Final answer:
To find the general solution in powers of x for the differential equation (x² - 1)y" + 4xy’ + 2y = 0, we use a power series method, yielding a recurrence relation Cn+2 = -Cn(4n + 2)/[(n+2)(n+1)]. The radius of convergence is at least up to x = ± 1 due to the singularity at x = 1.
Step-by-step explanation:
The differential equation given is (x² - 1)y" + 4xy’ + 2y = 0. To solve it, we typically look for a solution in the form of a power series y = ∑c_nx^n. Substituting this power series into the differential equation gives us a recurrence relation for the coefficients c_n.
When we derive the recurrence relation, it leads to Cn+2 = -Cn(4n + 2)/[(n+2)(n+1)], which allows us to calculate all the coefficients of the power series starting with an initial two, usually c0 and c1.
The guaranteed radius of convergence for the series solution can be determined from the fact that the differential equation has a regular singular point at x = 1. By Frobenius method, we can infer that the radius of convergence is at least up to the nearest singularity of the differential equation, which in this case is x = ± 1.