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.

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