Annihilator Method Differential Equations
Math 385 Supplement: the method of undetermined coefficients It is relatively easy to implement the method of undetermined coefficients as presented in the textbook, but not easy to understand why it works. This handout explains the idea behind the method, and is aimed at those students who want to understand mathematics at a deeper level.
Review of homogeneous equations
The homogeneous constant coefficient linear equation an y (n) +· · ·+a1 y +a0 y = 0 has the characteristic polynomial an rn +· · ·+a1 r+a0 = 0. From the roots r1 , . . . , rn of the polynomial we can construct the solutions y1 , . . . , yn , such as y1 = er1 x . We can also rewrite the equation in a weird-looking but useful way, using the symbol d D = dx .
Examples: equation: y − 5y + 6y = 0. polynomial: r2 − 5r + 6 = 0.
(factored): (r − 2)(r − 3) = 0. roots: 2, 3 weird-looking form of equation: (D − 2)(D − 3)y = 0 or (D2 − 5D + 6)y = 0. linearly independent solutions: y1 = e2x , y2 = e3x . general solution: y = c1 e2x + c2 e3x . equation: y + 10y + 25y = 0. polynomial: r2 + 10r + 25 = 0.
(factored): (r + 5)2 = 0. roots: −5, −5 weird-looking form of equation: (D + 5)2 y = 0 or (D2 + 10D + 25)y = 0. linearly independent solutions: y1 = e−5x , y2 = xe−5x . general solution: y = c1 e−5x + c2 xe−5x . equation: y − 4y + 8y = 0. polynomial: r2 − 4r + 8 = 0.
(factored): (r − 2 − 2i)(r − 2 + 2i) = 0. roots: 2 + 2i, 2 − 2i w-l. f. of equation: (D − 2 − 2i)(D − 2 + 2i)y = 0 or (D2 − 4D + 8)y = 0. linearly independent solutions: y1 = e2x cos 2x, y2 = e2x sin 2x. general solution: y = e2x (c1 cos 2x + c2 sin 2x). equation: (already in a weird-looking form) (D2 + 1)2 (D − 1)3 y = 0. polynomial: (r2 + 1)2 (r − 1)3 y = 0. roots: i, i, −i, −i, 1, 1, 1. general solution: y = (c1 + c2 x) cos x + (c3 + c4 x) sin x + (c5 + c6 x + c7 x2 )ex .
Annihilators
If f is a function, then the annihilator of f is a “differential operator”
˜
L = an Dn + · · · + an D + a0
˜
with the
Review of homogeneous equations
The homogeneous constant coefficient linear equation an y (n) +· · ·+a1 y +a0 y = 0 has the characteristic polynomial an rn +· · ·+a1 r+a0 = 0. From the roots r1 , . . . , rn of the polynomial we can construct the solutions y1 , . . . , yn , such as y1 = er1 x . We can also rewrite the equation in a weird-looking but useful way, using the symbol d D = dx .
Examples: equation: y − 5y + 6y = 0. polynomial: r2 − 5r + 6 = 0.
(factored): (r − 2)(r − 3) = 0. roots: 2, 3 weird-looking form of equation: (D − 2)(D − 3)y = 0 or (D2 − 5D + 6)y = 0. linearly independent solutions: y1 = e2x , y2 = e3x . general solution: y = c1 e2x + c2 e3x . equation: y + 10y + 25y = 0. polynomial: r2 + 10r + 25 = 0.
(factored): (r + 5)2 = 0. roots: −5, −5 weird-looking form of equation: (D + 5)2 y = 0 or (D2 + 10D + 25)y = 0. linearly independent solutions: y1 = e−5x , y2 = xe−5x . general solution: y = c1 e−5x + c2 xe−5x . equation: y − 4y + 8y = 0. polynomial: r2 − 4r + 8 = 0.
(factored): (r − 2 − 2i)(r − 2 + 2i) = 0. roots: 2 + 2i, 2 − 2i w-l. f. of equation: (D − 2 − 2i)(D − 2 + 2i)y = 0 or (D2 − 4D + 8)y = 0. linearly independent solutions: y1 = e2x cos 2x, y2 = e2x sin 2x. general solution: y = e2x (c1 cos 2x + c2 sin 2x). equation: (already in a weird-looking form) (D2 + 1)2 (D − 1)3 y = 0. polynomial: (r2 + 1)2 (r − 1)3 y = 0. roots: i, i, −i, −i, 1, 1, 1. general solution: y = (c1 + c2 x) cos x + (c3 + c4 x) sin x + (c5 + c6 x + c7 x2 )ex .
Annihilators
If f is a function, then the annihilator of f is a “differential operator”
˜
L = an Dn + · · · + an D + a0
˜
with the