The differential equation given is:

$xy' + 4y = 3x^2$

with the initial condition $y(1) = 9$.

Let's solve this step-by-step:

Step 1: Rewrite the equation

The equation is linear in standard form. Dividing the entire equation by $x$, we get:

$y' + \frac{4}{x}y = 3x$

This is now a first-order linear differential equation.

Step 2: Find the integrating factor

The integrating factor $\mu(x)$ is found by solving:

$\mu(x) = e^{\int \frac{4}{x} dx} = e^{4 \ln|x|} = |x|^4$

Since we are interested in positive $x$, we can drop the absolute value:

$\mu(x) = x^4$

Step 3: Multiply the equation by the integrating factor

Multiplying both sides of the equation $y' + \frac{4}{x}y = 3x$ by $x^4$, we get:

$x^4 y' + 4x^3 y = 3x^5$

Step 4: Simplify to a product rule

The left-hand side is the derivative of $x^4 y$:

$\frac{d}{dx}(x^4 y) = 3x^5$

Step 5: Integrate both sides

Integrating both sides with respect to $x$:

$x^4 y = \int 3x^5 dx = \frac{3x^6}{6} = \frac{x^6}{2}$

Step 6: Solve for $y$

Now, divide by $x^4$:

$y = \frac{x^6}{2x^4} + \frac{C}{x^4} = \frac{x^2}{2} + \frac{C}{x^4}$

Step 7: Apply the initial condition

We are given that $y(1) = 9$. Substituting $x = 1$ and $y = 9$ into the equation:

$9 = \frac{1^2}{2} + \frac{C}{1^4} = \frac{1}{2} + C$

Solving for $C$:

$9 - \frac{1}{2} = C \implies C = \frac{17}{2}$

Step 8: Final solution

Thus, the solution to the differential equation is:

$y(x) = \frac{x^2}{2} + \frac{17}{2x^4}$

Let me know if you have any questions or would like further details.

Here are 5 related questions:

1. What are first-order linear differential equations?
2. How do you find the integrating factor for a linear differential equation?
3. Why do we divide the original equation by $x$ to simplify it?
4. Can this method be applied to non-linear differential equations?
5. How does the initial condition help in finding the particular solution?

Tip: When solving first-order linear differential equations, always look for the integrating factor to simplify the process.