I'm studying mathematics at a Gymnasium level. This equals a High School/College level of mathematics. I'm now going to show you a part of a calculation to illustrate my problem.
The differential equation $(1)\;\;y''+4y'+3y=1-x^2$ is given. Calculate one particular solution.
Derivates $y=ax^2+bx+c$ two times. Inserting in $(1)$ gives
$2a+8ax+4b+3ax^2+3bx+3c=-1x^2+0x+1.$
So
$\left\{\begin{matrix} 2a+4b+3c=1\\ 8a+3b=0\\ 3a=-1 \end{matrix}\right.$
$\ldots$
This equation system is illustrating my problem. I understand that this is the correct method to calculate a particular solution to this differential equation, but I do not understand why. I made the clarification $-1x^2+0x+1$, and see that there exists connection(s) between this part and the other parts that I select from the differential equation to equal the parts in this (see the equation system). But I still do not understand why these parts are selected. Obviously (?) you would get the wrong answer if you don't use a method like this.
Q: Why is the differential equation split up as you see in the equation system, in order to be able to solve the differential equation? Is it possible for you to give me a general answer on how to think?
There are thousands of types of differential equations occurring in practice. Your ODE is linear, meaning that the unknown function and its derivatives occur as additive terms $a_k(x) y^{(k)}$ only, and there may be a term $b(x)$ which is free from $y$ or its derivatives. Now in systems of linear equations we are used to write all terms containing an unknown to the left of the equality sign and the constant terms on the right, like so: $$\eqalign{3x-7y&=20\cr 5x+8y&=-34\ .\cr}$$ In the case of your ODE the $1-x^2$ part is the $y$-free part, and therefore goes on the RHS.
Your ODE is of a type occurring in practice everyday. There is a complete theory for such equations. In order to find the general solution you would first have to solve the associated homogeneous equation $y''+4y'+3y=0$. I won't go into this here; but you should be aware that the "Ansatz" that you used in order to obtain a single particular solution does not always work. Had your ODE been $y''-y'=1-x^2$ you would be at a loss (try it!).