Jared Rosa
Math 480
Spring 2013
Some Notations & Definitions
• L(y) is the differential operator on y.
•
E.g. y”’ - 3y” + 2y = x ; can be rewritten as L(y) = x.
• Pc(r) is the characteristic polynomial of an
ordinary differential equation.
•
E.g. Pc(r) = r3 – 3r2 + 2 , is the characteristic polynomial
of the ode y”’ - 3y” + 2y.
So what is the Annihilator Method?
The Annihilator method is a faster method than
the variation of constants used to solve nonhomogenous differential equations with constant
coefficients.
How does the method work?
The annihilator method uses a second differential
operator denoted M, where the solution of the ode M(b) =
O is the right hand side of the first differential operator
L(y). Then using the Pc(r) of M(L(y)) we can determine the
solutions of L(y).
E.G.
L(y) = b(x) & M(b) = O
So we have
M(L(y)) = M(b) = O
Hence, M annihilates L.
When Can I use this Method?
The annihilator method only has two simple
requirement:
1) The differential operator that you want to
annihilate has constant coefficients. E.g. L(y) =
ay” + by’ + cy , where a, b, and c are constants
2) There exists a second differential operator M,
such that if L(y) = f(x), then M(f) = O. Or said
another way, the solution to the homogenous
equation M(y) is f(x).
List of Annihilators
Function
Pc(r) of an annihilator
• eax
• r–a
• xkeax
• (r-a)k+1
• sin(ax), cos(ax) (a real)
• r2 + a2
• xksin(ax), xkcos(ax) (a real)
• (r2 + a2)k+1
The Annihilator Vs. VOC
Annihilator Method
Variation Of Constants Method
L(y) = y”’ + y” + y’ + y = 1
L(y) = y”’ + y” + y’ + y = 1
y(0) = 0, y’(o) = 1, y”(0) = 0
y(0) = 0, y’(o) = 1, y”(0) = 0
M(y) = y’ = 0
First set L(y) = 0 to get Pc(r):
M(L(y)) = y(4) + y”’ + y” + y’ = 0
Pc(r) = r3 + r2 + r + 1 = (r2 + 1)(r + 1)
Pc(r) = r4 + r3 + r2 +r
So, y1 = cos(x), y2 = sin(x), y3 = e-x
0 = r4 + r3 +r2 + r = r(r3 + r2 + r + 1)
and yp = U1y1 + U2y2 + U3y3, where
0 = r(r2 + 1)(r + 1)
U1 = ½(cos(x) – sin(x))
yg = a + be-x + csin(x) + dcos(x)
U2 = ½(sin(x) + cos(x))
We assume a=1 since a has to be a
solution of L(y) to get:
y = 1 – ½ e-x + ½ sin(x) – ½ cos(x)
U3 = ½e-x
So yg = 1 + c1sin(x) + c2cos(x) + c3e-x
And y = = 1 – ½ e-x + ½ sin(x) – ½ cos(x)
Source
Coddington, Earl A. An introduction to ordinary differential equations.
Dover Publications, inc. New York: 1961. pg 86 – 92.