Solve the first-order linear differential equation $y' - \cot(x)y = \tan(x)$.

1. Problem Description

Solve the first-order linear differential equation $y' - \cot(x)y = \tan(x)$.

2. Solution Steps

This is a first-order linear differential equation of the form $y' + P(x)y = Q(x)$, where $P(x) = -\cot(x)$ and $Q(x) = \tan(x)$.

First, find the integrating factor $\mu(x)$:

$$\mu(x) = e^{\int P(x) dx} = e^{\int -\cot(x) dx} = e^{-\int \frac{\cos(x)}{\sin(x)} dx}$$

Let $u = \sin(x)$, then $du = \cos(x) dx$.

So, $\int \frac{\cos(x)}{\sin(x)} dx = \int \frac{1}{u} du = \ln|u| = \ln|\sin(x)|$.

Then, $\mu(x) = e^{-\ln|\sin(x)|} = e^{\ln|\sin(x)|^{-1}} = \frac{1}{|\sin(x)|}$.

Since we are looking for a solution, we can drop the absolute value and take $\mu(x) = \frac{1}{\sin(x)}$.

Multiply the original equation by the integrating factor:

$$\frac{1}{\sin(x)}y' - \frac{\cot(x)}{\sin(x)}y = \frac{\tan(x)}{\sin(x)}$$

$$\frac{1}{\sin(x)}y' - \frac{\cos(x)}{\sin^2(x)}y = \frac{\sin(x)}{\cos(x)\sin(x)}$$

$$\frac{1}{\sin(x)}y' - \frac{\cos(x)}{\sin^2(x)}y = \frac{1}{\cos(x)}$$

The left side is the derivative of $\frac{y}{\sin(x)}$:

$$\frac{d}{dx}\left(\frac{y}{\sin(x)}\right) = \frac{1}{\cos(x)} = \sec(x)$$

Integrate both sides with respect to $x$:

$$\int \frac{d}{dx}\left(\frac{y}{\sin(x)}\right) dx = \int \sec(x) dx$$

$$\frac{y}{\sin(x)} = \int \sec(x) dx = \ln|\sec(x) + \tan(x)| + C$$

$$y = \sin(x)(\ln|\sec(x) + \tan(x)| + C)$$

3. Final Answer

$y = \sin(x)(\ln|\sec(x) + \tan(x)| + C)$