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

Differential Equations · Axiom Academy

EXAMPLE Solving y'' - 5y' + 6y = 0 Using the characteristic equation to solve a second-order linear homogeneous ODE Find the general solution of the second-order linear homogeneous differential equation with constant coefficients y'' - 5y' + 6y = 0 . Excellent work! You've solved a second-order linear homogeneous ODE using the characteristic equation method — the real-distinct-roots case. The Characteristic Equation: Assuming y = e^ rx turns ay'' + by' + cy = 0 into the algebraic equation ar^2 + br + c = 0 , since the derivatives of e^ rx are just multiples of itself. Factoring: r^2 - 5r + 6 = 0 factors as (r-2)(r-3) = 0 , giving the two distinct real roots r_1 = 2 and r_2 = 3 . Distinct Real Roots: Whenever the characteristic equation has two distinct real roots r_1 and r_2 , the general solution is y = C_1e^ r_1x + C_2e^ r_2x — here, y = C_1e^ 2x + C_2e^ 3x . Always Verify: Substituting y = C_1e^ 2x + C_2e^ 3x back into y'' - 5y' + 6y gives exactly 0 for every choice of C_1 and C_2 , confirming the solution. This same characteristic-equation approach extends to the repeated-root and complex-root cases you'll see next — only the form of the general solution changes.

This is the written version of the interactive lesson above. See the full Differential Equations course.