Question

(1 point) Suppose that a fourth order differential equation has a solution y = 5e3*x cos(x). Find the initial conditions that this solution satisfies y(0)= 0 y'(0)=| | у" (0)- у" (0)-

Answer 1

"y = 5e^3x x cosx differentiating successively with respect to x we have y' = 15e^3x x cosx + 5e^3x cosx - 5e^3x x sinx then y'(0) = 5 again y"" = 15e^3x cosx + x d/dx(15e^3x cosx) + 15e^3x cosx - 5e^3x sinx - d/dx(5e^3x x sinx) y""(0) = 15 + 15 = 30 now y""' = 45e^3x cosx - 15e^3x sinx + x d^2/dx^2(15e^3x cosx) + d/dx(15e^3x cox) + 45e^3x cosx - 15e^3x sinx - 15e^3x sinx - 5e^3x cosx - d^2/dx^2(5e^3x x sinx) therefore y""'(0) = 45 + 45 + 45 - 5 - [d/dx{15e^3x sinx + 5e^3x sinx + 5e^3x x cosx}]_x = 0 = 130 - 5 + 5 = 130 therefore the tnitial conditions will be y(0) = 0, y'(0) = 5, y""(0) = 30, y""'(0) = 130"