Solving y'' - 5y' + 6y = 0

Calculus 2 · Axiom Academy

EXAMPLE Solving a Second-Order Differential Equation Build the characteristic equation, factor for its roots, then pin the constants with two initial conditions. Solve the second-order linear homogeneous equation y'' - 5y' + 6y = 0 subject to the two initial conditions below, finding the one particular solution that satisfies both. Nicely done. You solved a second-order linear homogeneous equation end to end — from trial solution to a single particular curve. Characteristic equation: For y'' + ay' + by = 0 , the trial solution y = e^ rx turns the differential equation into the algebraic equation r^2 + ar + b = 0 . General solution: Two distinct real roots r_1, r_2 give y = c_1 e^ r_1 x + c_2 e^ r_2 x . Initial conditions: y(0) and y'(0) produce two equations in c_1 and c_2 — solving the system fixes both constants. This problem: r^2 - 5r + 6 = 0 factors as (r-2)(r-3) , and the conditions give , so y = -4e^ 2x + 5e^ 3x . The same three moves — characterize, generalize, then apply conditions — drive the linear models behind oscillations, circuits, and mechanical systems.

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