Problem 81 - Thomas Calculus

Question

Solve the initial value problems in Exercises 81 and 82. $$ \frac{d y}{d x}=1+\frac{1}{x}, \quad y(1)=3 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Differential Equation and Initial Condition

We are given the differential equation \( \frac{dy}{dx} = 1 + \frac{1}{x} \) and the initial condition \( y(1) = 3 \). Our goal is to solve this differential equation and determine \( y(x) \) using the initial condition.

Step 2: Integrate the Differential Equation

To find \( y(x) \), we integrate both sides of the equation \( \frac{dy}{dx} = 1 + \frac{1}{x} \). Thus, we have:\[ y(x) = \int \left(1 + \frac{1}{x}\right) \, dx = \int 1 \, dx + \int \frac{1}{x} \, dx.\]The integration yields:\[y(x) = x + \ln|x| + C,\]where \( C \) is the constant of integration.

Step 3: Apply the Initial Condition

Using the initial condition \( y(1) = 3 \), plug \( x = 1 \) and \( y(x) = 3 \) into the equation:\[3 = 1 + \ln|1| + C.\]Since \( \ln|1| = 0 \), we simplify to:\[3 = 1 + 0 + C.\]Thus, \( C = 2 \).

Step 4: Write the General Solution Including the Constant

Now that we found the constant \( C \), substitute back into the function \( y(x) \):\[y(x) = x + \ln|x| + 2.\]This is the particular solution to the initial value problem.

Key Concepts

Differential Equations

Differential equations are mathematical expressions that relate a function with its derivatives. In this problem, we are dealing with the first-order differential equation \( \frac{dy}{dx} = 1 + \frac{1}{x} \). This equation describes how the rate of change of \( y \) with respect to \( x \) is related to the function itself.

• "dy/dx": Indicates the derivative of \( y \) with respect to \( x \).
• "1 + 1/x": Specifies the rate of change.

Differential equations are classified based on the order of their derivatives and whether they are linear or nonlinear. In this case, it is a first-order linear differential equation, as it depends on the first derivative and does not involve any products or powers of \( y \) or its derivative.

Integration

Integration is the process of finding the antiderivative or the original function from its derivative. When we integrate both sides of the given differential equation, we find \( y(x) \) from \( \frac{dy}{dx} = 1 + \frac{1}{x} \).The integration steps include:* Separating the integral into simpler parts: \( \int 1 \, dx + \int \frac{1}{x} \, dx \).* Solving each part: * \( \int 1 \, dx = x \) * \( \int \frac{1}{x} \, dx = \ln|x| \)As a result, the general solution to this integral is \( y(x) = x + \ln|x| + C \), where \( C \) is an unknown constant. This process helps us express the relationship described by the differential equation as a function.

Constant of Integration

The constant of integration, often denoted by \( C \), appears naturally when performing integration. It represents a family of curves that are solutions to a differential equation.When we integrate, we solve for the indefinite integral, which is not as precise. That's why we introduce \( C \) to account for any vertical shifts of the function graph. In our problem, on integrating \( \frac{dy}{dx} = 1 + \frac{1}{x} \), the constant \( C \) appears, and our solution becomes:* \( y(x) = x + \ln|x| + C \)The constant \( C \) will later be defined using the initial conditions to arrive at a particular solution.

Initial Conditions

Initial conditions help in finding the specific solution to a differential equation from the general solution derived after integration. Once we have integrated, the solution contains the constant of integration \( C \). This is where initial conditions come into play.The initial condition provided in this problem is \( y(1) = 3 \). By substituting \( x = 1 \) and \( y(x) = 3 \) into the equation \( y(x) = x + \ln|x| + C \), we solve for \( C \):

• Plugging in values: \( 3 = 1 + \ln|1| + C \)
• Since \( \ln|1| = 0 \), we simplify: \( 3 = 1 + 0 + C \)
• This gives us \( C = 2 \)

With this information, the particular solution to the differential equation becomes \( y(x) = x + \ln|x| + 2 \). Initial conditions are essential to find the exact function that satisfies both the differential equation and the given conditions at a specific point.