Question

I l Solution 0

Answer 1

"(x - 1) y"" - x y + y = 0 implies (x - 1) y"" + (1 - x) y = 0 (1) Now, y_1 = e^x so, y""_1 = e^x (x - 1) y""_1 + (1 - x) y_1 = (x - 1) e^x + (1 - x) e^x = 0 so it satisfy the differential equation (1) Let y_2 (x) = v (x) e^x be another solution of the differential equation (1) so, y'_2 (x) = (v + v') e^x y""_2 (x) = (v"" + 2 v' + v) e^x now, from (1), (x - 1) (v"" + 2 v' + v) e^x + (1 - x) v e^x = 0 implies (x - 1) (v"" + 2 v') = 0 implies v"" + 2 v' = 0 Let u = v', then u' = v"" so, u' + 2 u = 0 so, u = c e^- 2 x implies v' (x) = c e^- 2 x implies v (x) = -1/2 c e^-"