Problem 25 - Calculus Early Transcendentals

Question

Find a solution to the initial-value problem. \(x^{2} y^{\prime}+2 x y=0, y(1)=2 \quad[\text {Hint}:\) Interpret the left-hand side of the equation as the derivative of a product of two functions. \(]\)

Step-by-Step Solution

Answer

The solution to the initial-value problem is \( y(x) = \frac{2}{x^2} \).

Step 1: Identify the Structure of the Equation

The given differential equation is \( x^2 y' + 2xy = 0 \). The hint suggests interpreting the left-hand side as the derivative of a product of two functions. Notice that it resembles a derivative of \( x^2 y \).

Step 2: Rewrite Using Product Rule

Recall the product rule: if \( u = x^2 \) and \( v = y \), then the derivative is \( u'v + uv' = (x^2)'y + x^2 y' = 2xy + x^2 y' \). Thus, the equation can be rewritten as \( \frac{d}{dx}(x^2 y) = 0 \).

Step 3: Solve the Differential Equation

Since \( \frac{d}{dx}(x^2 y) = 0 \), it means \( x^2 y \) is constant. Therefore, \( x^2 y = C \), where \( C \) is a constant determined by the initial condition.

Step 4: Apply Initial Condition

Use the initial condition \( y(1) = 2 \) to find \( C \). Substituting \( x = 1 \) and \( y = 2 \) into \( x^2 y = C \), we get \( 1^2 \times 2 = C \), so \( C = 2 \).

Step 5: Express y(x)

Substitute \( C = 2 \) back into the equation \( x^2 y = 2 \) to find \( y(x) \). Divide both sides by \( x^2 \) to solve for \( y \), giving \( y(x) = \frac{2}{x^2} \).

Key Concepts

Understanding Initial-Value Problems

An initial-value problem is a type of differential equation that comes with specified values, called initial conditions, which are used to find a particular solution. In this problem, we have the differential equation \( x^2 y' + 2xy = 0 \) with the initial condition \( y(1) = 2 \). The initial condition is crucial because it helps specify which solution of the differential equation to choose from many possible solutions.

The general solution of a differential equation typically contains an arbitrary constant. The initial condition fixes this constant to ensure the solution matches the given specific starting value of the function. In our example, the initial condition \( y(1) = 2 \) is used to determine the constant in the solution, making it uniquely dependent on this condition.

The Product Rule in Calculus

The product rule is a fundamental rule for differentiation used when dealing with the product of two functions. It is stated as follows: if you have two functions \( u(x) \) and \( v(x) \), the derivative of their product is \( u'v + uv' \).

In the context of our problem, the product rule plays a critical role in simplifying the original differential equation. By recognizing that the left-hand side of the equation resembles the derivative of \( x^2 y \), solved using the product rule, the equation can be rewritten in a more manageable form.

• Given: \( u = x^2 \) and \( v = y \).
• Then by the product rule: \( (x^2 y)' = x^2 y' + 2xy \).

This transformation allows us to observe that the equation can be expressed as a derivative, which simplifies solving it tremendously.

Solutions to Differential Equations

A solution to a differential equation is a function that satisfies the given equation. Solving a differential equation often involves finding an expression for the unknown function \( y(x) \) that satisfies both the equation and any initial conditions.

• In our problem, once the equation is rewritten as \( \frac{d}{dx}(x^2 y) = 0 \), it becomes clear that \( x^2 y \) is a constant because its derivative is zero.
• The general solution is \( x^2 y = C \), where \( C \) is an arbitrary constant.

To find this constant, we use the initial condition \( y(1) = 2 \) to calculate \( C \), resulting in \( C = 2 \).
The particular solution is then found by solving for \( y(x) \), giving \( y(x) = \frac{2}{x^2} \). This solution satisfies both the differential equation and the initial condition and thus is the sought-after specific solution of the initial-value problem.