I have been reading R. Kent Nagle's Fundamental's of Differential Equations textbook and I'm really confused as to the meaning of the terms of "Particular Solution" and "General Solution", specifically as they change from being used in a first-order equation to a second-order equation.
So in a first-order differential equation, your answer should have a "C" somewhere resulting from integrating somewhere. This would be called your general solution because you haven't specified any initial conditions. If you did, and you incorporated that information into your solution, then it would be called your particular solution. Alright, so far so good.
But when I started learning about second-order differential equations, I got really confused because I understood that if you solve a homogeneous second-order linear differential equation with no initial condition information then you would get a general solution with two "C"s because of its second order. This made sense, but when I learned about the Method of Undetermined Coefficients and the Variation of Parameters, I learned that the general solution to a non-homogeneous second-order linear differential equation involved a particular solution AND the general solution to the homogeneous diff eq. But I don't see why or how this could make sense. Does this occur because the terms particular and general are redefined for second-order diff EQs? I'm overall just very confused about the terminology. General clarification would be greatly appreciated.
Edit 1: Thank you both Professor @Robert Israel and @K.defaoite
It makes a bit more sense, but I'm still overall confused. To be a little more explicit, why does the Method of Undetermined Coefficients give you a particular solution? Again, I'm very used to the idea that a particular solution is a general solution with initial conditions applied to it (from first-order differential equations) but I don't see the particular-ness that the Method of Undetermined Coefficients gives you in the way that solving an initial value problem for a first-order differential equation does. Thanks again to you both.