To solve the differential equation \((D^2 + 3D + 2)y = xe^x \sin 2x\), we can proceed as follows:

1. **Rewrite the differential operator \(D\)**: Recall that \(D\) represents the derivative operator with respect to \(x\). So, \(D^2\) represents the second derivative, \(D^2 y = y''\), and \(D y = y'\).

2. **Find the characteristic equation**: Solve the homogeneous part of the differential equation first:
   \[
   (D^2 + 3D + 2)y = 0.
   \]
   This can be rewritten as:
   \[
   (D + 1)(D + 2)y = 0.
   \]
   The characteristic equation is:
   \[
   r^2 + 3r + 2 = 0.
   \]
   Factoring this gives:
   \[
   (r + 1)(r + 2) = 0.
   \]
   Therefore, the roots are \(r = -1\) and \(r = -2\).

3. **General solution of the homogeneous equation**:
   \[
   y_h = C_1 e^{-x} + C_2 e^{-2x},
   \]
   where \(C_1\) and \(C_2\) are constants.

4. **Particular solution for the non-homogeneous equation**: We now need to find a particular solution \(y_p\) for the non-homogeneous equation \((D^2 + 3D + 2)y = xe^x \sin 2x\).

   Since the non-homogeneous term is \(xe^x \sin 2x\), we can use the method of undetermined coefficients. The appropriate form for \(y_p\) would be:
   \[
   y_p = e^x \left( (Ax + B)\cos 2x + (Cx + D)\sin 2x \right).
   \]

5. **Substitute \(y_p\) into the differential equation**: Compute the first and second derivatives of \(y_p\):
   \[
   y_p = e^x \left( (Ax + B)\cos 2x + (Cx + D)\sin 2x \right).
   \]
   \[
   y_p' = e^x \left( ((A+1)x + B) \cos 2x + ((C+1)x + D) \sin 2x \right) + 2 e^x \left( (Ax + B) \sin 2x - (Cx + D) \cos 2x \right).
   \]
   \[
   y_p'' = e^x \left( ((A+2)x + B) \cos 2x + ((C+2)x + D) \sin 2x \right) + 4 e^x \left( (A x + B) \sin 2x - (C x + D) \cos 2x \right) - 4 e^x \left( (Ax + B) \cos 2x + (Cx + D) \sin 2x \right).
   \]

6. **Collect coefficients and solve for \(A\), \(B\), \(C\), and \(D\)**:
   \[
   y_p'' + 3y_p' + 2y_p = xe^x \sin 2x.
   \]

   Substitute the derivatives into the differential equation and equate coefficients for \(\cos 2x\) and \(\sin 2x\).

7. **Combine the homogeneous and particular solutions**: The general solution is:
   \[
   y = y_h + y_p = C_1 e^{-x} + C_2 e^{-2x} + e^x \left( (Ax + B)\cos 2x + (Cx + D)\sin 2x \right).
   \]

8. **Determine the coefficients \(A\), \(B\), \(C\), and \(D\)** by equating coefficients on both sides of the equation. This step involves a lot of algebra and solving a system of linear equations.

Would you like a detailed step-by-step process for solving for \(A\), \(B\), \(C\), and \(D\) in the particular solution, or do you have any other questions?

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**Related Questions**:
1. How do we solve second-order differential equations using the characteristic equation method?
2. What are the different methods to solve non-homogeneous differential equations?
3. Can you explain the method of undetermined coefficients?
4. How do you solve a system of linear equations that arise from comparing coefficients?
5. What are the common pitfalls when applying the method of undetermined coefficients?
6. How do you handle resonance in differential equations?
7. How does the presence of \(xe^x \sin 2x\) influence the choice of the particular solution's form?
8. How can we verify the solution to a differential equation?

**Tip**: Always verify your particular solution by substituting it back into the original differential equation to ensure it satisfies the equation.

Learn how to solve the differential equation (D^2 + 3D + 2)y = xe^x sin 2x step-by-step. This problem covers differential equations, characteristic equations, and the method of undetermined coefficients. Suitable for college-level students.

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