Problem 83 - Thomas Calculus in SI Units

Question

Solve the initial value problems $$\frac{d y}{d x}=1+\frac{1}{x}, \quad y(1)=3$$

Step-by-Step Solution

Answer

The solution is \( y(x) = x + \ln |x| + 2 \).

Step 1: 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 \( y(1) = 3 \).

Step 2: 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: \( y(x) = \int \left( 1 + \frac{1}{x} \right) \, dx = \int 1 \, dx + \int \frac{1}{x} \, dx \).

Step 3: 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.

Step 4: 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 \( C = 2 \).

Step 5: 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.

Key Concepts

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.

• 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.

Understanding this basic form is essential for solving differential equations and applies to many initial value problems.

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.

• \( \int 1 \, dx \) 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 \( x \).

This process of integration is crucial for reconstructing the original function and is foundational in calculus and engineering.

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.

• 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.

Understanding the purpose of this constant helps in neatly solving initial value and boundary condition problems.

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 \( C \).

• 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 \( C = 2 \).

An initial condition is vital as it makes the solution uniquely identifiable, leading to the understanding and solving of specific, real-world problems.