Problem 10 - Thomas Calculus

Question

Solve the differential equations in Exercises \(1-14\) $$ x \frac{d y}{d x}=\frac{\cos x}{x}-2 y, \quad x>0 $$

Step-by-Step Solution

Answer

The solution is \( y = \frac{\sin x + C}{x^2} \), where \( C \) is a constant.

Step 1: Identify the Type of Differential Equation

The given differential equation is \( x \frac{dy}{dx} = \frac{\cos x}{x} - 2y \). This is a first-order linear differential equation of the form \( x \frac{dy}{dx} + 2y = \frac{\cos x}{x} \).

Step 2: Rearrange the Equation

Rearrange the equation into the standard format for a linear differential equation: \( \frac{dy}{dx} + \frac{2}{x}y = \frac{\cos x}{x^2} \).

Step 3: Find the Integrating Factor

The integrating factor \( \mu(x) \) is determined by \( e^{\int P(x) \, dx} \), where \( P(x) = \frac{2}{x} \). Thus, \( \mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2\ln x} = x^2 \).

Step 4: Multiply through by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( x^2 \): \( x^2 \frac{dy}{dx} + 2xy = \cos x \).

Step 5: Convert Left-Hand Side into a Derivative

Realize that the left-hand side is the derivative of a product: \( \frac{d}{dx}(x^2 y) = \cos x \).

Step 6: Integrate Both Sides

Integrate both sides with respect to \( x \): \( \int \frac{d}{dx}(x^2 y) \, dx = \int \cos x \, dx \). This results in \( x^2 y = \sin x + C \), where \( C \) is the integration constant.

Step 7: Solve for \( y \)

Rearrange the equation \( x^2 y = \sin x + C \) to solve for \( y \): \( y = \frac{\sin x + C}{x^2} \).

Key Concepts

Integrating Factor

In solving first-order linear differential equations, the integrating factor is a powerful tool. It's essentially a function that simplifies the equation and enables us to find its solution more easily. The basic idea is to transform a non-exact equation into an exact one by multiplying every term by this integrating factor. To find the integrating factor, we take the coefficient of the unknown function, referred to as \( P(x) \), in its standard linear form \( \frac{dy}{dx} + P(x) y = Q(x) \).The integrating factor, \( \mu(x) \), is given by the expression \( e^{\int P(x) \, dx} \). In our exercise, the differential equation has \( P(x) = \frac{2}{x} \), which leads to an integrating factor \( \mu(x) = x^2 \). By multiplying the entire equation by \( x^2 \), we effectively simplify the equation, allowing the left-hand side to be expressed as the derivative of a product.

Linear Differential Equation

A linear differential equation, in the context of first-order equations, typically involves a function and its derivative. The structure is linear because each term is either the function itself or its derivative, with no other functional dependencies. It generally takes the form \( \frac{dy}{dx} + P(x) y = Q(x) \).This form makes solving these equations manageable, as they can be systematically approached using integrating factors. In our given exercise, after rearranging, the differential equation is expressed as \( \frac{dy}{dx} + \frac{2}{x} y = \frac{\cos x}{x^2} \). Recognizing this form allows us to apply the integrating factor method, which is key to finding the solution.

Integration Constant

When solving differential equations, integrating both sides will typically introduce an integration constant, often represented as \( C \). This constant emerges from the indefinite integral, reflecting the fact that antiderivatives are determined up to a constant.In practice, the integration constant represents the general nature of solutions to differential equations, encompassing a family of curves. When more information (like initial conditions) is provided, we can determine a specific value for \( C \), thus specifying a unique solution. In our exercise, after integrating both sides, we reach the expression \( x^2 y = \sin x + C \). Here, \( C \) is the integration constant, illustrating the generality of the solution \( y = \frac{\sin x + C}{x^2} \). Without additional conditions, \( C \) remains undetermined, offering a wide range of potential solutions.