You have provided a linear differential equation of the form:

\[
(D^2 + 6D + 1) y = x e^x \sin(x)
\]

Here, \( D \) represents the differential operator \( \frac{d}{dx} \). 

### Step 1: Characteristic Equation
First, solve the homogeneous equation:

\[
(D^2 + 6D + 1) y = 0
\]

This gives the characteristic equation:

\[
r^2 + 6r + 1 = 0
\]

Using the quadratic formula:

\[
r = \frac{-6 \pm \sqrt{36 - 4}}{2} = \frac{-6 \pm \sqrt{32}}{2} = \frac{-6 \pm 4\sqrt{2}}{2} = -3 \pm 2\sqrt{2}
\]

So, the roots are \( r_1 = -3 + 2\sqrt{2} \) and \( r_2 = -3 - 2\sqrt{2} \).

The general solution to the homogeneous equation is:

\[
y_h(x) = C_1 e^{(-3 + 2\sqrt{2})x} + C_2 e^{(-3 - 2\sqrt{2})x}
\]

### Step 2: Particular Solution
To find the particular solution \( y_p(x) \) to the non-homogeneous equation, we use the method of undetermined coefficients. 

For the right-hand side \( x e^x \sin(x) \), a trial solution could be of the form:

\[
y_p(x) = (Ax + B) e^x \sin(x) + (Cx + D) e^x \cos(x)
\]

### Step 3: Apply Differential Operator

Apply the operator \( D^2 + 6D + 1 \) to \( y_p(x) \), and equate it to \( x e^x \sin(x) \) to find the coefficients \( A \), \( B \), \( C \), and \( D \).

### Step 4: General Solution
Combine the homogeneous and particular solutions:

\[
y(x) = y_h(x) + y_p(x)
\]

The general solution will be:

\[
y(x) = C_1 e^{(-3 + 2\sqrt{2})x} + C_2 e^{(-3 - 2\sqrt{2})x} + y_p(x)
\]

If you would like, I can compute the exact values for \( A \), \( B \), \( C \), and \( D \) for the particular solution. Let me know if you want that detailed calculation.

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Do you want further details or have any questions about this solution? Here are some related questions:

1. How do you solve differential equations using the method of undetermined coefficients?
2. What are the steps to derive the characteristic equation from a homogeneous differential equation?
3. How does the choice of trial solution depend on the right-hand side of the differential equation?
4. What is the significance of the characteristic roots in the solution to differential equations?
5. How can variation of parameters be used as an alternative to find particular solutions?

**Tip:** For complex roots of the characteristic equation, the solution often involves exponential functions combined with trigonometric functions.

Learn how to solve the differential equation (D^2 + 6D + 1)y = xe^x*sin(x) using the method of undetermined coefficients and characteristic equation. This problem covers advanced topics in differential equations and is suitable for advanced undergraduate students and beyond.

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