Problem 22
Question
Find the solution of the given initial value problem. $$ y^{\prime}(x)=\sec ^{2}(x) \quad y(\pi / 4)=3 $$
Step-by-Step Solution
Verified Answer
The solution to the initial value problem is \( y(x) = \tan(x) + 2 \).
1Step 1: Identify the Differential Equation and Initial Condition
The exercise gives us a first-order differential equation \( y'(x) = \sec^2(x) \) with an initial condition \( y(\pi/4) = 3 \). The goal is to find \( y(x) \) that satisfies both the differential equation and the initial condition.
2Step 2: Integrate the Differential Equation
Recognize that the right-hand side \( \sec^2(x) \) is the derivative of \( \tan(x) \). Thus, to solve for \( y(x) \), integrate both sides with respect to \( x \): \[ y(x) = \int \sec^2(x) \, dx = \tan(x) + C . \] This gives us an expression with an unknown constant \( C \).
3Step 3: Apply the Initial Condition
To find \( C \), use the initial condition \( y(\pi/4) = 3 \). Substitute \( x = \pi/4 \) and \( y = 3 \) into the general solution: \[ 3 = \tan(\pi/4) + C . \] Since \( \tan(\pi/4) = 1 \), we have: \[ 3 = 1 + C . \] Solving for \( C \), we find \( C = 2 \).
4Step 4: Write the Final Solution
Substitute the value of \( C \) back into the general solution to get the specific solution:\[ y(x) = \tan(x) + 2 . \] This function satisfies both the differential equation and the initial condition.
Key Concepts
IntegrationInitial Value ProblemFirst-order Differential Equation
Integration
Integration is a fundamental mathematical technique used to find a function when its derivative is known. In this type of problem, we reverse the process of differentiation by integrating the expression.
In our exercise, the function we needed to integrate is related to the right-hand side of the differential equation, specifically \( \sec^2(x) \). Recognizing this, we used the integration of \( \sec^2(x) \), which is the derivative of \( \tan(x) \). The integral of a function often includes an arbitrary constant denoted by \( C \).
It's important to understand that integrating \( \sec^2(x) \) not only gives us \( \tan(x) \), but also accounts for any constant change that impacts the function, which is why we add \( C \). This constant is crucial, as it represents an infinite number of potential solutions that we narrow down using initial conditions. Through integration, we transform a function derived from change into one that represents the accumulated total of that change.
In our exercise, the function we needed to integrate is related to the right-hand side of the differential equation, specifically \( \sec^2(x) \). Recognizing this, we used the integration of \( \sec^2(x) \), which is the derivative of \( \tan(x) \). The integral of a function often includes an arbitrary constant denoted by \( C \).
It's important to understand that integrating \( \sec^2(x) \) not only gives us \( \tan(x) \), but also accounts for any constant change that impacts the function, which is why we add \( C \). This constant is crucial, as it represents an infinite number of potential solutions that we narrow down using initial conditions. Through integration, we transform a function derived from change into one that represents the accumulated total of that change.
Initial Value Problem
An initial value problem in the context of differential equations provides specific conditions that allow us to find a unique solution. By specifying an initial condition, such as \( y(\pi/4) = 3 \), we are able to determine the particular constant \( C \) in our integrated solution.
Without the initial condition, we could only derive the general solution \( y(x) = \tan(x) + C \). However, knowing \( y(\pi/4) = 3 \) allows us to pinpoint the exact solution that fits the given conditions. By substituting \( x = \pi/4 \) into the equation and solving for \( C \), we find \( C = 2 \).
This concept is essential because it ensures that we do not just find a solution, but the exact solution pertinent to the scenario described in the initial condition. The initial value condition effectively serves as a guide, anchoring our mathematical exploration with a real-world reference point.
Without the initial condition, we could only derive the general solution \( y(x) = \tan(x) + C \). However, knowing \( y(\pi/4) = 3 \) allows us to pinpoint the exact solution that fits the given conditions. By substituting \( x = \pi/4 \) into the equation and solving for \( C \), we find \( C = 2 \).
This concept is essential because it ensures that we do not just find a solution, but the exact solution pertinent to the scenario described in the initial condition. The initial value condition effectively serves as a guide, anchoring our mathematical exploration with a real-world reference point.
First-order Differential Equation
A first-order differential equation involves derivatives of the function and is characterized by having the highest derivative of one. In this problem, our differential equation is \( y'(x) = \sec^2(x) \), which is of first order since the highest derivative present is \( y' \), or \( \frac{dy}{dx} \).
First-order differential equations can often be solved directly by integration, as in our example. These equations frequently describe various dynamic processes in physics, chemistry, and even economics, where the rate of change of a quantity is considered.
Solving these equations generally involves rearranging and integrating both sides. However, the presence of an initial condition, like \( y(\pi/4) = 3 \), helps us find a specific solution. The process exemplifies how these equations model real phenomena, providing insights into how a change in one variable affects another, and how initial states influence future behavior.
First-order differential equations can often be solved directly by integration, as in our example. These equations frequently describe various dynamic processes in physics, chemistry, and even economics, where the rate of change of a quantity is considered.
Solving these equations generally involves rearranging and integrating both sides. However, the presence of an initial condition, like \( y(\pi/4) = 3 \), helps us find a specific solution. The process exemplifies how these equations model real phenomena, providing insights into how a change in one variable affects another, and how initial states influence future behavior.
- Recognize the order of the equation.
- Determine whether it's suitable for integration.
- Apply initial value conditions to find a specific solution.
Other exercises in this chapter
Problem 22
In each of Exercises \(17-28,\) solve the given initial value problem. $$ \frac{d y}{d x}+2 x y=\exp \left(-x^{2}\right) \quad y(0)=5 $$
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Find the center of \(\operatorname{mass}(\bar{x}, \bar{y})\) of the given region \(\mathcal{R}\), assuming that it has uniform mass density. \(\mathcal{R}\) is
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In each of Exercises 19-24, use the method of washers to calculate the volume \(V\) obtained by rotating the given planar region \(\mathcal{R}\) about the \(y\)
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Calculate the length of the given parametric curve. $$ x=\exp (2 t) \quad y=\exp (3 t) \quad-1 \leq t \leq 1 $$
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