# Help needed understanding solution-forms of non-homogeneous differential equations

On my lecture material on solving non-homogeneous differential equations, I have a chart that I'm supposed to consult that tells me the the form of the attempt-particular solution. For example: If the differential equation is of the form $$y''+ay'+by=sin(kx),$$, my particular solution takes the form $$Acos(kx)+Bsin(kx)$$ if the characteristic polynomial $$P(ik) \neq 0$$. If the $$P(ik) =0$$, then the particular solution takes the form $$Axcos(kx)+Bxsin(kx)$$.

Why is this result true in general? I can see after doing some examples that I don't find a solution with the first form if the characteristic polynomial has the complex root involving $$k$$, but can we show in a simple manner why this is true, and why the addition of $$x$$ to the form fixes the situation?

Also, there is another curiosity: If the differential equation is of the form $$y''+ay'+by=e^{kx}$$,the attempted particular solution involves $$Ae^{kx}$$ if $$P(k)\neq 0$$, $$Axe^{kx}$$ if $$P(k)=0$$ but $$P'(k)\neq0$$, and $$Ax^2e^{kx}$$ if $$P(k)=P'(k)=0$$. Again, I can see from problem examples that these attempted solution forms do work, but what is the reason behind it?

So to summarize: Can we show, preferably by using the properties of the characteristic equation, the reasons why we don't find solutions when the characteristic equation or its derivatives take on certain values?

Basically one takes the higher-order homogeneous ODE and rewrites it as a linear system of first-order ODEs $$u'=Au$$ whose solution is essentially $$e^{A x}$$. The characteristic polynomial of the original ODE is in fact the characteristic polynomial of $$A$$, which determines its structure, hence that of the space of solutions.