reduction of order
Suppose you already know one solution of a second-order equation — perhaps you guessed it, perhaps it came from a table — but you need a second, independent one to complete the picture. Reduction of order is a systematic recipe that squeezes that second solution out of the first, turning a hard second-order problem into an easy first-order one.
The trick is to look for a second solution of the form y2 = v(x) y1(x), where y1 is the known solution and v(x) is a function to be found — you let the unknown multiplier v carry the new information. Substitute y2 = v y1 into y'' + p y' + q y = 0. Because y1 already satisfies the equation, all the terms without a derivative of v cancel, and the v term itself drops out too; what survives is an equation containing only v'' and v', with no plain v. That is the whole point: setting w = v' turns it into a FIRST-order linear equation for w, which you solve with an integrating factor and integrate once more to get v. The named result is v' proportional to (1/y1^2) e^(-integral of p dx), so y2 = y1 times the integral of that.
Reduction of order is the workhorse behind several headline results. It is exactly how the repeated-root second solution x e^(r x) is derived. It is the standard way to find a partner solution for variable-coefficient equations where one solution is known (for instance from a series or a lucky guess). And it shows, structurally, why one known solution genuinely is 'half the battle': the remaining work is only first-order. The catch is that you must already possess one nonzero solution to start — reduction of order extends a foothold, it does not create one from nothing.
For y'' - 6 y' + 9 y = 0 with known solution y1 = e^(3x), set y2 = v e^(3x). The reduced equation becomes v'' = 0 (since p = -6, the e^(-integral p) factor and 1/y1^2 combine to a constant), so v = x and y2 = x e^(3x) — recovering the repeated-root second solution.
Substituting y2 = v y1 reduces the second-order equation to a first-order one for v' — here just v'' = 0.
Reduction of order requires one solution you already know; it cannot generate the first solution from scratch. Also remember to put the equation in standard form y'' + p y' + q y = 0 (leading coefficient 1) before reading off p for the formula.