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

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

A: \( y = Ce^{3x} \)

### Worksheet

**Step 1: Recognize the Form**

The given differential equation is of the form \( \frac{dy}{dx} = ky \), where \( k \) is a constant (in this case, \( k = 3 \)). This is a separable differential equation.

**Step 2: Separate the Variables**

To separate the variables, we rewrite the equation so that all terms involving \( y \) are on one side and all terms involving \( x \) are on the other side:

\[ \frac{dy}{y} = 3 \, dx \]

**Step 3: Integrate Both Sides**

Integrate both sides of the equation:

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

**Step 4: Perform the Integration**

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

\[ \ln|y| = 3x + C \]

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

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

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

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

Since \( e^{\ln|y|} = |y| \), we can write:

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

We can simplify \( e^{3x + C} \) as \( e^{3x} \cdot e^C \). Let \( e^C = C' \), where \( C' \) is a new constant (which can be positive or negative). Thus:

\[ |y| = C' e^{3x} \]

Therefore, \( y = \pm C' e^{3x} \). To simplify the notation, we can combine the \(\pm\) and \(C'\) into a single constant \( C \), which can be any real number. Hence:

\[ y = C e^{3x} \]

Final Solution

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

\[ y = C e^{3x} \]

where \( C \) is an arbitrary constant.