Problem 93 - University Calculus: Early Transcendentals

Question

Solve the initial value problems in Exercises. $$\frac{d y}{d x}=\frac{1}{x^{2}}+x, \quad x > 0 ; \quad y(2)=1$$

Step-by-Step Solution

Answer

The solution is \( y(x) = -\frac{1}{x} + \frac{x^2}{2} - \frac{1}{2} \).

Step 1: Identify the Differential Equation

We have the differential equation \( \frac{d y}{d x} = \frac{1}{x^{2}} + x \) with the initial condition \( y(2) = 1 \). This differential equation is a first-order separable equation that we can integrate directly.

Step 2: Integrate the Right Side

Integrate both terms of the right side: \( \int \left( \frac{1}{x^2} + x \right) \, dx = \int \frac{1}{x^2} \, dx + \int x \, dx \). The first integral is \(-\frac{1}{x} + C_1 \) and the second is \( \frac{x^2}{2} + C_2 \).

Step 3: Solve for the General Solution

Combine the integrals to get the general solution: \( y(x) = -\frac{1}{x} + \frac{x^2}{2} + C \), where \( C = C_1 + C_2 \) is an arbitrary constant.

Step 4: Apply the Initial Condition

Use the initial condition \( y(2) = 1 \) to find the value of \( C \). Substitute \( x = 2 \) and \( y = 1 \) into the general solution: \( 1 = -\frac{1}{2} + \frac{4}{2} + C \). Simplify to find \( C = -\frac{1}{2} \).

Step 5: Write the Particular Solution

Substitute \( C = -\frac{1}{2} \) back into the general solution to get the particular solution: \( y(x) = -\frac{1}{x} + \frac{x^2}{2} - \frac{1}{2} \).

Key Concepts

Understanding Initial Value Problems

In the study of differential equations, an initial value problem (IVP) is a problem where we seek a solution to a differential equation that satisfies a specific condition, known as the initial condition. In general, an IVP involves finding a function that not only solves the differential equation but also takes on a given value at a particular point. For example, in the exercise, the differential equation is given by \( \frac{d y}{d x} = \frac{1}{x^{2}} + x \) with an initial condition of \( y(2) = 1 \).

The initial condition is important because it helps in determining a unique solution from the family of solutions that a differential equation might have. Without it, solutions remain arbitrary and are represented with a constant \( C \) until particular conditions are known. This makes the problem within the exercise an initial value problem, as it requires both solving the differential equation and applying the initial condition \( y(2)=1 \) to find the specific solution.

Exploring First-Order Equations

First-order differential equations are among the simplest types of differential equations and involve only the first derivative of the unknown function. Such equations take the form \( \frac{dy}{dx} = f(x, y) \). In our example, the equation \( \frac{d y}{d x} = \frac{1}{x^{2}} + x \) is first-order because it contains the first derivative of \( y \) with respect to \( x \), and no higher derivatives.

These equations are often separable, meaning that we can manipulate them such that all terms involving \( y \) are on one side of the equation and all terms involving \( x \) are on the other side. This allows for straightforward integration to find the solution. Understanding first-order equations is crucial as they lay the groundwork for more complex differential equations encountered in advanced studies.

Mastering Integration Techniques

Integration is a vital technique used to solve differential equations, especially separable ones. In the given problem, the integration of the right-hand side, \( \frac{1}{x^{2}} + x \), involves two integrals: \( \int \frac{1}{x^{2}} \, dx \) and \( \int x \, dx \).

• The integral \( \int \frac{1}{x^{2}} \, dx \) results in \( -\frac{1}{x} + C \), a standard antiderivative familiar from algebra.
• The integral \( \int x \, dx \) yields \( \frac{x^{2}}{2} + C \), again following basic integration rules.

By combining these results, we derive the general solution. To obtain a particular solution, we leverage the initial condition to determine the constant \( C \). Often, mastering such integration techniques is essential for solving and understanding not just first-order equations but many other types of differential equations.