Introduction:
In this problem, we are given a differential equation (D3 - 1)y = excos(x)sin3(x) and we need to find its particular integral.

Solution:
To find the particular integral of the given differential equation, we will use the method of undetermined coefficients.

Step 1: Find the complementary function
First, we need to find the complementary function of the given differential equation. We can do this by assuming that y = e^(mx), where m is a constant. Substituting this into the differential equation, we get:
(D3 - 1)(e^(mx)) = 0
m^3e^(mx) - e^(mx) = 0
(m^3 - 1)e^(mx) = 0

This gives us the roots of the characteristic equation:
m = 1, -1/2 + i√3/2, -1/2 - i√3/2

Therefore, the complementary function is given by:
y_c = c1e^x + c2e^(-1/2x)cos(√3/2x) + c3e^(-1/2x)sin(√3/2x)

Step 2: Guess the form of the particular integral
Next, we need to guess the form of the particular integral based on the form of the non-homogeneous term. Since the non-homogeneous term is a product of excos(x) and sin3(x), we can guess that the particular integral will be of the form:
y_p = Axcos(x)sin3(x) + Bxsin(x)sin3(x) + Ccos(x)sin3(x) + Dsin(x)sin3(x)

Step 3: Find the coefficients
Substituting the guessed form of the particular integral into the differential equation, we get:
(D3 - 1)(Axcos(x)sin3(x) + Bxsin(x)sin3(x) + Ccos(x)sin3(x) + Dsin(x)sin3(x)) = excos(x)sin3(x)

Expanding the differential operator, we get:
-8Asin(x)sin3(x) - 6Bcos(x)sin3(x) + (A + 9C)cos(x)sin3(x) + (3A - B)xcos(x)cos3(x) + (3B + A)xsin(x)cos3(x) + (D - C)cos(x)cos3(x) + (C + D)xsin(x)cos3(x) = excos(x)sin3(x)

Equating the coefficients of like terms, we get the following system of equations:
3A - B = 0
A + 9C = 0
B + 3A = 0
C + D = 0
D - C = e

Solving these equations, we get:
A = -1/3
B = -1
C = 1/27
D = -1/27
E = 0

Therefore, the particular integral is given by:
y_p = (-1/3)xcos(x)sin3(x) - xsin(x)sin3(x)/