if the DE x^2.y"+xy'-y=0 has a general solution c1x+c2.x^-1, use the variation of parameters to find the particular solution x^2.y"+xy'-y=2-2/x

To find the particular solution of the differential equation x^2.y'' + xy' - y = 2 - 2/x using the variation of parameters method, we first need to find the complementary solution.

The differential equation x^2.y'' + xy' - y = 0 is a second-order linear homogeneous differential equation. It has characteristic roots r1 = 1 and r2 = -1, since the auxiliary equation is r^2 - 1 = 0. This means that the complementary solution is given by the linear combination of the two fundamental solutions: y_c(x) = c1x + c2/x.

Next, we need to find the particular solution using the variation of parameters method. The particular solution is given by y_p(x) = u1(x)y1(x) + u2(x)y2(x), where u1(x) and u2(x) are functions to be determined.

1. Find the Wronskian:
The Wronskian W(x) = y1(x)y2'(x) - y1'(x)y2(x) is the determinant of the matrix formed by the two fundamental solutions' derivatives. In our case, y1(x) = x and y2(x) = 1/x, so:
W(x) = x * (-1/x^2) - 1 * (1/x) = -1/x - 1/x = -2/x.

2. Determine u1(x) and u2(x):
To find u1(x) and u2(x), we use the following formulas:
u1(x) = -∫(f(x) * y2(x))/W(x) dx
u2(x) = ∫(f(x) * y1(x))/W(x) dx,

where f(x) = 2 - 2/x in our case.

Calculating the integrals, we have:
u1(x) = -∫((2 - 2/x) * (1/x)) / (-2/x) dx
u1(x) = ∫(1 - 1/x^2) dx
u1(x) = x + 1/x + C1,

and
u2(x) = ∫((2 - 2/x) * x) / (-2/x) dx
u2(x) = ∫(2x - 2) dx
u2(x) = x^2 - 2x + C2.

Therefore, the particular solution is:
y_p(x) = (x + 1/x)(x) + (x^2 - 2x)(1/x)
y_p(x) = x^2 + 1 + x - 2 + x
y_p(x) = x^2 + 3x - 1.

Finally, the general solution of the differential equation x^2y'' + xy' - y = 2 - 2/x is given by:
y(x) = y_c(x) + y_p(x)
y(x) = c1x + c2/x + x^2 + 3x - 1.