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

Question

$$y''-2y'+3y=\sin x$$

$s^2-2s+3=0$
$s=1 \pm i\sqrt {2}$
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

Answer 1

The starred part consists in basic algebraic manipulations of complex numbers, with $i^2 = -1$ and $e^{ix} = \cos x + i \sin x$. Indeed, \begin{aligned} \frac{1}{i^2 -2i + 3} & = \frac{1}{2 -2i}\\ & = \frac{1}{2} \frac{1}{1 - i} \frac{1+i}{1+i} \\ & = \frac{1}{2} \frac{1 + i}{2} \, . \end{aligned} Now, multiplying by $e^{ix}$ and taking the imaginary part gives the equations of the OP.

Answer 2

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.