Problem 73 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(71-90\) . $$\frac{d y}{d x}=\frac{1}{x^{2}}+x, \quad x>0 ; \quad y(2)=1$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Differential Equation

The given differential equation to solve is \( \frac{dy}{dx} = \frac{1}{x^2} + x \). Additionally, we are provided with the initial condition \( y(2) = 1 \). This initial condition will help us find the constant of integration after solving the differential equation.

Step 2: Integrate the Differential Equation

To solve the differential equation, integrate both sides with respect to \( x \). The left side becomes \( y \) and the right side becomes the integral \( \int \left( \frac{1}{x^2} + x \right) \, dx \). Split the integral: \( \int \frac{1}{x^2} \, dx + \int x \, dx \).

Step 3: Solve Each Integral Separately

1. For \( \int \frac{1}{x^2} \, dx \): Rewrite as \( \int x^{-2} \, dx \). This integrates to \( -x^{-1} = -\frac{1}{x} \).2. For \( \int x \, dx \): This integrates to \( \frac{x^2}{2} \).

Step 4: Write the General Solution

Combine the results of the integration: \[ y = -\frac{1}{x} + \frac{x^2}{2} + C \]Here, \( C \) is the constant of integration.

Step 5: Apply the Initial Condition

Use the initial condition \( y(2) = 1 \) to find \( C \): \[ 1 = -\frac{1}{2} + \frac{4}{2} + C \]Simplify to find \( C \):\[ 1 = -\frac{1}{2} + 2 + C \]\[ 1 = \frac{3}{2} + C \]\[ C = 1 - \frac{3}{2} = -\frac{1}{2} \]

Step 6: Write the Particular Solution

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

Key Concepts

Integration

Integration is a fundamental concept in calculus used to solve differential equations. When solving a first-order differential equation like \( \frac{dy}{dx} = \frac{1}{x^2} + x \), integration helps find the antiderivative which reverses the differentiation process.
It does this by accumulating a function's rate of change to restore the original function.

• The integral \( \int \frac{1}{x^2} \, dx \) can be rewritten using the power rule for integration as \( \int x^{-2} \, dx \). This results in \(-x^{-1}\) or \(-\frac{1}{x}\).
• The integral \( \int x \, dx \) is straightforward, resulting in \( \frac{x^2}{2} \) by applying the power rule \((n=1)\).

Integration allows us to find the general solution of a differential equation. This solution includes a constant of integration, denoted as \( C \), which is determined later using initial conditions.

Initial Value Problem

An Initial Value Problem (IVP) is a differential equation accompanied by a specific condition called the initial condition. This unique condition defines the value of the function at a particular point.
In our problem, the differential equation \( \frac{dy}{dx} = \frac{1}{x^2} + x \) is paired with the initial condition \( y(2) = 1 \).

• The purpose of the initial condition is crucial as it allows us to solve for the constant \( C \) in the general solution.
• Without an initial condition, we would have an infinite number of solutions since \( C \) could take any value.

Thus, the initial value problem (IVP) leads us to a solution that is specific to the given conditions.

Constant of Integration

When integrating an indefinite integral, the process introduces a term known as the constant of integration, \( C \). This constant represents any possible value that could have been lost during the differentiation step.
In the context of the differential equation \( \frac{dy}{dx} = \frac{1}{x^2} + x \), the integration produces the general solution:

• \( y = -\frac{1}{x} + \frac{x^2}{2} + C \).

The general solution includes \( C \), reflecting that without additional information, the set of solutions is infinite. Thankfully, using initial conditions allows us to determine \( C \) specifically by substituting a known point on the solution curve. In this example, applying the initial condition \( y(2) = 1 \) let us calculate the constant to be \( C = -\frac{1}{2} \).
This specificity facilitates deriving a single, particular solution satisfying both the differential equation and initial condition.

Particular Solution

Once we have the general solution from integrating the differential equation, the next step is using the initial condition to find the particular solution. This solution is tailored specifically to the initial condition provided.
The general solution for our exercise is:

• \( y = -\frac{1}{x} + \frac{x^2}{2} + C \).

By substituting the initial condition \( y(2) = 1 \) into the general solution, we found the constant \( C = -\frac{1}{2} \).
Plugging \( C \) back into the equation gives us:

• \( y = -\frac{1}{x} + \frac{x^2}{2} - \frac{1}{2} \).

This particular solution represents the unique function that not only solves the differential equation but also fulfills the initial condition, illustrating the specific behavior of the system described.