Solution:

Given function:
The given function is y(x) = x sin x.

Form of the differential equation:
We are given that y(x) = x sin x is a solution of an nth order linear differential equation with constant coefficients. Let's assume the general form of the differential equation to be:
a_n * y^(n) + a_(n-1) * y^(n-1) + ... + a_2 * y'' + a_1 * y' + a_0 * y = 0

Differentiating y(x):
To find the differential equation, we need to differentiate y(x) multiple times.

First derivative:
y'(x) = (d/dx)(x sin x) = sin x + x cos x

Second derivative:
y''(x) = (d/dx)(sin x + x cos x) = cos x - x sin x + cos x = 2 cos x - x sin x

Third derivative:
y'''(x) = (d/dx)(2 cos x - x sin x) = -2 sin x - sin x - x cos x = -3 sin x - x cos x

Substituting into the differential equation:
Now, let's substitute y(x), y'(x), y''(x), and y'''(x) back into the general form of the differential equation.

a_n * y^(n) + a_(n-1) * y^(n-1) + ... + a_2 * y'' + a_1 * y' + a_0 * y = 0
a_n * (2 cos x - x sin x) + a_(n-1) * (-3 sin x - x cos x) + ... + a_2 * (2 cos x - x sin x) + a_1 * (sin x + x cos x) + a_0 * (x sin x) = 0

Simplifying the equation:
Now, let's simplify the equation by collecting like terms:

[(a_n + a_2) * 2 cos x + (a_1 + a_0) * sin x] + [(-a_(n-1) - a_2) * x sin x + (a_1 - 3a_(n-1)) * cos x] = 0

For this equation to hold true for all values of x, the coefficients of each term must individually be zero.

Setting coefficients to zero:
Setting the coefficients of each term to zero, we get:

(a_n + a_2) = 0
(a_1 + a_0) = 0
(-a_(n-1) - a_2) = 0
(a_1 - 3a_(n-1)) = 0

Relations between coefficients:
From the above equations, we can deduce the following relations between the coefficients:

a_n = -a_2
a_0 = -a_1
a_(n-1) = -a_2
3a_(n-1) = a_1

Determining the minimum value of n:
From the relation 3a_(n-1) = a_1, we can see that the