5.1: 34, 36, 38, 40, 42, 46, 48 5.1-34 ====== y'' + 2y' - 15y = 0 r^2 + 2r -15=0, char equation (r+1)^2 -16=0, complete the square (r+1-4)(r+1+4)=0, factor difference of squares r=3, r= -5, find the two roots [could have used the quadratic formula] exp(3x), exp(-5x), the two atoms implies by Euler's theorems y = linear combination of the atoms y = c1 exp(3x) + c2 exp(-5x) 5.1-36 ====== 2y'' + 3y' = 0 2r^2+3r=0, char equation r(2r+3)=0, factored r=0, r=-3/2, roots exp(0x), exp(-3x/2), two atoms implied by Euler's theorems y=linear combination of the atoms y=linear combination of 1, exp(-3x/2) y = c1 + c2 exp(-3x/2) 5.1-38 ====== 4y'' + 8y' + 3y = 0 4r^2+8r+3=0, char equation (2r+1)(2r+3)=0, factored by inverse FOIL r=-1/2, r=-3/2, roots exp(-x/2), exp(-3x/2), atoms by Euler's theorems y=linear combination of the atoms 5.1-40 ====== 9y'' - 12y' + 4y = 0 9r^2-12r+4=0, char equation (3r-2)(3r-2)=0, factored by inverse FOIL r=2/3, r=2/3, double root) exp(2x/3), x exp(2x/3), two atoms by Euler's theorems y=linear combination of the atoms 5.1-42 ====== 35y'' - y' - 12y = 0 35r^2-r-12=0, char equation (7r+4)(5r-3)=0, factored by inverse FOIL r=-4/7, r=3/5, roots exp(-4x/7), exp(3x/5), two atoms by Euler's theorems y=linear combination of the atoms 5.1-46 ====== y = c1 exp(10x) + c2 exp(100x), find the DE Take partials on c1, c2 across the general solution to find the basis functions: exp(10x), exp(100x) These atoms correspond to Euler roots r=10, r=100 Then (r-10)(r-100)=0 must replicate the characteristic equation. Expand, r^2-110r+1000=0, the charactistic equation Then y'' -110y' + 1000y=0 is the differential equation 5.1-48 ====== y = exp(x)(c1 exp(sqrt(2)x) + c2 exp(-sqrt(2)x)), find the DE y = c1 exp(r1 x) + c2 exp(r2 x) where exp(r1 x)=exp(x)exp(sqrt(2)x)=exp([1+sqrt(2)]x), exp(r2 x)=exp(x)exp(-sqrt(2)x)=exp([1-sqrt(2)]x), imply r1=1+sqrt(2), r2=1-sqrt(2). Like the preceding problem, use the factor-root theorem of college algebra to reconstruct the characteristic equation (r - r1)(r - r2)=0 r^2 - (r1+r2)r + r1 r2=0 r^2 - 2r + 5 = 0 Then the differential equation must be y'' - 2y' + 5y = 0