Answer:
The given differential equation is
y(4) - 4y'''' + 4y'' = 0
Let
w = y''
Then
y'''' = w'
Substituting, we get
w - 4w' + 4w = 0
or
w' - 5w = 0
This equation factors as
(w - 5)(w + 1) = 0
so
w = 5 or w = -1
If w = 5, then y'' = 5, so y' = 5x + C1 and y = 5x^2 + C1x + C2.
If w = -1, then y'' = -1, so y' = -x + C1 and y = -x^2/2 + C1x + C2.
Therefore, the general solution of the given differential equation is
y = (5x^2 + C1x + C2) + (-x^2/2 + C1x + C2) = 3x^2 + Cx + C2
where C and C2 are arbitrary constants