homogeneous second-order equation
Imagine a swing that nobody is pushing. Once you let go, it keeps moving — slowing, oscillating, settling — purely under its own internal rules, gravity and friction, with no outside hand. The homogeneous second-order equation describes exactly this kind of free, unforced behaviour: the system's natural motion when nothing external is driving it.
It is the second-order linear equation with the forcing term set to zero: y'' + p(x) y' + q(x) y = 0. The right-hand side being 0 is what makes it 'homogeneous' in this context — every term involves the unknown y (or its derivatives), nothing stands alone. This has a beautiful consequence. If y1 and y2 are both solutions, then so is c1 y1 + c2 y2 for any constants c1 and c2, because plugging a sum in just adds the two zeros you already had. The solution set is closed under addition and scaling — it is a vector space — and for a well-posed second-order equation that space turns out to be exactly two-dimensional.
Solving the homogeneous equation is the indispensable first half of every linear problem. Even when you really want to solve the forced equation y'' + p y' + q y = g(x), you first find the homogeneous (complementary) solutions, then add one particular solution. So the homogeneous case is not a warm-up to be skipped; it is the backbone, and the rest of the chapter is built on top of it.
For y'' - y = 0, both y1 = e^x and y2 = e^(-x) work. So does 5 e^x - 2 e^(-x), and indeed c1 e^x + c2 e^(-x) for any c1, c2. Substituting any such combination gives 0 = 0, confirming the family.
Two basic solutions plus all their linear combinations sweep out the entire two-parameter family of solutions.
'Homogeneous' here means only 'right-hand side equals zero'. Do not confuse it with the first-order sense of a 'homogeneous-degree equation' (where dy/dx depends only on y/x) — same word, unrelated idea.