Question

(4) Determine the general solution of the given differential equation that is valid in any interval not including the singular point T- 22

Answer 1

"Given (x - 2)^2 y"" + 5 (x - 2) y' + 8y = 0 solution let (x - 1) = e^z = > z = log (x - 2) Therefore dt/dx = 1/(x - 2) Therefore dy/dx = dy/dt dt/dx = 1/(x - 1) dy/dt (1) Therefore d^2 y/dx^2 = d/dx (1/x - 2 dy/dt) = 1/(x - 2)^2 dy/dt + 1/(x - 2) d/dt (dy/dt) dz/dx = -1/(x - 2)^2 dy/dt + 1/(x - 1)^2 d^2 y/dt^2 = 1/(x - 1)^2 [d^2 y/dz^2 - dy/dt] (2) from (1) & (2) we have (x - 2) dy/dx = dy/dt (x - 1)^2 d^2y/dx^2 = (d^2y/dz^2 - dy/dz) Therefore The given equation becomes (d^2 y/dz^2 - dy/dt) + 5 (dy/dt) + 8y = 0 \r\n which can be written as (D^2 + 4D + 8) y = 0, D = d/dt The auxiliary equation is (m^2 + 4m + 6) = 0 m = -4 "