second order

2010-12-03 2:53 am

plz show me the steps~
thx in advance* =)


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更新1:

a typo above: should be (ln |sec x + tan x|)

更新2:

thx^^ im surprised u found it...!!* could u explain a bit on:: ∫- sin x tan x dx =∫( cos x - sec x )dx if i dont know those certain trig identities.. thx alot*=)

回答 (1)

2010-12-03 4:17 am
✔ 最佳答案
y'' + y = tan x   yh = c1cos x + c2sin x ~ homogenous solution  Let Q = tan x    yp = u1y1 + u2y2 = u1 cos x + u2 sin x  W =│ cos x sin x │= 1    │- sin x cos x│  u1 =∫( - y2Q/W )dx    =∫- sin x tan x dx =∫( cos x - sec x )dx    = sin x - ln│sec x + tan x│  u2 =∫( y1Q/W )dx    =∫cos x tan xdx =∫sin x dx    = - cos x  yp = u1 cos x + u2 sin x    = sin x cos x - cos x ln│sec x + tan x│- sin x cos x    = - cos x ln│sec x + tan x│  → yp = - cos x ln│sec x + tan x│~ particular solution

2010-12-02 22:05:52 補充:
tanx=sinx/cosx


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