This is a first-order linear differential equation:

$\frac{dy}{dx} + 3y = 3x^3 - 4x^2 - 5x$

We can solve it using an integrating factor. The general form of a linear differential equation is:

$\frac{dy}{dx} + P(x)y = Q(x)$

Here, $P(x) = 3$ and $Q(x) = 3x^3 - 4x^2 - 5x$.

Step 1: Find the integrating factor

The integrating factor, $\mu(x)$, is given by:

$\mu(x) = e^{\int P(x) \, dx} = e^{\int 3 \, dx} = e^{3x}$

Step 2: Multiply the entire equation by the integrating factor

We multiply both sides of the equation by $e^{3x}$:

$e^{3x} \frac{dy}{dx} + 3e^{3x}y = \left( 3x^3 - 4x^2 - 5x \right) e^{3x}$

The left-hand side simplifies to:

$\frac{d}{dx} \left( y e^{3x} \right) = \left( 3x^3 - 4x^2 - 5x \right) e^{3x}$

Step 3: Integrate both sides

We now integrate both sides with respect to $x$:

$y e^{3x} = \int \left( 3x^3 - 4x^2 - 5x \right) e^{3x} \, dx$

This is an integration by parts problem, and to solve it, we can break down the integrals for each term on the right-hand side.

Step 4: Solve the integral

We'll solve each term separately:

1. For the term $3x^3 e^{3x}$,
2. For $-4x^2 e^{3x}$,
3. For $-5x e^{3x}$.

The complete solution involves repeated integration by parts. To simplify things, the result of this integration is:

[ y e^{3x} = \frac{x^3}{e^{3x}} - \frac{4x^2