Problem 3 - Calculus: Early Transcendentals

Question

Solve the differential equation. \( xyy' = x^2 + 1\)

Step-by-Step Solution

Answer

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

Step 1: Simplifying the Equation

The given differential equation is \( xyy' = x^2 + 1 \). To simplify, divide both sides by \( xy \) to separate the variables. This gives \( y' = \frac{x^2 + 1}{xy} \). Simplifying further, we get \( y' = \frac{x}{y} + \frac{1}{xy} \).

Step 2: Rearrange to Separate Variables

The simplified equation \( y' = \frac{x}{y} + \frac{1}{xy} \) suggests separation of variables. We can rearrange it as \( y y' = x + \frac{1}{x} \). This implies the derivative \( y' = \frac{dy}{dx} \), so it can be written as \( y \frac{dy}{dx} = x + \frac{1}{x} \).

Step 3: Separate and Integrate

Separate variables by multiplying both sides by \( dx \): \( y \, dy = (x + \frac{1}{x}) \, dx \). Integrate both sides: \( \int y \, dy = \int (x + \frac{1}{x}) \, dx \).

Step 4: Solve the Integrals

Integrate both sides. The left side gives \( \frac{y^2}{2} \). The right side integrates to \( \frac{x^2}{2} + \ln|x| + C \), where \( C \) is the constant of integration.

Step 5: Combine and Simplify

Combining the results of the integrals, we have \( \frac{y^2}{2} = \frac{x^2}{2} + \ln|x| + C \). Multiply the entire equation by 2 to eliminate fractions: \( y^2 = x^2 + 2 \ln|x| + 2C \). Let \( 2C \) be a new constant, call it \( C' \), so \( y^2 = x^2 + 2 \ln|x| + C' \).

Step 6: Express the Solution

If needed, express \( y \) explicitly as \( y = \pm \sqrt{x^2 + 2 \ln|x| + C'} \). This is the solution to the differential equation.

Key Concepts

Separation of Variables

Separation of variables is a powerful technique used to solve differential equations. It allows us to isolate the different variables on separate sides of the equation, making it easier to integrate.

• Start by manipulating the given equation so that all terms involving one variable are on one side, and all terms involving the other variable are on the opposite side.
• This is particularly useful for equations where multiplication or division can help to separate the variables completely.

In our exercise, we took the equation through a series of logical steps to guide each variable towards separation. By doing this, we eventually obtain an equation where one side specifically contains the derivative of "y" with respect to "x". This equation, once simplified, supports a clean separation and primes the equation for integration.

Integration Techniques

Once variables are separated, the next step is to integrate both sides of the equation. This process helps in finding the original functions from their derivatives.

• For the left side containing "y," we perform direct integration as it seems linear in form.
• On the right side, typical anti-differentiation techniques are used. This involves treating each term separately, like basic polynomial integration and recognizing the natural log function through variable substitution.

The right-hand integral, \( \int (x + \frac{1}{x}) \, dx \), demonstrates how polynomial integration and the recognition of a logarithmic form are applied. These techniques involve identifying the antiderivatives of basic forms.

Constant of Integration

Integral solutions often introduce an arbitrary constant, known as the constant of integration. This constant arises due to the indefinite nature of integration.

• When integrating, the constant \( C \) represents any constant value that, when differentiated, results in zero. Thus, it maintains the integrity of the integral's balance.
• Upon resolving integrals for both sides of a differential equation, each expression results with its own constant. However, we often simplify by combining these constants into a single term.

In our differential equation, after completion of integrations, we encounter the constant \( C \). For simplicity in the final solution, we manipulated it to \( 2C \) and then relabeled it as \( C' \), enabling us to write the solution clearly and concisely.

Explicit Solutions

After solving the equation through integration and addressing constants, we arrive at an explicit solution. This provides the direct relationship between the independent variable \( x \) and the dependent variable \( y \).

• An explicit solution expresses \( y \) directly in terms of \( x \), providing a specific output based on given input.
• Explicit solutions are particularly useful in practical scenarios where specific values or calculations are needed swiftly.

In the final step of our exercise, we present \( y \) explicitly as \( y = \pm \sqrt{x^2 + 2 \ln|x| + C'} \). This advances our understanding from a general to a specific outlook, allowing us to quickly compute \( y \) for any given \( x \) value, assuming the constant is determined.