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Derivatives of Cosecant and Secant in Differential Equations

In calculus and differential equations, understanding the derivatives of cosecant (\\( \csc x \\)) and secant (\\( \sec x \\)) functions is essential, as these often appear in various ODEs. Here, I'll illustrate three examples where derivatives of these functions play a role in solving ordinary differential equations (ODEs). Each example will cover different cases and types of ODEs that frequently incorporate \\( \csc x \\) and \\( \sec x \\) derivatives.

Key Derivatives

Before diving into the examples, let's review the derivatives of \\( \csc x \\) and \\( \sec x \\):

1. Derivative of \\( \csc x \\): \[ \frac{d}{dx}(\csc x) = -\csc x \cot x \]

2. Derivative of \\( \sec x \\): \[ \frac{d}{dx}(\sec x) = \sec x \tan x \]

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Example 1: Solving a Simple First-Order ODE with \\( \csc x \\) Function

Consider the differential equation:

\[ \frac{dy}{dx} = y \cdot \csc x \]

Solution:

1. Separate Variables: \[ \frac{1}{y} \, dy = \csc x \, dx \]

2. Integrate Both Sides: \[ \int \frac{1}{y} \, dy = \int \csc x \, dx \] The integral on the left side is \\( \ln |y| \\). The integral of \\( \csc x \\) is \\( \ln | \csc x - \cot x | \\): \[ \ln |y| = \ln | \csc x - \cot x | + C \]

3. Solve for \\( y \\): \[ y = e^{\ln |\csc x - \cot x| + C} = A (\csc x - \cot x) \]

Thus, the solution is:

\[ y = A (\csc x - \cot x) \]

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Example 2: Solving a Second-Order Linear ODE Involving \\( \sec x \\)

Consider the second-order ODE:

\[ \frac{d^2y}{dx^2} - \sec x \, \frac{dy}{dx} = 0 \]

Solution:

1. Set \\( v = \frac{dy}{dx} \\) and rewrite the ODE: \[ \frac{dv}{dx} - \sec x \, v = 0 \]

2. Separate Variables: \[ \frac{1}{v} \, dv = \sec x \, dx \]

3. Integrate Both Sides: \[ \ln |v| = \ln | \sec x + \tan x | + C \]

4. Solve for \\( v \\): \[ v = A (\sec x + \tan x) \]

Thus, the general solution is:

\[ y = A \ln | \sec x + \tan x | + B \]

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Example 3: Nonhomogeneous First-Order ODE with \\( \csc x \\) as a Forcing Function

Consider the first-order ODE:

\[ \frac{dy}{dx} + y \cdot \cot x = \csc x \]

Solution:

1. Identify the Integrating Factor: \[ \mu(x) = e^{\int \cot x \, dx} = |\sin x| \]

2. Multiply Through by the Integrating Factor: \[ \frac{d}{dx} \left( y \cdot \sin x \right) = 1 \]

3. Integrate Both Sides: \[ y \cdot \sin x = x + C \]

4. Solve for \\( y \\): \[ y = \frac{x + C}{\sin x} \]