Answer to Question #145226 in Differential Equations for Nikhil Singh

Question

By the method of variation of parameters, solve the following differential equation:
d^2y/dx^2+4y =4 tan2x

Answer 1

$\dfrac{d^2y}{dx^2} +4y = 4tan \ 2x\\ y''+4y = 4 tan \ 2x\\ y'' + 4y = 0\\ k^2+4 = 0\\ k_1 = 2i, \ k_2 = -2i\\ Therefore, \ an \ additional \ function \ is \ specified\\ y_c = C_1cos2x +C_2sin2x\\ y = Acos2x +Bsin2x\\ be \ the \ complete\ solution\\ of\ the\ given\ equation\ where\\ A\ and\ B\ are\ to\ be\ found\\ We\ have\ y_1 = cos2x, \ y_2 = sin2x\\ y_1'=-2sin2x\\ y_2' = 2cos2x\\ Then\ W = y_1*y_2' -y_2*y_1' = \\ = 2cos^2\ 2x + 2sin^2\ 2x = 2\\ A' = \dfrac{-y_2 * 4tan\ 2x}{W}\\ B' = \dfrac{y_1 * 4tan\ 2x}{W}\\ A' = \dfrac{-sin\ 2x * 4tan\ 2x}{2}\\ B' = \dfrac{-cos\ 2x* 4tan\ 2x}{2}\\ A =\int \dfrac{-2sin^2\ 2x }{cos2x} dx\\ \ \\ B = \int 2sin\ 2x\ dx\\ A = -log(sec\ 2x +tan\ 2x) + sin\ 2x+C_1\\ B = -cos\ 2x + C_2\\ y = Acos2x +Bsin2x\\ y = C_1cos2x +C_2sin2x -cos\ 2x*log(sec\ 2x +tan\ 2x)\\ answer: \\ y = C_1cos2x +C_2sin2x -cos\ 2x*log(sec\ 2x +tan\ 2x)\\$