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? $$ x y^{\prime}=2 \frac{\cos x}{x}-3 y $$

Short answer

Solutions vary with initial conditions due to the constant \( c \). Behavior changes at singularities or for different initial values.

Step-by-step solution

Rewrite the Differential Equation

The given differential equation is \( x y^{\prime} = 2 \frac{\cos x}{x} - 3y \). To solve it, we rewrite it in a standard form. Divide all terms by \( x \):\[ y^{\prime} = \frac{2 \cos x}{x^2} - \frac{3y}{x}. \]

Identify the Type of Differential Equation

Observe that the equation can be treated as a linear first-order differential equation of the form \( y^{\prime} + P(x)y = Q(x) \) where:- \( P(x) = \frac{3}{x} \)- \( Q(x) = \frac{2 \cos x}{x^2} \)

Find the Integrating Factor

The integrating factor \( \mu(x) \) is given by:\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{3}{x} \, dx} = e^{3 \ln |x|} = |x|^3. \]

Solve the Differential Equation

Multiply the differential equation by the integrating factor:\[ |x|^3 y^{\prime} + 3|x|^2 y = 2 \cos x |x|. \]This becomes:\[ \frac{d}{dx}(|x|^3 y) = 2 \cos x |x|. \]Integrate both sides:\[ |x|^3 y = \int 2 \cos x |x| \, dx. \]

Perform Integration

To solve \( |x|^3 y = \int 2 \cos x |x| \, dx \):Depending on whether \( x \) is positive or negative, simplify as needed. For positive \( x \), this integral simplifies and results in a solution involving sine and cosine identities. Computing yields:\[ y = \frac{\sin x}{x} + c|x|^{-3}, \]where \( c \) is the constant of integration.

Sketch a Family of Solutions

Each value of \( c \) results in a different curve in the family of solutions. Use graphing software or a calculator to plot these curves. Observe how they behave for different initial conditions.

Analyze Effect of Initial Conditions

Different values of the constant \( c \) in the general solution \( y = \frac{\sin x}{x} + c|x|^{-3} \) correspond to different initial conditions. Some values of \( c \) might lead to singularities or unstable behavior for large or small \( x \).

Key concepts

first-order linear differential equations

A first-order linear differential equation is a type of equation that involves a function and its first derivative. These equations are expressed in the form \( y' + P(x)y = Q(x) \). In this form:

• \( y' \) represents the derivative of the function \( y \) with respect to \( x \).
• \( P(x) \) and \( Q(x) \) are functions of \( x \).

These equations are called 'linear' because they involve the first power of the function and its derivative. The solution typically involves finding a function \( y(x) \), that satisfies the equation across some domain of \( x \). To solve these equations, methods like integrating factors are often used to simplify the process, allowing you to find the general form of the solutions, or a 'family of solutions,' depending on the initial conditions applied.

integrating factor

One of the most efficient tools for solving first-order linear differential equations is the integrating factor. This method involves multiplying the entire differential equation by a specially chosen function that simplifies the process.To find the integrating factor, you compute:\[e^{\int P(x) \, dx}\]where \( P(x) \) is the coefficient of \( y \) in the standard form of the equation. Once you have the integrating factor, you multiply it throughout the differential equation. This transforms it into a form that can be easily integrated on both sides, helping you to solve for \( y \). The integrating factor is crucial because it helps the differential equation to fit into the format needed to apply these integration techniques effectively. This step can significantly simplify the path to finding a solution, connecting the derivative and the function in a single, integrable expression.

initial conditions

When solving differential equations, initial conditions are specific values chosen for the function or its derivatives at a particular point. These conditions determine a unique solution from the general solution or the family of solutions that a differential equation can have. For a first-order equation, an initial condition could be something like \( y(x_0) = y_0 \), where the function \( y \) equals \( y_0 \) at \( x = x_0 \).Initial conditions are vital because they allow you to find the specific curve out of an infinite set of possible solutions. The general solution includes a constant of integration, \( c \), which can be adjusted by substituting the initial condition into the solution. This pinpointing transforms the family of solutions into a unique solution that satisfies both the equation and the condition. Without this specific information, you would only know a general potential set of solutions without knowing which one applies to a particular scenario.

family of solutions

A family of solutions refers to the infinite set of solutions to a differential equation that differ only by a constant multiple. In first-order linear differential equations, the general solution often involves a constant of integration, \( c \). This constant represents the family characteristic, as it can take on any real number value.Exploring a family of solutions means understanding how different values of \( c \) affect the shape and behavior of the solution curve. For instance, in our specific exercise, the solution to the differential equation includes a term with \( c \, |x|^{-3} \). Varying \( c \) will shift and adjust each curve, showing the range of possible solutions. This helps visually and analytically depict how different scenarios or initial conditions lead to different behaviors in the solution. By examining the family, you get a comprehensive view of how the solution behaves across different integrals and conditions.