Please explain \\ show your work If the ODE y\" + 9y\' - 36y = tan x is solved b

Question

Please explain show your work

If the ODE y" + 9y' - 36y = tan x is solved by the variation of parameters technique, with A(x) and B(x) representing the unknown functions, then a pair of linear equations in A' and B' will be A'e3x + B'e-12x = 0 3A'e3x - 12B'e-12x = tan x A'e3x + B'e12x = 0 3A'e3x + 12B'e12x = tan x A'e-3x + B'e12x = 0 -3A'e-3x + 12B'e12x = tan x A'e-3x + B'e-l2x = 0 -3A'e-3x - 12B'e-12x = tan x A'e4x + B'e-9x = 0 4A'e4x - 9B'e-9x = tan x A'e4x + B'e9x = 0 4A'e4x + 9B'e9x = tan x A'e-4x + B'e9x = 0 4A'e-4x + 9B'e9x = tan x A'e-4x + B'e-9x = 0 -4A'e-4x - 9B'e-9x = tan x A'e2x + B'e18x = 0 2A'e2x + 18B'e18x = tan x A'e-2x + B'e18x = 0 -2A'e-2x + 18B'e18x = tan x A'e-2x + B'e-18x = 0 -2A'e-2x - 18B'e-18x = tan x A'e2x + B'e-l8x = 0 2A'e2x - 18B'e-18x = tan x none of these is possible

Explanation / Answer

Answer is A) We can find the complementary solution of the equation y'' + 9y' -36 =0 Yc = Ae^(4x) + Be^(-12x) Now using Lagrange's identity, we can find the particular solution by below two equations ; A'e^(4x) + B'e^(-12x) = 0 and A'[e^(4x)]' + B'[e^(-12x)]' = 4A'e^(4x) - 12B'e^(-12x) = tanx (RHS of the nonhomogenous equation)

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