# Solve the first-order ODE: \(\frac{dy}{dx} = 3y\)

**1. Write down the given differential equation:**

### \[

\frac{dy}{dx} = 3y

\]

**2. Separate the variables:**

To separate the variables, we need to get all the \(y\)-terms on one side and the \(x\)-terms on the other side. Divide both sides by \(y\) (assuming \(y \neq 0\)) and multiply both sides by \(dx\):

\[

\frac{1}{y} \, dy = 3 \, dx

\]

**3. Integrate both sides:**

Integrate the left side with respect to \(y\) and the right side with respect to \(x\):

\[

\int \frac{1}{y} \, dy = \int 3 \, dx

\]

**4. Evaluate the integrals:**

The integral of \(\frac{1}{y}\) with respect to \(y\) is \(\ln|y|\), and the integral of \(3\) with respect to \(x\) is \(3x\):

\[

\ln|y| = 3x + C

\]

Here, \(C\) is the constant of integration.

**5. Solve for \(y\):**

To solve for \(y\), we exponentiate both sides to remove the natural logarithm:

\[

e^{\ln|y|} = e^{3x + C}

\]

Since \(e^{\ln|y|} = |y|\) and \(e^{3x + C} = e^{3x} \cdot e^C\), we get:

\[

|y| = e^{3x} \cdot e^C

\]

Let \(e^C = C’\), where \(C’\) is a new constant (still a constant, but written in a simpler form):

\[

|y| = C’ e^{3x}

\]

Since \(C’\) can be any positive constant, we write it as a general constant \(C\) which can be positive or negative:

\[

y = Ce^{3x}

\]

where \(C\) is an arbitrary constant.

**Final Answer**

The general solution to the differential equation \(\frac{dy}{dx} = 3y\) is:

### \[

y = Ce^{3x}

\]

where \(C\) is an arbitrary constant.