Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, solve each differential equation. $$ \frac{d y}{d x}-\frac{y}{x}=3 x^{3} ; y=3 \text { when } x=1 $$

Step-by-Step Solution

Answer

The solution is \( y = x^4 + 2x \).

Step 1: Identify Equation Type

The given differential equation \( \frac{d y}{d x}-\frac{y}{x}=3x^{3} \) is in the linear form \( \frac{dy}{dx} + P(x)y = Q(x) \), with \( P(x) = -\frac{1}{x} \) and \( Q(x) = 3x^3 \).

Step 2: Find the Integrating Factor

The integrating factor is calculated as \( \mu(x) = e^{\int P(x) \, dx} = e^{-\int \frac{1}{x} \, dx} = e^{- ext{ln}|x|} = \frac{1}{x} \).

Step 3: Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \frac{1}{x} \) to obtain: \( \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = 3x^2 \).

Step 4: Rewrite Equation as the Derivative of a Product

Recognize the left-hand side as the derivative of a product: \( \frac{d}{dx} \left( \frac{y}{x} \right) = 3x^2 \).

Step 5: Integrate Both Sides

Integrate both sides: \( \int \frac{d}{dx} \left( \frac{y}{x} \right) \, dx = \int 3x^2 \, dx \). This results in \( \frac{y}{x} = x^3 + C \), where \( C \) is the constant of integration.

Step 6: Solve for y

Multiply through by \( x \) to isolate \( y \): \( y = x^4 + Cx \).

Step 7: Use Initial Condition to Find C

Use the initial condition \( y = 3 \) when \( x = 1 \) to solve for \( C \): \( 3 = 1^4 + C \times 1 \). Thus, \( 3 = 1 + C \), leading to \( C = 2 \).

Step 8: Write the Final Solution

Substitute \( C = 2 \) back into the equation for \( y \): \( y = x^4 + 2x \).

Key Concepts

Linear Differential Equations

A linear differential equation is an equation that involves a dependent variable and its derivatives, where the dependent variable is linear in all its terms. The general form of a first-order linear differential equation is:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

In this exercise, we identified the given equation \( \frac{d y}{d x}-\frac{y}{x}=3 x^{3} \) as a linear differential equation by rewriting it in the standard form. Here, the function \( P(x) \) is \(-\frac{1}{x}\), and \( Q(x) \) is \(3x^{3}\).
Understanding this form is crucial because it allows us to utilize specific techniques like finding an integrating factor to solve the equation.

Integrating Factor

The concept of an integrating factor is a powerful tool for solving linear differential equations. It is a function that, when multiplied with the differential equation, transforms it into an exact equation which is easier to solve. The integrating factor \( \mu(x) \) is calculated as follows:

• \( \mu(x) = e^{\int P(x) \, dx} \)
• For the exercise, with \( P(x) = -\frac{1}{x} \), the integrating factor becomes:
• \( \mu(x) = e^{-\int \frac{1}{x} \, dx} = e^{-\ln|x|} = \frac{1}{x} \)

Once obtained, we multiply the entire differential equation by this integrating factor. This process allows us to convert the equation into the derivative of a product form, simplifying further integration.

Initial Value Problem

An initial value problem (IVP) not only involves solving a differential equation but also requires the solution to satisfy an initial condition, such as \( y = 3 \) when \( x = 1 \). The initial condition helps us determine the specific solution curve out of the family of possible solutions.

During the example, after integrating and solving for \( y \) in terms of \( x \) and a constant \( C \), we input the values from the initial condition to find \( C \). So we obtained:
\( 3 = 1^4 + C \) leads to \( C = 2 \).

This step is essential to tailor the general solution to satisfy particular real-world situations or specific scenarios outlined by the problem.

Solving Differential Equations

Solving differential equations often involves several steps, especially for linear types. In our case, after identifying the type and finding the integrating factor, we:

• Multiplied the equation by the integrating factor.
• Recognized and rewrote it as the derivative of a product, \( \frac{d}{dx} \left( \frac{y}{x} \right) = 3x^2 \).
• Integrated both sides to find \( \frac{y}{x} = x^3 + C \).
• Used the initial condition to solve for the constant \( C \).

Finally, substituting back \( C = 2 \), we get the particular solution \( y = x^4 + 2x \).
This approach ensures we methodically address each part of the equation and find a solution that fits all given conditions.