John777
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Could someone please explain the y2 solution for repeated roots in Frobenius method where y2=y1lnx+xs [tex]\Sigma[/tex] CnxnI am struggling to figure out how to solve this
The discussion focuses on the Frobenius method for solving differential equations with repeated roots, specifically addressing the solution form for the second solution, y2, when the indicial root is a double root. The equation discussed is the Euler-Cauchy equation, represented as x^2y'' + αxy' + βy = 0, with the context of a regular singular point at x=0. The solution for y2 is given as y2 = y1ln(x) + x^s Σ C_n x^n, where y1 is the first solution and C_n are coefficients derived from the series expansion.
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