Final answer:
The general solution of the differential equation y(4) - 2y'' + y = 0 is a combination of exponentials y = (C1 + C2x)e^x + (C3 + C4x)e^-x with constants C1, C2, C3, and C4.
Step-by-step explanation:
Finding the General Solution of a Higher-Order Differential Equation
To find the general solution of the given higher-order differential equation y (4) - 2y '' + y = 0, we start by assuming a solution of the form y = erx. Substituting this into the differential equation and simplifying, we arrive at a characteristic equation: r4 - 2r2 + 1 = 0. This factors into (r2 - 1)2 = 0, which further factors into (r - 1)2(r + 1)2 = 0. The roots are r = 1 and r = -1, both with multiplicity two.
The general solution is thus a combination of these roots, considering their multiplicity: y = (C1 + C2x)ex + (C3 + C4x)e-x, where C1, C2, C3, and C4 are arbitrary constants determined by initial conditions.