Annihilator method

f(x)	Annihilator table
${\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}\!}$	${\displaystyle D^{n+1}\!}$
${\displaystyle e^{kx}\!}$	${\displaystyle D-k\!}$
${\displaystyle x^{n}.e^{kx}\!}$	${\displaystyle (D-k)^{n+1}\!}$
${\displaystyle \cos(bx)\;\;\mathrm {or} \;\;\sin(bx)\!}$	${\displaystyle D^{2}+b^{2}\!}$
${\displaystyle x^{n}\cos(bx)\;\;\mathrm {or} \;\;x^{n}\sin(bx)\!}$	${\displaystyle (D^{2}+b^{2})^{n+1}\!}$
${\displaystyle e^{ax}\cos(bx)\;\;\mathrm {or} \;\;e^{ax}\sin(bx)\!}$	${\displaystyle (D-a)^{2}+b^{2}=D^{2}-2aD+a^{2}+b^{2}\!}$
${\displaystyle x^{n}e^{ax}\cos(bx)\;\;\mathrm {or} \;\;x^{n}e^{ax}\sin(bx)\!}$	${\displaystyle \left[(D-a)^{2}+b^{2}\right]^{n+1}=\left[D^{2}-2aD+a^{2}+b^{2}\right]^{n+1}\!}$
${\displaystyle a_{n}x^{n}+\cdots +a_{1}x+a_{0}+b_{1}e^{\pm c_{1}x}+\cdots +b_{k}e^{\mp c_{k}x}\!}$	${\displaystyle D^{n+1}(D\mp c_{1}).\cdots .(D\pm c_{k})\!}$

If ${\displaystyle f(x)}$ consists of the sum of the expressions given in the table, the annihilator is the product of the corresponding annihilators.

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

Setting ${\displaystyle y=c_{1}y_{1}+c_{2}y_{2}+c_{3}y_{3}+c_{4}y_{4}}$ we find

${\displaystyle {\begin{aligned}\sin(kx)&=P(D)y\\[8pt]&=P(D)(c_{1}y_{1}+c_{2}y_{2}+c_{3}y_{3}+c_{4}y_{4})\\[8pt]&=c_{1}P(D)y_{1}+c_{2}P(D)y_{2}+c_{3}P(D)y_{3}+c_{4}P(D)y_{4}\\[8pt]&=0+0+c_{3}(-k^{2}-4ik+5)y_{3}+c_{4}(-k^{2}+4ik+5)y_{4}\\[8pt]&=c_{3}(-k^{2}-4ik+5)(\cos(kx)+i\sin(kx))+c_{4}(-k^{2}+4ik+5)(\cos(kx)-i\sin(kx))\end{aligned}}}$

giving the system

${\displaystyle i=(k^{2}+4ik-5)c_{3}+(-k^{2}+4ik+5)c_{4}}$

${\displaystyle 0=(k^{2}+4ik-5)c_{3}+(k^{2}-4ik-5)c_{4}}$

which has solutions

${\displaystyle c_{3}={\frac {i}{2(k^{2}+4ik-5)}}}$, ${\displaystyle c_{4}={\frac {i}{2(-k^{2}+4ik+5)}}}$

giving the solution set

${\displaystyle {\begin{aligned}y&=c_{1}y_{1}+c_{2}y_{2}+{\frac {i}{2(k^{2}+4ik-5)}}y_{3}+{\frac {i}{2(-k^{2}+4ik+5)}}y_{4}\\[8pt]&=c_{1}y_{1}+c_{2}y_{2}+{\frac {4k\cos(kx)-(k^{2}-5)\sin(kx)}{(k^{2}+4ik-5)(k^{2}-4ik-5)}}\\[8pt]&=c_{1}y_{1}+c_{2}y_{2}+{\frac {4k\cos(kx)+(5-k^{2})\sin(kx)}{k^{4}+6k^{2}+25}}.\end{aligned}}}$

This solution can be broken down into the homogeneous and nonhomogeneous parts. In particular, ${\displaystyle y_{p}={\frac {4k\cos(kx)+(5-k^{2})\sin(kx)}{k^{4}+6k^{2}+25}}}$ is a particular integral for the nonhomogeneous differential equation, and ${\displaystyle y_{c}=c_{1}y_{1}+c_{2}y_{2}}$ is a complementary solution to the corresponding homogeneous equation. The values of ${\displaystyle c_{1}}$ and ${\displaystyle c_{2}}$ 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 ${\displaystyle y_{1}=e^{(2+i)x}}$ and ${\displaystyle y_{2}=e^{(2-i)x}}$ can be further rewritten using Euler's formula:

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

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

Then ${\displaystyle c_{1}y_{1}+c_{2}y_{2}=c_{1}e^{2x}(\cos x+i\sin x)+c_{2}e^{2x}(\cos x-i\sin x)=(c_{1}+c_{2})e^{2x}\cos x+i(c_{1}-c_{2})e^{2x}\sin x}$, and a suitable reassignment of the constants gives a simpler and more understandable form of the complementary solution, ${\displaystyle y_{c}=e^{2x}(c_{1}\cos x+c_{2}\sin x)}$.