Introduction
In this problem, we are given a differential equation and we have to find its solution. The given differential equation is a second-order linear homogeneous differential equation with variable coefficients. The method we will use to solve this equation is the method of undetermined coefficients.

Solution

Step 1: Find the characteristic equation
The characteristic equation is obtained by assuming a solution of the form y=e^(mx). Substituting this in the differential equation, we get the characteristic equation:
m^2- cosx m - (1-cotx) = 0

Step 2: Find the roots of the characteristic equation
Solve the characteristic equation to obtain the roots. The roots of the characteristic equation are given by:
m1 = (cosx + sqrt(cos^2x + 4(1-cotx))) / 2
m2 = (cosx - sqrt(cos^2x + 4(1-cotx))) / 2

Step 3: Find the homogeneous solution
The homogeneous solution of the differential equation is given by:
y_h = c1e^(m1x) + c2e^(m2x)

Step 4: Find the particular solution
To find the particular solution, we assume that y_p = A(x)e^(mx), where A(x) is a function of x. We differentiate y_p twice with respect to x and substitute it in the differential equation. Then we solve for A(x).

The derivative of y_p is:
y_p' = A'(x)e^(mx) + Ame^(mx)
y_p'' = A''(x)e^(mx) + 2A'(x)me^(mx) + A(m^2)e^(mx)

Substituting these in the differential equation, we get:
A''(x)e^(mx) + 2A'(x)me^(mx) + A(m^2)e^(mx) - cosx(A'(x)e^(mx) + Ame^(mx)) - (1-cotx)A(x)e^(mx) = e^xsinx

Simplifying and grouping the terms, we get:
e^mx(A''(x) + 2m A'(x) + (m^2 - cosx)A(x)) = e^xsinx + (cotx - 1)A(x)e^mx

Since the left side is a function of x only and the right side is a function of x and e^x, the only way this equation can be true for all values of x is if the coefficient of e^mx on both sides is equal. Therefore, we set:
A''(x) + 2m A'(x) + (m^2 - cosx)A(x) = 0

Step 5: Solve for A(x)
The solution of the above differential equation is found using the method of undetermined coefficients. We assume that A(x) is of the form:
A(x) = B(x)sinx + C(x)cosx

Substituting this in the differential equation, we get:
B''(x)sinx + C''(x)cosx + 2m(B'(x)cosx - C'(x)sinx)