To solve the given second-order linear differential equation:

\[
y'' + 4y' + 9y = x^2 + 3x
\]

We'll use the method of undetermined coefficients, which involves two main steps: solving the homogeneous equation and then finding a particular solution.

### Step 1: Solve the Homogeneous Equation
The homogeneous equation associated with the given differential equation is:

\[
y'' + 4y' + 9y = 0
\]

Assume a solution of the form \(y_h = e^{rx}\). Substituting into the homogeneous equation gives:

\[
r^2 e^{rx} + 4r e^{rx} + 9 e^{rx} = 0
\]

Factoring out \(e^{rx}\) (which is never zero), we get the characteristic equation:

\[
r^2 + 4r + 9 = 0
\]

Solve this quadratic equation using the quadratic formula:

\[
r = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 - 36}}{2} = \frac{-4 \pm \sqrt{-20}}{2} = \frac{-4 \pm 2i\sqrt{5}}{2} = -2 \pm i\sqrt{5}
\]

So, the roots are complex: \(r = -2 + i\sqrt{5}\) and \(r = -2 - i\sqrt{5}\).

The general solution to the homogeneous equation is:

\[
y_h(x) = e^{-2x} \left(C_1 \cos(\sqrt{5}x) + C_2 \sin(\sqrt{5}x)\right)
\]

### Step 2: Find a Particular Solution
To find a particular solution \(y_p\) for the non-homogeneous equation \(y'' + 4y' + 9y = x^2 + 3x\), assume a solution of the form:

\[
y_p(x) = Ax^2 + Bx + C
\]

Substitute \(y_p(x)\), \(y_p'(x)\), and \(y_p''(x)\) into the differential equation:

\[
y_p''(x) = 2A, \quad y_p'(x) = 2Ax + B, \quad y_p(x) = Ax^2 + Bx + C
\]

Substitute into the original equation:

\[
2A + 4(2Ax + B) + 9(Ax^2 + Bx + C) = x^2 + 3x
\]

Expanding and collecting like terms gives:

\[
(9A)x^2 + (8A + 9B)x + (2A + 4B + 9C) = x^2 + 3x + 0
\]

Equating coefficients of like terms on both sides:

1. \(9A = 1 \implies A = \frac{1}{9}\)
2. \(8A + 9B = 3 \implies 8\left(\frac{1}{9}\right) + 9B = 3 \implies \frac{8}{9} + 9B = 3 \implies 9B = \frac{19}{3} \implies B = \frac{19}{27}\)
3. \(2A + 4B + 9C = 0 \implies 2\left(\frac{1}{9}\right) + 4\left(\frac{19}{27}\right) + 9C = 0 \implies \frac{2}{9} + \frac{76}{27} + 9C = 0 \implies C = -\frac{26}{81}\)

So, the particular solution is:

\[
y_p(x) = \frac{1}{9}x^2 + \frac{19}{27}x - \frac{26}{81}
\]

### Final Solution
The general solution to the differential equation is the sum of the homogeneous and particular solutions:

\[
y(x) = e^{-2x} \left(C_1 \cos(\sqrt{5}x) + C_2 \sin(\sqrt{5}x)\right) + \frac{1}{9}x^2 + \frac{19}{27}x - \frac{26}{81}
\]

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

### Related Questions:
1. How would you find the particular solution if the non-homogeneous term were \(e^{2x}\) instead of \(x^2 + 3x\)?
2. What changes in the solution method if the roots of the characteristic equation are real and distinct?
3. How do you solve a differential equation with repeated roots in the characteristic equation?
4. What is the method of variation of parameters, and when is it used instead of undetermined coefficients?
5. How would the solution change if the non-homogeneous term had a sine or cosine function?

### Tip:
When solving differential equations, always check the form of the non-homogeneous term to choose the appropriate method for finding a particular solution.

Learn how to solve the second-order linear differential equation y'' + 4y' + 9y = x^2 + 3x step-by-step. This problem involves the method of undetermined coefficients and covers key concepts in differential equations and characteristic equations. Suitable for college-level students.

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