The given differential equation is:

\[
\tan(x) \frac{dy}{dx} = y
\]

This is a first-order separable differential equation. To solve it, follow these steps:

### Step 1: Rearrange the equation
Rearrange the equation to separate the variables \(y\) and \(x\):

\[
\frac{1}{y} \, dy = \frac{1}{\tan(x)} \, dx
\]

Since \(\frac{1}{\tan(x)} = \cot(x)\), we can rewrite it as:

\[
\frac{1}{y} \, dy = \cot(x) \, dx
\]

### Step 2: Integrate both sides
Now, integrate both sides:

\[
\int \frac{1}{y} \, dy = \int \cot(x) \, dx
\]

The integral of \(\frac{1}{y}\) is \(\ln|y|\), and the integral of \(\cot(x)\) is \(\ln|\sin(x)|\), so we get:

\[
\ln|y| = \ln|\sin(x)| + C
\]

where \(C\) is the constant of integration.

### Step 3: Solve for \(y\)
Exponentiate both sides to eliminate the logarithms:

\[
|y| = e^{\ln|\sin(x)| + C} = e^C \cdot |\sin(x)|
\]

Let \(e^C = C_1\), where \(C_1\) is a new constant. Therefore, we have:

\[
|y| = C_1 |\sin(x)|
\]

Thus, the general solution is:

\[
y = C_1 \sin(x)
\]

### Final Answer:
The general solution to the differential equation \(\tan(x) \frac{dy}{dx} = y\) is:

\[
y = C_1 \sin(x)
\]

Would you like further details or clarifications?

Here are 5 related questions you might find helpful:
1. How would the solution change if the equation were \(\tan(x) \frac{dy}{dx} = y^2\)?
2. What are separable differential equations, and how do you identify them?
3. How do you solve non-separable differential equations?
4. Can we solve this differential equation using another method like integrating factors?
5. How do boundary conditions affect the solution to a differential equation?

**Tip:** When solving separable differential equations, always focus on isolating terms involving \(y\) on one side and terms involving \(x\) on the other before integrating.

Learn how to solve the separable differential equation (tan x)dy/dx = y with a step-by-step explanation. This problem involves logarithmic integration and covers the use of trigonometric functions in differential equations. Suitable for undergraduate calculus students.

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