) Solve: y (4) − 5y ′′ + 4y = 0 .

This is a linear homogeneous differential equation with constant coefficients. The characteristic equation is r^4 - 5r^2 + 4 = 0, which can be factored as (r^2 - 4)(r^2 - 1) = 0. This gives us four roots: r = ±2 and r = ±1.

The general solution is a linear combination of the solutions corresponding to these four roots. The solutions are:

y1(x) = e^(2x)
y2(x) = e^(-2x)
y3(x) = e^x
y4(x) = e^(-x)

Therefore, the general solution of the differential equation is:

y(x) = c1 e^(2x) + c2 e^(-2x) + c3 e^x + c4 e^(-x)

where c1, c2, c3, and c4 are constants that can be determined from initial or boundary conditions.