The problem is to solve the second-order linear non-homogeneous ordinary differential equation: $y'' - 6y' + 9y = xe^{3x}$.

1. Problem Description

The problem is to solve the second-order linear non-homogeneous ordinary differential equation: $y'' - 6y' + 9y = xe^{3x}$.

2. Solution Steps

First, we solve the homogeneous equation $y'' - 6y' + 9y = 0$. The characteristic equation is $r^2 - 6r + 9 = 0$. This factors as $(r - 3)^2 = 0$, so we have a repeated root $r = 3$. Therefore, the general solution to the homogeneous equation is $y_h = c_1e^{3x} + c_2xe^{3x}$, where $c_1$ and $c_2$ are arbitrary constants.

Next, we find a particular solution to the non-homogeneous equation $y'' - 6y' + 9y = xe^{3x}$. Since $e^{3x}$ and $xe^{3x}$ are solutions to the homogeneous equation, we try a particular solution of the form $y_p = (Ax^3 + Bx^2)e^{3x}$.

We compute the derivatives:

$y_p' = (3Ax^2 + 2Bx)e^{3x} + 3(Ax^3 + Bx^2)e^{3x} = (3Ax^3 + (3A + 2B)x^2 + 2Bx)e^{3x}$

$y_p'' = (9Ax^2 + (6A + 4B)x + 2B)e^{3x} + 3(3Ax^3 + (3A + 2B)x^2 + 2Bx)e^{3x} = (9Ax^3 + (9A + 6B + 9A)x^2 + (6A + 4B + 6B)x + 2B)e^{3x} = (9Ax^3 + (18A + 6B)x^2 + (6A + 10B)x + 2B)e^{3x}$

Substituting into the differential equation, we get:

$(9Ax^3 + (18A + 6B)x^2 + (6A + 10B)x + 2B)e^{3x} - 6(3Ax^3 + (3A + 2B)x^2 + 2Bx)e^{3x} + 9(Ax^3 + Bx^2)e^{3x} = xe^{3x}$

$(9A - 18A + 9A)x^3 + (18A + 6B - 18A - 12B + 9B)x^2 + (6A + 10B - 12B)x + 2Be^{3x} = xe^{3x}$

$(0)x^3 + (3B)x^2 + (6A - 2B + 10B - 12B)x + 2B = x$

$(3B)x^2 + (6A - 2B)x + 2B = x$

$(6A - 2B)x + (6A - 2B)x = xe^{3x}$

$(6A-2B)x = x$

Comparing coefficients, we must have $3B=0$ and $6A-2B = 1$.

So $B=0$ and $6A = 1$, which means $A = \frac{1}{6}$.

Thus, $y_p = \frac{1}{6}x^3e^{3x}$.

Therefore, the general solution is $y = y_h + y_p = c_1e^{3x} + c_2xe^{3x} + \frac{1}{6}x^3e^{3x}$.

3. Final Answer

$y = c_1e^{3x} + c_2xe^{3x} + \frac{1}{6}x^3e^{3x}$