Given the ODE y\'\' + y = tan x, the general solution is yc + yp, where yc = C1c

Question

Given the ODE y'' + y = tan x, the general solution is yc + yp, where

yc = C1cos x + C2sin x. For this ODE, yp =Question 2 options:
A) x(cos x) + (sin x)ln|sec x|
B) -x(sin x) - (cos x)ln|sin x|
C) -x(cos x) + (sin x)ln|sin x|
D) (1/2)sec x
E) -(cos x)ln|sec x + tan x|
F) -2 + (sin x)ln|sec x + tan x|
G) -(cos x)ln|sec x + tan x| - (sin x)ln|csc x + cot x|
H) -(sin x)ln|csc 2x + cot 2x|
I) (1/2)tan x + (1/2)(cos x)ln|sec x + tan x|
J) -(sin x)ln|csc x + cot x|
K) none of these
I will only award points for the correct answer

Explanation / Answer

E

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