Problem 10 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Solve the given differential equation. \(\frac{d y}{d x}=\frac{\cos x}{-\sin y}\)

Step-by-Step Solution

Answer

The solution to the differential equation \(\frac{d y}{d x}=\frac{\cos x}{-\sin y}\) is given by the equation: |-\csc y| = A|\tan x + \sec x|.

Step 1 Separation of Variables

First, write the equation as a product of a function of \(x\) and function of \(y\). This would be done by multiplying both sides by \(-\sin y\). The equation will then appear as: \(-\sin y \frac{d y}{d x}=\cos x\). Then, divide each side by their respective variables, \(-\sin y\) and \(\cos x\). This results in: \(\frac{dy}{-\sin y} = \frac{dx}{\cos x} \). This allows for the process of separation of variables, leaving a function of \(y\) on one side and a function of \(x\) on the other.

Step 2 Integration

The next step is to integrate both sides with respect to their respective variables. Note that the integral of \( \frac{1}{\cos x} \) is \( \ln|\sec x + \tan x| + C \) and the integral of \( \frac{1}{-\sin y} \) is \( -\ln|\csc y| \). After integration, the equation would thus be: \( \ln|-\csc y| = \ln|\sec x + \tan x| + C \).

Step 3 Simplification

Apply the property of logarithms where if \( \ln a = \ln b + \ln c \), then \( \ln a - \ln b = \ln c \). This effectively removes the logarithms from the equation. Then the equation would be: |-\csc y| = |\tan x + \sec x|e^C. Since C is an arbitrary constant, let \( \frac{1}{|e^C|} \) be another constant \( \frac{1}{A} \). This transforms the equation into |-\csc y| = A|\tan x + \sec x|. This is the general solution to the given differential equation in implicit form. To solve for y explicitly, we can further manipulate this equation.

Key Concepts

Separation of Variables

Separation of variables is a fundamental technique used to solve differential equations, especially when they can be written as the product of two separate functions of different variables. In the problem presented, the differential equation is initially given as \( \frac{dy}{dx} = \frac{\cos x}{-\sin y} \). The goal of separation of variables is to rearrange this equation so that all terms involving \( y \) are on one side and all terms involving \( x \) are on the other.
To achieve this:

• We multiply both sides of the equation by \(-\sin y\).
• This results in \(-\sin y \frac{dy}{dx} = \cos x\).
• By dividing each side by their respective functions, \( -\sin y \) and \( \cos x \), we achieve separation: \( \frac{dy}{-\sin y} = \frac{dx}{\cos x} \).

This transformation allows us to integrate each side independently with respect to its own variable, facilitating the solution process through integration in the next steps.

Integration

Integration is a crucial step in solving a differential equation once the variables are separated. It involves finding the antiderivatives or integrals of the functions on each side of the equation. For our separated equation, \( \frac{dy}{-\sin y} = \frac{dx}{\cos x} \), we perform integration as follows:

• The left side, \( \int \frac{dy}{-\sin y} \), integrates to \( -\ln|\csc y| \).
• The right side, \( \int \frac{dx}{\cos x} \), results in \( \ln|\sec x + \tan x| + C \), where \(C\) is the constant of integration.

Integration changes the form of the equation into \( -\ln|\csc y| = \ln|\sec x + \tan x| + C \).
The constant of integration \(C\) arises because when integrating, any constant could have been differentiated to zero, so we include an arbitrary constant to account for all potential solutions. This is integral to representing the family of all possible solutions.

Trigonometric Identities

Trigonometric identities are mathematical tools that help simplify and solve equations involving trigonometric functions. In this exercise, recognizing and using identities allows us to simplify the integration results and rewrite the equation in a more familiar format. Here's how some identities come into play:

• The integral \( \int \frac{1}{\cos x} \, dx = \ln|\sec x + \tan x| + C \) stems from recognizing that \( \frac{1}{\cos x} \) is \( \sec x \).
• The identity \( \csc y = \frac{1}{\sin y} \) helps us express the integral \( \int \frac{dy}{-\sin y} \) as \( -\ln|\csc y| \).

Using these identities allows for a clearer translation of the integration process results into a form that is easier to manage, aiding in further simplifications or checks if needed.
Ultimately, these trigonometric relationships are pivotal in navigating the complexities of the solution and ensuring that the algebraic manipulations add up unmistakably.