Final answer:
The general solution to the fourth-order differential equation y(4) − 2y'' + y = 0 is y(x) = (A1 + B1x)e^(x) + (A2 + B2x)e^(-x), where A1, A2, B1, B2 are constants determined by initial conditions.
Step-by-step explanation:
The question asks us to find the general solution of the differential equation y(4) − 2y'' + y = 0. We start by looking for solutions of the form y = e^(rx), where r is a constant. Substituting this into the equation gives us a characteristic equation: r^4 − 2r^2 + 1 = 0. This factors as (r^2 - 1)^2 = 0, which has roots r = 1 and r = -1, each with multiplicity 2.
The general solution for repeated roots is of the form y(x) = (A + Bx)e^(rx) for each repeated root r. Therefore, the general solution to the differential equation is: y(x) = (A1 + B1x)e^(x) + (A2 + B2x)e^(-x), where A1, A2, B1, B2 are constants determined by initial conditions.