Question

(1 point) Find y as a function of x if y(x) = -2sin(2x)+7cos(2x)+8

Answer 1

Y'' + 4y' = 0 A linear homogeneous ODE with constant coefficients has the form of a_n y(n) + --- + a_1 y' + a_0 y = 0. For an equation a_n y^(n) + --- + a_1 y^1 + a+ 0 y = 0, assume a solutionof the form y = e%gamma x. So, y'' + 4y' = 0 So, (e^gamma x)mn + 4(e^gamma x)^1 = 0 So, (gamma^3 e%n) + 4(gamma e^n) = 0 So, e^n(gamma^3 + 4 gamma) = 0 So, gamma(gamma^2 + 4) = 0 So, gamma = 0: gamma = plusminus 2i So, the general solution of the one real root and two complex root has taken the form is y = C_1 e^gamma x + e^ax (C_2 cos beta x + C_3 sin beta x) so, y = C_1 e^0x + e^ox(C_2 cos (2x) + C_3 sin (2x)) so, y = C_1 + C_2 cos (2x) + C_3 sin (2x)