Annihilator method

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In mathematics, the annihilator method is a procedure used to find a particular solution to certain types of inhomogeneous ordinary differential equations. It is similar to the method of undetermined coefficients, but instead of guessing the particular solution in the method of undetermined coefficients, the particular solution is determined systematically in this technique. 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 combination to satisfy the ODE.

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

Example[edit]

Given y''-4y'+5y=\sin(kx), P(D)=D^2-4D+5. The simplest annihilator of \sin(kx) is A(D)=D^2+k^2. The zeros of A(z)P(z) are \{2+i,2-i,ik,-ik\}, 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^{ikx},e^{-ikx}\}.

Setting y=c_1y_1+c_2y_2+c_3y_3+c_4y_4 we find


\begin{align}
\sin(kx) & = P(D)y \\[8pt]
& = P(D)(c_1y_1+c_2y_2+c_3y_3+c_4y_4) \\[8pt]
& =c_1P(D)y_1+c_2P(D)y_2+c_3P(D)y_3+c_4P(D)y_4 \\[8pt]
& =0+0+c_3(-k^2-4ik+5)y_3+c_4(-k^2+4ik+5)y_4 \\[8pt]
& =c_3(-k^2-4ik+5)(\cos(kx)+i\sin(kx))
+c_4(-k^2+4ik+5)(\cos(kx)-i\sin(kx))
\end{align}

giving the system

i=(k^2+4ik-5)c_3+(-k^2+4ik+5)c_4
0=(k^2+4ik-5)c_3+(k^2-4ik-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


\begin{align}
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 \\[8pt]
& =c_1y_1+c_2y_2+\frac{4k\cos(kx)-(k^2-5)\sin(kx)}{(k^2+4ik-5)(k^2-4ik-5)} \\[8pt]
& =c_1y_1+c_2y_2+\frac{4k\cos(kx)+(5-k^2)\sin(kx)}{k^4+6k^2+25}.
\end{align}

This solution can be broken down into the homogeneous and nonhomogeneous parts. In particular, y_p = \frac{4k\cos(kx)+(5-k^2)\sin(kx)}{k^4+6k^2+25} is a particular integral for the nonhomogeneous differential equation, and y_c = c_1y_1 + c_2y_2 is a complementary solution to the corresponding homogeneous equation. The values of c_1 and c_2 are determined usually through a set of initial conditions. Since this is a second-order equation, two such conditions are necessary to determine these values.

The fundamental solutions y_1 = e^{(2+i)x} and y_2 = e^{(2-i)x} can be further rewritten using Euler's formula:

e^{(2+i)x} = e^{2x} e^{ix} = e^{2x} (\cos x + i \sin x)
e^{(2-i)x} = e^{2x} e^{-ix} = e^{2x} (\cos x - i \sin x)

Then c_1 y_1 + c_2 y_2 = c_1e^{2x} (\cos x + i \sin x) + c_2 e^{2x} (\cos x - i \sin x) = (c_1 + c_2) e^{2x} \cos x + i(c_1 - c_2) e^{2x} \sin x, and a suitable reassignment of the constants gives a simpler and more understandable form of the complementary solution,  y_c = e^{2x} (c_1 \cos x + c_2 \sin x).