In Exercises 10-23 use the method suggested by Exercise 9 to find a second solut
ID: 3109725 • Letter: I
Question
In Exercises 10-23 use the method suggested by Exercise 9 to find a second solution y_2 that isn't a constant a second multiple of the solution y_1. Choose K conveniently to simplify y_2. y" - 2y' - 3y = 0; y_1 = e^3x y" - 6y' + 9y = 0; y_1 = e^3x y" - 2ay' + a^2 y = 0; (a = constant); y_1 = e^ax x^2 y" + xy' - y = 0; y_1 = x x^2 y" - xy' + y = 0; y_1 = x x^2 y" - (2a - 1)xy' + a^2 y = 0 (a = nonzero constant): x > 0; y_1 = x^a 4x^2 y" - 4xy' + (3 - 16x^2)y = 0; y_1 = x^1/2 e^2x (x - 1)y" - xy' + y = 0; y_1 = e^x x^2 y" - 2xy' + (x^2 + 2) y = 0; y_1 = x cos x 4x^2 (sin x) y" - 4x(x cos x + sin x)y' + (2x cos x + 3 sin x)y = 0; y_1 = x^1/2 (3x - 1)y" - (3x + 2)y' - (6x - 8)y = 0; y_1 = e^2x (x^2 - 4)y" + 4xy' + 2y = 0; y_1 = 1/x - 2 (2x + 1)xy" - 2(2x^2 - 1)y' - 4(x + 1)y = 0: y_1 = 1/x (x^2 - 2x)y" + (2 -x)y' = 0; y_1 = e^xExplanation / Answer
LET K=-4
P(X)=-2X
U'=(-4)E2X/E6X=-4E-4X
INTEGRATE BOTH SIDES WE GET U=E-4X
Y2=UY1
=E-4XE3X=E-X