1) (D3-3D2-5D+6)y. We can find the zeroes of r3-3r2-5r+6 using Newton's iterative method:
r=-1.76372, 0.87475, 3.88897. Therefore, the characteristic solution would be:
y=Ae-1.76372x+Be0.87475x+Ce3.88897, where A, B, C are constants. This can also be written:
y=A(0.1714x)+B(2.3983x)+C(48.8605x).
2) (D3-11D+20)y. If we create the cubic equation r3-11r+20=0, we can factorise. If we try rational zeroes first we discover that r=-4 is a factor and the co-factor is r2-4x+5, which has complex zeroes:
r2-4x+5=0, r2-4x=-5, r2-4x+4=-1, (r-2)2=i2, r=2±i. The complete factorisation is:
(r+4)(r-2+i)(r-2-i).
If (D3-11D+20)y=0, then the solution of the DE would be:
y=Ae-4x+Be(2-i)x+Ce(2+i)x, which becomes:
y=Ae-4x+Be2x(cos(x)-isin(x))+Ce2x(cos(x)+isin(x)), which can be written:
y=ae-4x+be2xcos(x)+ce2xsin(x), where a, b and c are constants (which may be complex). This is the characteristic solution of the DE.