Question

Solve the following differential equations. Use your calculator to draw a family of solutions. Are there certain initial conditions that change the behavior of the solution? $$ x^{3} y^{\prime}+2 x^{2} y=x+1 $$

Short answer

The initial conditions, expressed through constant \( C \), affect the asymptotic solutions and behavior.

Step-by-step solution

Identify the Type of the Differential Equation

The given differential equation is \( x^{3} y^{\prime} + 2x^{2} y = x + 1 \). This is a first-order linear differential equation in standard form \( y' + P(x)y = Q(x) \) after dividing the entire equation by \( x^3 \).

Rewrite the Equation in Standard Form

Divide every term by \( x^3 \) to obtain the standard form: \( y' + \frac{2}{x} y = \frac{1}{x^3} + \frac{1}{x^3}x \). Simplifying gives \( y' + \frac{2}{x} y = \frac{1}{x^2} + \frac{1}{x^3} \).

Find the Integrating Factor

The integrating factor \( \mu(x) \) is found using the formula \( \mu(x) = e^{\int P(x) \, dx} \), where \( P(x) = \frac{2}{x} \). Calculating \( \int \frac{2}{x} \, dx = 2 \ln |x| = \ln x^2 \), hence \( \mu(x) = e^{\ln x^2} = x^2 \).

Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by \( x^2 \) to get: \( x^2 y' + 2xy = \frac{x}{x} + \frac{1}{x} \). This simplifies the equation to the derivative form \( \frac{d}{dx}(x^2 y) = x + \frac{1}{x} \).

Integrate Both Sides

Integrate both sides of the equation \( \frac{d}{dx}(x^2 y) = x + \frac{1}{x} \): the left side integrals to \( x^2 y \), and the right side evaluates to \( \frac{x^2}{2} + \ln |x| + C \), where \( C \) is the constant of integration.

Solve for y

To solve for \( y \), divide through by \( x^2 \) to obtain \( y = \frac{x^2}{2x^2} + \frac{\ln |x|}{x^2} + \frac{C}{x^2} = \frac{1}{2} + \frac{\ln |x| + C}{x^2} \).

Analyze the Solution and Initial Conditions

The family of solutions is \( y = \frac{1}{2} + \frac{\ln |x| + C}{x^2} \). Initial conditions choosing specific values of \( C \) can change the behavior, especially affecting asymptotic behavior as \( x \to \infty \) or near \( x = 0 \). For example, if \( C \) is large, the \( \frac{C}{x^2} \) term dominates.

Key concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a specific type of differential equation involving the first derivative of a function. In our case, it appears as: - The general form is \( y' + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \).- Our exercise involves transforming the given equation into this standard form. Initially, the equation is given as \( x^{3} y^{\prime} + 2 x^{2} y = x + 1 \), which becomes linear after rewriting.By dividing every term by \( x^3 \), we reshape it to fit the required formula. Thus, we understand that the solution process involves several transformations to simplify the equation's form, making it approachable for sequential resolution. Linear differential equations like this have various solution methods, including the integrating factor technique.

Integrating Factor

The integrating factor is a crucial tool when solving first-order linear differential equations. It simplifies the differential equation into a single integrated form.- To find the integrating factor, denoted as \( \mu(x) \), use the formula: \( \mu(x) = e^{\int P(x) \, dx} \).- Here, \( P(x) = \frac{2}{x} \) from the rewritten equation. Integrating this gives \( 2 \ln|x| \), and when transformed, \( \mu(x) = e^{\ln x^2} = x^2 \).The integrating factor makes our equation manageable by ensuring the left side becomes an exact derivative. Once the equation is multiplied by this factor, both sides are integrated, allowing us to find an explicit solution for \( y \). Understanding the role of the integrating factor is pivotal in solving and simplifying linear differential equations.

Initial Conditions

Initial conditions are specific values that allow us to find a unique solution from a family of solutions in differential equations.- In our solution, we get the general solution \( y = \frac{1}{2} + \frac{\ln |x| + C}{x^2} \), which involves a constant \( C \).- The initial conditions are used to solve for \( C \), hence determining the particular solution for a given problem.Choosing different initial values of \( y \) when \( x \) is known allows you to calculate \( C \) explicitly. By doing so, you ensure that the solution adheres to a specific curve out of the many possible solutions provided by the general equation. Initial conditions tailor the solutions to specific real-world scenarios, making them practically relevant.

Asymptotic Behavior

Asymptotic behavior refers to how a solution behaves as the independent variable approaches extrema, like infinity or zero, in differential equations.- In our exercise, we observe how \( y = \frac{1}{2} + \frac{\ln |x| + C}{x^2} \) behaves for large or small \( x \).- When \( x \to \infty \), the \( \frac{\ln |x| + C}{x^2} \) term approaches zero, making \( y \) tend towards \( \frac{1}{2} \).- As \( x \) approaches zero, the behavior is dominated by \( \frac{\ln |x| + C}{x^2} \), which can drastically change depending on \( C \).By analyzing the asymptotic behavior, we can predict long-term trends or potential instabilities in solutions. It's crucial for understanding the implications or predictions made by differential equations, especially in complex systems.