To find the particular integral of the equation $\left(D^2+1\right)y=\sin x $,I converted this equation to its inverse $y=\frac{1}{D^2+1}\sin x$ . The solution to this can be obtained by taking the imaginary part of $y=\frac{1}{D^2+1}e^{ix}$
From here, I used 2 different methods to proceed.
First method: I directly used the formula $$\frac{e^{ax}}{P\left(D\right)}=\frac{xe^{ax}}{P'\left(D\right)},P\left(D\right)=0,P'\left(D\right)\ \neq 0 $$ which gave me $\frac{xe^{ix}}{2i}$ . The imaginary part of this is -$\frac{x\cos x}{2} $, which becomes my particular integral.
Second method: I first converted the$ \frac{1}{D^2+1}$ into partial fractions,which gave me $\frac{1}{2i}\left(\frac{1}{D-i}-\frac{1}{D+i}\right)$ . Then I took $ e^{ix}$ inside, which gave me $\frac{1}{2i}\left(xe^{ix}-\frac{e^{ix}}{2i}\right)$ The first term was calculated using the formula in method 1, and the second term using the formula $$\frac{e^{ax}}{P\left(D\right)}=\frac{e^{ax}}{P\left(a\right)},P\left(a\right)\neq 0 $$ This is clearly not equal to the equation obtained using method 1(without splitting into partial fractions). What am I doing wrong here?
The difference between both, $\frac{e^{ix}}4$, is a homogeneous solution, both are valid solutions of the inhomogeneous equation and can serve as particular solutions.