Thank you for the clarification! Let's solve the corrected differential equation:

### Corrected Differential Equation:
\[
\frac{dy}{dx} + 3y = e^{mx}
\]
where \(m\) is a given real number.

This is a **first-order linear differential equation** of the form:
\[
\frac{dy}{dx} + P(x)y = Q(x),
\]
where \(P(x) = 3\) and \(Q(x) = e^{mx}\).

### Steps to Solve:

1. **Find the Integrating Factor (IF):**

   The integrating factor for a linear differential equation is given by:
   \[
   \mu(x) = e^{\int P(x) \, dx}.
   \]

   In our case, \(P(x) = 3\), so:
   \[
   \mu(x) = e^{\int 3 \, dx} = e^{3x}.
   \]

2. **Multiply Through by the Integrating Factor:**

   Multiply the entire differential equation by the integrating factor \(e^{3x}\):
   \[
   e^{3x} \frac{dy}{dx} + 3e^{3x} y = e^{3x} e^{mx}.
   \]

   This simplifies to:
   \[
   e^{3x} \frac{dy}{dx} + 3e^{3x} y = e^{(m+3)x}.
   \]

3. **Rewrite the Left Side as a Derivative:**

   Notice that the left side is the derivative of \((e^{3x} y)\):
   \[
   \frac{d}{dx} (e^{3x} y) = e^{(m+3)x}.
   \]

4. **Integrate Both Sides:**

   Integrate both sides with respect to \(x\):
   \[
   e^{3x} y = \int e^{(m+3)x} \, dx.
   \]

5. **Compute the Integral:**

   The integral of \(e^{(m+3)x}\) is \(\frac{e^{(m+3)x}}{m+3}\), provided \(m \neq -3\). Thus:
   \[
   e^{3x} y = \frac{e^{(m+3)x}}{m+3} + C,
   \]
   where \(C\) is the constant of integration.

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

   Divide both sides by \(e^{3x}\) to isolate \(y\):
   \[
   y = \frac{e^{(m+3)x}}{(m+3)e^{3x}} + \frac{C}{e^{3x}}.
   \]

   Simplify further:
   \[
   y = \frac{e^{(m+3)x - 3x}}{m+3} + Ce^{-3x}.
   \]

   This reduces to:
   \[
   y = \frac{e^{mx}}{m+3} + Ce^{-3x}.
   \]

   This is the general solution for \(m \neq -3\).

7. **Special Case when \(m = -3\):**

   If \(m = -3\), the equation becomes:
   \[
   \frac{dy}{dx} + 3y = e^{-3x}.
   \]

   The integrating factor is still \(e^{3x}\), so:
   \[
   \frac{d}{dx} (e^{3x} y) = e^{3x} e^{-3x} = 1.
   \]

   Integrating both sides:
   \[
   e^{3x} y = \int 1 \, dx = x + C.
   \]

   Solving for \(y\), we get:
   \[
   y = e^{-3x} (x + C).
   \]

### Final Answer:

The one-parameter family of solutions for the differential equation \(\frac{dy}{dx} + 3y = e^{mx}\) is:

\[
y = 
\begin{cases} 
\frac{e^{mx}}{m+3} + Ce^{-3x}, & \text{if } m \neq -3 \\
e^{-3x}(x + C), & \text{if } m = -3
\end{cases}
\]
where \(C\) is an arbitrary constant.

Would you like more details on any step, or do you have any specific questions?

### Further Exploration:

1. How do you apply the integrating factor method to other linear differential equations?
2. What happens to the solution when \(m = -3\) in terms of its behavior at infinity?
3. Can you use an alternative method, such as variation of parameters, to solve this differential equation?
4. How does the choice of initial conditions affect the value of the constant \(C\)?
5. What is the physical interpretation of this solution if it were to model a real-world scenario?

### Tip:

Always check for special cases in differential equations, like when parameters might lead to division by zero or other undefined behaviors.

Learn how to solve the differential equation dy/dx + 3y = e^mx step-by-step. This problem involves advanced concepts in differential equations, including the integrating factor method. Suitable for advanced college-level students and anyone interested in differential equations.

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