Solve differential equation $y''-2y'+3y=\sin x$ using invers operator method. 
$$y''-2y'+3y=\sin x$$

$s^2-2s+3=0$
$s=1 \pm i\sqrt {2}$ 
uisng operator $(D^2-2D+3)y_1=\sin x$ 
\begin{aligned}
y_1 &= \frac{1}{D^2-2D+3}\sin x \\
&= \Im\left(\frac{1}{D^2-2D+3} e^{ix}\right) \\
&= \Im \left(\frac{1}{i^2-2i+3}e^{ix}\right) \\
&= \Im \left(\frac{1}{2-2i}e^{ix}\right) && (***) \\
&= \frac{1}{2} \Im \left( \frac{1+i}{1+1} e^{ix}\right) && (***)\\
& = \frac{1}{4} \Im \left((1+i)(\cos x+i \sin x)\right) && (***) \\
& = \dots
\end{aligned}
i tried to solve problem using differential operator, but i dont understand what im means here, and also the formula too. i dont understand the star part, can someone explain what is this? is this relate to complex number? and does anyone know useful link about this material? and also i tried to solve using undetermined coef or variation of parameter, but compare to this it really takes lots of time.

2. and for differential operator:L(D) ->L'(D) so how to differentiate a differential operator?
$(D-1)(D-2)^3 => (D-2)^2(4D-5)$ can someone give me hint thanks

Another way to look this equation is to do the following:
$$(D^2-2D+3)y=\sin(x) \implies (D-1+i\sqrt{2})(D-1-i\sqrt{2})y=\sin(x)$$
Now, substitute $w=(D-1-i\sqrt{2})y$ to obtain
$$(D-1+i\sqrt{2})w=\sin(x)$$
This is a linear first order ODE, which can be easily solved by the closed formula for the first order equation. After having obtained this solution you just have to solve
$$w=(D-1-i\sqrt{2})y$$
by using the general formula for the first order linear ODE.
