Abel's identity
/ AH-bel /
Here is a small miracle: you can find out exactly how the Wronskian of two solutions behaves — whether it stays nonzero, where it might vanish, its precise formula — WITHOUT knowing the solutions themselves. All you need are the coefficients of the equation. Abel's identity is the formula that performs this trick.
For y'' + p(x) y' + q(x) y = 0, Abel's identity says the Wronskian satisfies W(x) = W(x0) e^(-integral from x0 to x of p(t) dt). It is derived in one elegant step: differentiate W = y1 y2' - y2 y1' to get W' = y1 y2'' - y2 y1'', then substitute y1'' = -p y1' - q y1 and y2'' = -p y2' - q y2 from the equation; the q-terms cancel and you are left with the simple first-order equation W' = -p W, whose solution is the exponential above. Notice what dropped out: q(x) plays no role at all, and the solutions y1, y2 vanished from the final formula.
Two consequences make this identity indispensable. First, the exponential e^(-integral of p) is never zero, so Abel proves the all-or-nothing law for the Wronskian: it is either zero everywhere (when W(x0) = 0) or nonzero everywhere — there is no in-between. That is the deep reason the Wronskian test for independence is reliable. Second, in reduction of order, knowing W from Abel and knowing one solution y1 lets you recover the second solution by a single integration, since W = y1 y2' - y2 y1' is itself a first-order equation for y2. Abel's identity quietly underwrites much of the homogeneous theory.
For y'' + 3 y' + 2 y = 0 (here p = 3), Abel gives W(x) = W(0) e^(-3x). The actual solutions e^(-x), e^(-2x) have Wronskian e^(-x)(-2 e^(-2x)) - e^(-2x)(-e^(-x)) = -e^(-3x), matching the e^(-3x) shape Abel predicted from p alone.
Abel predicts the Wronskian's shape from the coefficient p alone, before any solution is found.
Abel's identity assumes the equation is in standard form with leading coefficient 1, so that p is the actual coefficient of y'. For a y'' + b y' + c y = 0 first divide by a; the exponent uses b/a, not b.