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Magazine Chapter 7: Problem 83
Solve the initial value problems $$\frac{d y}{d x}=1+\frac{1}{x}, \quad y(1)=3$$
Short Answer
Expert verified
The solution is \( y(x) = x + \ln |x| + 2 \).
Step by step solution
01
Identify the Differential Equation
The given differential equation is \( \frac{d y}{d x} = 1 + \frac{1}{x} \). Our task is to derive the function \( y(x) \) from this equation and meet the initial condition \( y(1) = 3 \).
02
Integrate Both Sides
Integrate both sides of the equation with respect to \( x \). The left side becomes \( \int \frac{d y}{d x} \, dx = y(x) \), and the right side becomes \( \int \left( 1 + \frac{1}{x} \right) \, dx \). This gives us: \( y(x) = \int \left( 1 + \frac{1}{x} \right) \, dx = \int 1 \, dx + \int \frac{1}{x} \, dx \).
03
Solve the Integrals
Compute each integral separately: \( \int 1 \, dx = x \) and \( \int \frac{1}{x} \, dx = \ln |x| \). Thus, \( y(x) = x + \ln |x| + C \) where \( C \) is the constant of integration.
04
Apply Initial Condition
Apply the initial condition \( y(1) = 3 \) to determine \( C \). Substitute \( x = 1 \) and \( y = 3 \) into the equation: \( 3 = 1 + \ln |1| + C \). Since \( \ln |1| = 0 \), we have \( C = 2 \).
05
Write the Final Solution
Substitute \( C = 2 \) back into the equation for \( y(x) \): \( y(x) = x + \ln |x| + 2 \). This is the solution to the initial value problem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical equations that involve the derivatives of a function. They describe how a particular quantity changes with time or another variable. In many cases, these equations model real-world phenomena such as motion, growth, or decay.
In our exercise, the equation given is a simple first-order linear differential equation: \( \frac{d y}{d x} = 1 + \frac{1}{x} \). The goal is to find the function \( y(x) \) whose derivative matches this equation.
In our exercise, the equation given is a simple first-order linear differential equation: \( \frac{d y}{d x} = 1 + \frac{1}{x} \). The goal is to find the function \( y(x) \) whose derivative matches this equation.
- The term \( \frac{d y}{d x} \) indicates the rate of change of \( y \) concerning \( x \).
- This particular equation suggests that \( y \) changes at a rate that is always a little more than \( 1 \) and decreases as \( x \) increases due to the \( \frac{1}{x} \) part.
Integration
Integration is the process of finding a function given its derivative, and it plays a crucial role in solving differential equations. Simply put, it's the reverse of differentiation.
In our initial value problem, we integrate both sides of \( \frac{d y}{d x} = 1 + \frac{1}{x} \), to recover \( y \(x\) \). This process involves calculating the antiderivative of each term.
In our initial value problem, we integrate both sides of \( \frac{d y}{d x} = 1 + \frac{1}{x} \), to recover \( y \(x\) \). This process involves calculating the antiderivative of each term.
- \( \int 1 \, dx \) gives us \( x \), representing a constant rate of change with respect to \( x \).
- \( \int \frac{1}{x} \, dx \) results in \( \ln |x| \), capturing the naturally occurring rate of change inversely proportional to \( x \).
Constant of Integration
The constant of integration is an important concept when working with indefinite integrals. It accounts for the infinite number of possible antiderivatives a function can have due to the constant difference between any two constants.
After integrating \( \frac{d y}{d x} = 1 + \frac{1}{x} \), we arrive at \( y(x) = x + \ln |x| + C \), where \( C \) is the constant of integration.
After integrating \( \frac{d y}{d x} = 1 + \frac{1}{x} \), we arrive at \( y(x) = x + \ln |x| + C \), where \( C \) is the constant of integration.
- Each specific value of \( C \) offers a different solution curve, all of which are parallel translations of one another.
- The constant allows us to satisfy any additional condition or requirement set by a problem, like the initial value condition.
Initial Condition
An initial condition provides specific values for a function at a particular point. It simplifies a general solution to a differential equation by specifying one of the infinite solutions provided by an indefinite integration.
In our problem, the initial condition \( y(1) = 3 \) was given to determine the constant of integration \( C \).
In our problem, the initial condition \( y(1) = 3 \) was given to determine the constant of integration \( C \).
- To find \( C \), substitute \( x = 1 \) and \( y = 3 \) into the integrated equation \( y(x) = x + \ln |x| + C \).
- Since \( \ln |1| = 0 \), the equation simplifies to \( 3 = 1 + 0 + C \), resulting in \( C = 2 \).
