The given differential equation is:

\[
y' + \ln(2) \cdot y = 0
\]

This is a first-order linear differential equation of the form \( y' + P(x)y = 0 \), where \( P(x) \) is a constant, \( \ln(2) \). The general solution can be found using separation of variables.

### Step 1: Rewrite the equation
We can rewrite the equation as:

\[
\frac{dy}{dx} = -\ln(2) \cdot y
\]

### Step 2: Separate variables
Separate \( y \) and \( x \):

\[
\frac{dy}{y} = -\ln(2) \, dx
\]

### Step 3: Integrate both sides
Integrate both sides:

\[
\int \frac{1}{y} \, dy = -\ln(2) \int 1 \, dx
\]

This gives:

\[
\ln|y| = -\ln(2) \cdot x + C
\]

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

### Step 4: Solve for \( y \)
Exponentiate both sides to solve for \( y \):

\[
|y| = e^{-\ln(2) \cdot x + C}
\]

\[
y = \pm e^C \cdot e^{-\ln(2) \cdot x}
\]

Let \( e^C = C_1 \), where \( C_1 \) is a new constant, so the solution becomes:

\[
y = C_1 e^{-\ln(2) \cdot x}
\]

Thus, the general solution is:

\[
y(x) = C_1 e^{-\ln(2) \cdot x}
\]

---

Would you like more details on any specific step? Here are some related questions:

1. How do we apply separation of variables to more complex differential equations?
2. What is the significance of the constant of integration in differential equations?
3. How does the logarithmic function influence the behavior of the solution?
4. Can this method be extended to nonlinear differential equations?
5. What other types of first-order differential equations exist?

**Tip:** Always verify that the general solution satisfies the original equation by substituting it back in.

Learn how to solve the first-order linear differential equation y' + ln(2)y = 0 using separation of variables. This solution involves key concepts like exponential functions and logarithmic properties. Suitable for advanced high school or undergraduate math students.

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