I do not know the eg I am using can be on the site before. Actually, I was learning differential operators using a book given by our prof , and there was an example in it says like this> $\frac {d^2y}{dx^2}-3\frac{dy}{dx}+2y=e^{5x}$ now I have solved these kinds of question using the characteristic equation but how do I solve it using inverse operation of differential. And how these two methods are physically different in general using characteristic equation two roots will $y=c_1e^x+c_2e^{2x}$ but how do I solve it using the inverse operator, please can anybody share me the process of solving it. There is no proper site or papers to learn it so I will be very much happy if anyone can do an analysis for this question.
edit:I have posted the same question in chat room for an answer,and my problems were critically understood by user @semiclassical and what he says is Notationally you’d write your ODE as $(D^2−3D+2)y(x)=e^{5x}$, with $D=d/dx$. Or w/e your prof uses for differentiation One then can factor the left side as $(D−1)(D−2)y$, which makes evident that the characteristic polynomial has roots 1,2 The less-obvious part is how to get the particular solution The simplest way is to guess it (method of undetermined coefficients) Formally one does have $y=(D−1)−1(D−2)−1e^{5x}$ but figuring out how to apply that is the problem? this what I want to understand and how can I generalize it for the equation which has n roots (all are different) n roots(all are same) n roots(some are imaginary)
$$\frac {d^2y}{dx^2}-3\frac{dy}{dx}+2y=e^{5x}$$ $$(D^2-3D+2)y=e^{5x}$$ $$(D-1)(D-2)y=e^{5x}$$ $$y=\dfrac 1 {(D-1)(D-2)}e^{5x}$$ By fraction decomposition: $$y=\dfrac 1{(D-2)}{e^{5x}}-\dfrac 1{(D-1)}{e^{5x}}$$ Now I use the well known result: $$\dfrac 1{D-a}e^b=e^b\dfrac 1 {D+b-a}1$$ $$y=\dfrac 1{(D-2)}{e^{5x}}-\dfrac 1{(D-1)}{e^{5x}}$$ $$y=\dfrac 1{(D-2)}{e^{2x}e^{3x}}-\dfrac 1{(D-1)}{e^{4x}e^x}$$ $$y=e^{2x}\dfrac 1{D}{e^{3x}}-e^x\dfrac 1{D}{e^{4x}}$$ Integrate the exponentials: $$y=e^{2x}(\dfrac {e^{3x}}3+A)-e^x(\dfrac {e^{4x}}4+B)$$ Finally: $$y(x)=C_1e^x+C_2e^{2x}+\dfrac {e^{5x}}{12}$$ This lead to the same result you get with classical methods.