Characteristic Equations with Distinct Real Roots

A Simplification... Ideally we would like to be able to solve lots of different kinds of higher-order ODEs, like we did for first-order ODEs, however this turns out to be REALLY HARD. So we will make some simplifications. I We will restrict ourselves to discussing only higher-order LINEAR ODEs and IVPs. I We will assume (at least initially) that all of the coefficient functions a1 (x), a2 (x), . . . , an (x) are actually just constants. We will call them just a1 , a2 , . . . , an . I We will (initially) restrict ourselves to talking about homogeneous linear ODEs. I Later, we will discuss nonhomogeneous linear ODEs. I We MAY talk about linear ODEs with nonconstant coefficient functions as well. Homogeneous Linear ODEs with Constant Coefficients A homogeneous linear ODE with constant coefficients has the following form, an y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 y = 0, where the ai for 0 ≤ i ≤ n are real numbers. What kinds of functions y (x) could possibly satisfy such an equation? Homogeneous Linear ODEs with Constant Coefficients We GUESS y (x) = e rx for some constant r since a linear combination of the derivatives must be identically zero. y = e rx y 0 = re rx y 00 = r 2 e rx .. . y (n) = r n e rx Let’s plug this into the ODE. Homogeneous Linear ODEs with Constant Coefficients an y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 y n rx an r e + an−1 r rx n n−1 rx e + · · · + a1 re e (an r + an−1 r n−1 rx + a0 e rx = 0 = 0 + · · · + a1 r + a0 ) = 0 Note that it is impossible for e rx to be equal to 0 since it is always positive for any real number x. Hence, if y (x) = e rx to be a solution to the ODE, we would need for r to satisfy the polynomial equation an r n + an−1 r n−1 + · · · + a1 r + a0 = 0. Definition Consider the nth order homogeneous linear ODE with constant coefficients, an y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 y = 0. The characteristic equation (or auxiliary equation of the ODE is given by an r n + an−1 r n−1 + · · · + a1 r + a0 = 0. Fact 1 If r is a real number satisfying the characteristic equation of some homogeneous linear ODE with constant coefficients, then y (x) = e rx is a solution to the ODE. Fact 2: Follows from Principal of Superposition (Homogeneous Version) If r1 , r2 , . . . , rn are DISTINCT real roots of the characteristic equation to some homogeneous linear ODE with constant coefficients (i.e. they all satisfy the characteristic equation and ri 6= rj for all 1 ≤ i < j ≤ n), then a general solution of the ODE is given by y (x) = C1 e r1 x + C2 e r2 x + · · · + Cn e rn x . In other words, every solution to the ODE is of the above form for some constants C1 , C2 , . . . , Cn . Special Note about this Chapter In this chapter, we are only considering second-order homogeneous linear ODEs with constant coefficients, hence... I ODE: ay 00 + by 0 + cy = 0 I Char. Eq.: ar 2 + br + c = 0 I Distinct Real Roots: r = r1 , r = r2 I General Solution: y (x) = C1 e r1 x + C2 e r2 x Example Find a general solution to the ODE. y 00 + 2y 0 − 3y = 0 Example Find a general solution to the ODE. y 00 + 9y 0 − 10y = 0 Example Find a general solution to the ODE. y 00 + 5y 0 + 6y = 0 Example Find a general solution to the ODE. y 00 − 6y 0 + 9y = 0 HINT: You can’t with current knowledge.