Final answer:
The minimum radius of convergence of the power series solution for the differential equation at x_0 = 1 is the distance to the nearest singular point. Factoring the quadratic portion of the differential equation yields singular points at x = 2 and x = 3, making the minimum radius of convergence 1, which is choice e.
Step-by-step explanation:
The differential equation given is (x² - 5x + 6)y" - 3xy' - y = 0, where we are looking to find the minimum value for the radius of convergence ρ of a power series solution about x0 = 1.
To find the radius of convergence of the power series, we need to identify the singular points of the differential equation. A singular point is where the coefficients of the derivatives in the differential equation become undefined or infinite. In our equation, (x² - 5x + 6) becomes zero for the values of x that turn the quadratic equation to zero.
We factor the quadratic equation x² - 5x + 6:
(x - 2)(x - 3) = 0, which has roots at x = 2 and x = 3. These are the singular points of the differential equation.
Since we are considering a power series solution about x0 = 1, the radius of convergence will be at least the distance to the nearest singular point. The distances to the singular points from x0 = 1 are 1 and 2, respectively.
Therefore, the minimum radius of convergence ρ is 1, corresponding to the nearest singular point.
Thus, the correct answer is e. ρ = 1.