We are asked to solve the first-order linear initial value problem:

$\frac{dy}{dx} + 3y = 3x^3 - 4x^2 - 5x, \quad y(0) = 0$

Step 1: Write the equation in standard linear form

The given equation is already in standard form:

$\frac{dy}{dx} + P(x)y = Q(x)$ where $P(x) = 3$ and $Q(x) = 3x^3 - 4x^2 - 5x$.

Step 2: Find the integrating factor

The integrating factor is found using the formula:

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

Step 3: Multiply through by the integrating factor

Multiply both sides of the differential 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 becomes:

$\frac{d}{dx} \left( y e^{3x} \right)$

Thus, the equation becomes:

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

Step 4: Integrate both sides

Now integrate both sides with respect to $x$:

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

We need to integrate the right-hand side, which involves integrating by parts. I'll compute this step.The integral of $\left(3x^3 - 4x^2 - 5x\right)e^{3x}$ is:

$\int \left(3x^3 - 4x^2 - 5x\right)e^{3x} \, dx = \frac{(27x^3 - 63x^2 - 3x + 1)}{27} e^{3x}$

Step 5: Write the general solution

Now we have:

$y e^{3x} = \frac{(27x^3 - 63x^2 - 3x + 1)}{27} e^{3x} + C$

Dividing both sides by $e^{3x}$, we get:

$y(x) = \frac{(27x^3 - 63x^2 - 3x + 1)}{27} + Ce^{-3x}$

Step 6: Apply the initial condition $y(0) = 0$

Substitute $x = 0$ and $y(0) = 0$ into the solution:

$0 = \frac{(27(0)^3 - 63(0)^2 - 3(0) + 1)}{27} + C e^{0}$ $0 = \frac{1}{27} + C$ $C = -\frac{1}{27}$

Step 7: Final solution

Substitute $C = -\frac{1}{27}$ into the general solution:

$y(x) = \frac{(27x^3 - 63x^2 - 3x + 1)}{27} - \frac{1}{27}e^{-3x}$

Simplifying:

$y(x) = \frac{27x^3 - 63x^2 - 3x}{27} + \frac{1}{27}(1 - e^{-3x})$

Thus, the solution to the differential equation is:

$y(x) = x^3 - \frac{7x^2}{3} - \frac{x}{9} + \frac{1}{27}(1 - e^{-3x})$

Let me know if you would like more details or have questions!

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Here are 5 related questions:

1. How is the integrating factor method applied to general linear differential equations?
2. What is the significance of the initial condition $y(0) = 0$ in determining the constant?
3. Can integrating factors be applied to non-linear differential equations?
4. How does one perform integration by parts for more complex functions?
5. What is the general solution form for homogeneous first-order linear differential equations?

Tip: Always check that your final solution satisfies the original equation and any given initial conditions.