I would like to find out the particular integral (P.I.) of the differential equation :
$$(D^2-5D+6)y=e^{2x}$$
Result:
P.I. of the D.E., $F(D)y=X$ is
(i) $\dfrac{1}{F(D)}e^{ax}=\dfrac{1}{F(a)}e^{ax}$ provided $F(a)\neq 0$, here $X=e^{ax}$
(ii) $\dfrac{1}{F(D)}e^{ax}V=e^{ax}\dfrac{1}{F(D+a)}V$, here $X=e^{ax}V$
Now,for the given D.E. the P.I.=$\dfrac{1}{D^2-5D+6}e^{2x}=\dfrac{1}{(D-2)(D-3)}e^{2x}=e^{2x}\dfrac{1}{(D+2-2)(D+2-3)}(1)=e^{2x}\dfrac{1}{D(D-1)}(1)=e^{2x}\dfrac{1}{D}(1-D)^{-1}(1)=e^{2x}\dfrac{1}{D}(1+D+\dots)(1)=e^{2x}\dfrac{1}{D}(1)=xe^{2x}$
If we integrate first and then differentiate,we get different result, like below :
P.I.=$\dfrac{1}{D^2-5D+6}e^{2x}=e^{2x}(1-D)^{-1}\dfrac{1}{D}(1)=e^{2x}(1-D)^{-1}(x)=e^{2x}(1+D+\dots)(x)=e^{2x}(x+1)$
In the above two methods gives two different results.Why?
The function $e^{2x}$ is a solution of the homogeneous equation. So in the second case you are finding the particular solution from the first case, plus a solution of the homogeneous equation: $$ e^{2x}(x+1)=xe^{2x}+e^{2x}. $$
"Particular solutions" are not unique: as your example shows, if you add a solution of the homogeneous solution to a particular solution, you get another particular solution.