s)}{ds} \, ds = 4 \sec s + C$$ Now, we have the general solution: $$f(s) = 4 \sec s + C$$

Use the initial condition to find the particular solution

The given initial condition is: $$f(\pi / 4) = 1$$ Plugging this into our general solution, we get: $$1 = 4 \sec (\pi / 4) + C$$ Using the fact that \(\sec(\pi / 4) = \sqrt{2}\), we can solve for C: $$1 = 4\sqrt{2} + C$$ So, the constant \(C = 1 - 4\sqrt{2}\). Now, the particular solution is: $$f(s) = 4 \sec s + 1 - 4\sqrt{2}$$

Graph the particular solution

To graph the particular solution: $$f(s) = 4 \sec s + 1 - 4\sqrt{2}$$ we can use graphing software like Desmos, Geogebra, or any other software that plots graphs of functions. When plotting the particular solution, we will observe its behavior and its relationship with the initial condition given \(f(\pi / 4) = 1\).

Key concepts

Integration of Differential Equations

Solving differential equations often starts with the process of integration. Integration allows us to find a function whose rate of change, or derivative, is described by a given differential equation.

Consider the given problem, where we have a first-order differential equation involving the trigonometric functions secant, denoted as \( \sec s \), and tangent, \( \tan s \). To find the general solution, we integrate the differential equation \( f'(s) = 4 \sec s \tan s \). Recognizing that the integral of \( \sec s \tan s \), is \( \sec s \), simplifies our task:
\[ \int 4 \sec s \tan s \, ds = 4\sec s + C \]
Here, \( C \) is the constant of integration and represents an infinite number of solutions, as any constant we add will not affect the derivative. It is important to understand that when we integrate a differential equation, we are generally finding a family of possible solutions, as we are undoing the process of differentiation. Without additional information, such as an initial value, we can't determine the exact solution.

Initial Value Problems

An initial value problem adds specific conditions to a differential equation that allows us to determine the exact solution out of the family of general solutions obtained through integration. These conditions are often provided as the value of the function, or its derivatives at a particular point, referred to as the 'initial value'.

In our exercise, the initial value given is \( f(\pi / 4) = 1 \). With this information, we can find the particular constant \( C \) that will satisfy this condition:
\[ 1 = 4\sec(\pi / 4) + C \] After computing \( 4\sec(\pi / 4) \), we can isolate \( C \) to obtain the specific solution to our problem. This particular solution is the only one out of the infinite general solutions that will pass through the point \( (\pi / 4, 1) \) on its graph. It's crucial to understand that while the process of integration provides us with a general formula, the initial value helps us fine-tune that formula to meet specific conditions.

Graphing Solutions of Differential Equations

Graphing solutions is a visual way to understand the behavior of differential equations. It also plays a pivotal role in verifying solutions, especially when dealing with initial value problems. The graph of the particular solution depicts how the function behaves for all values of \( s \) and helps identify any peculiarities such as asymptotes or discontinuities.

In this scenario, we seek to graph the particular solution \( f(s) = 4\sec s + 1 - 4\sqrt{2} \). Tools like Desmos or Geogebra make plotting straightforward, and students can input the function to instantly visualize the solution. By graphing, we can confirm that the curve indeed passes through the point \((\pi / 4, 1)\), confirming our algebraic solution. An effective graph will showcase all relevant features of the function, including its intercepts and behavior as \( s \) approaches any discontinuities or boundaries in the function's domain. By mastering the graphing of differential equations, students can transform abstract equations into tangible functions that they can see and analyze.