To solve the differential equation $y' = 3e^x - 3y$, we can use an integrating factor approach since this is a first-order linear differential equation of the form $y' + p(x)y = q(x)$.

1. Rewrite the equation:
   $y' + 3y = 3e^x$ where $p(x) = 3$ and $q(x) = 3e^x$.

2. Find the integrating factor:
   The integrating factor $\mu(x)$ is given by $e^{\int p(x) \, dx} = e^{\int 3 \, dx} = e^{3x}$.

3. Multiply through by the integrating factor:
   Multiply both sides of the differential equation by $e^{3x}$: $e^{3x} y' + 3e^{3x} y = 3e^{4x}$ This simplifies to: $\frac{d}{dx}(e^{3x} y) = 3e^{4x}$

4. Integrate both sides:
   Integrate with respect to $x$: $e^{3x} y = \int 3e^{4x} \, dx$ The integral on the right is: $\int 3e^{4x} \, dx = \frac{3}{4}e^{4x} + C$ where $C$ is the constant of integration.

   So we have: $e^{3x} y = \frac{3}{4}e^{4x} + C$

5. Solve for $y$:
   Divide by $e^{3x}$ to isolate $y$: $y = \frac{3}{4}e^x + Ce^{-3x}$

Solution

The general solution to the differential equation is: $y = \frac{3}{4}e^x + Ce^{-3x}$ where $C$ is an arbitrary constant.

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

Here are some related questions to explore:

1. How would you find a particular solution if given an initial condition?
2. What is the role of the integrating factor in solving linear differential equations?
3. How does the solution behavior change as $x \to \infty$?
4. Can we use a different method, like separation of variables, for this equation?
5. How would you solve this if the equation were non-linear?

Tip: Always check if your differential equation can be simplified or transformed before applying a specific method.