SUBJECT IS D.E

Question

the subject in this question DIFFERENTIAL EQUATION, * FACTOR THE FOLLOWING OPERATOR 1. D quib - 3D squred - 5D +6 2 d quib - 11D +20

Answer 1

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+Be^0.87475x+Ce^3.88897, where A, B, C are constants. This can also be written:

y=A(0.1714^x)+B(2.3983^x)+C(48.8605^x).

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+Be^2x(cos(x)-isin(x))+Ce^2x(cos(x)+isin(x)), which can be written:

y=ae^-4x+be^2xcos(x)+ce^2xsin(x), where a, b and c are constants (which may be complex). This is the characteristic solution of the DE.