Annihilator method

From Wikipedia, the free encyclopedia

In mathematics, the annihilator method is a procedure used to find a particular solution to certain types of inhomogeneous ordinary differential equations. It is equivalent to the method of undetermined coefficients, and the two names are sometimes used interchangeably, although they can describe different techniques. The phrase undetermined coefficients can also be used to refer to the step in the annihilator method in which the coefficients are calculated.

The annihilator method is used as follows. Given the ODE $P (D) y = f (x)$ , find another differential operator $A (D)$ such that $A (D) f (x) = 0$ . This operator is called the annihilator, thus giving the method its name. Applying $A (D)$ to both sides of the ODE gives a homogeneous ODE $\big(A(D)P(D)\big)y = 0$ for which we find a solution basis $\{y_1,\ldots,y_n\}$ as before. Then the original inhomogeneous ODE is used to construct a system of equations restricting the coefficients of the linear combinations to satisfy the ODE.

This method is not as general as variation of parameters in the sense that an annihilator does not always exist.

[edit] Example

Given $y'' - 4 y' + 5 y = sin(k x)$ , $P (D) = D 2 - 4 D + 5$ . The simplest annihilator of $sin(k x)$ is $A (D) = D 2 + k 2$ . The zeros of $A (z) P (z)$ are ${2 + i,2 - i, i k, - i k}$ , so the solution basis of $A (D) P (D)$ is ${y 1, y 2, y 3, y 4} = {e (2 + i) x, e (2 - i) x, e i k x, e - i k x}.$

Setting $y = c 1 y 1 + c 2 y 2 + c 3 y 3 + c 4 y 4$ we find

$sin(k x)$	$= P (D) y$
	$= P (D)(c 1 y 1 + c 2 y + c 3 y 3 + c 4 y 4)$
	$= c 1 P (D) y 1 + c 2 P (D) y 2 + c 3 P (D) y 3 + c 4 P (D) y 4$
	$= 0 + 0 + c 3 ( - k 2 - 4 i k + 5) y 3 + c 4 ( - k 2 + 4 i k + 5) y 4$
	$= c 3 ( - k 2 - 4 i k + 5)(cos(k x) + i sin(k x)) + c 4 ( - k 2 + 4 i k + 5)(cos(k x) - i sin(k x))$

giving the system

i = (k 2 + 4 i k - 5) c 3 + ( - k 2 + 4 i k + 5) c 4

0 = (k 2 + 4 i k - 5) c 3 + (k 2 - 4 i k - 5) c 4

which has solutions

$c_3=\frac i{2(k^2+4ik-5)}$ , $c_4=\frac i{2(-k^2+4ik+5)}$

giving the solution set

$y\,$	$=c_1y_1+c_2y_2+\frac i{2(k^2+4ik-5)}y_3+\frac i{2(-k^2+4ik+5)}y_4$
	$=c_1y_1+c_2y_2+\frac{4k\cos(kx)-(k^2-5)\sin(kx)} {(k^2+4ik-5)(k^2-4ik-5)}$
	$=c_1y_1+c_2y_2 +\frac{4k\cos(kx)+(5-k^2)\sin(kx)}{k^4+6k^2+25}.$

Categories: Differential equations

Annihilator method

From Wikipedia, the free encyclopedia

[edit] Example

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