Re: [理工] [工數]-O.D.E.

[kagato (包)]

※ 引述《smallprawn (水中瑕)》之銘言：
:        2
: 1.(x-1) y''+4(x-1)y'+2y=1

let x-1 = exp^t , t = ln(x-1) , D=d/dt

[D(D-1) + 4D + 2 ]y = 1

[D^2+3D+2]y = 1

yh = c1exp^(-t) + c2exp^(-2t)

yp = 1/2

y = yh + yp = c1exp^(-t) + c2exp^(-2t) + 1/2

            = c1/(x-1) + c2/(x-1)^2 + 1/2


:                             2
: 答案為y=(c1/(x-1))+(c2/(x-1) )+1/2
:                              2   2
: 但我解出來的答案為 y=(1/(x-1) )[x +c1x+c2]

: 是我解錯嗎= =!?請高手解答!!

你的答案如果x^2係數是1/2應該就對@@
:                2
: 2.xy''+y'=(y'')

題目應該是 xy'' + y' = (y')^2 吧!?

let y'=p → xp' + p  = p^2

            (xp)'    = (xp)^2 / x^2


         → d(xp)/(xp)^2 = dx/x^2

               1/xp = 1/x + c1

         →  1 = p +c1xp = p(c1x+1) = y' (c1x+1)


              dy = dx/(c1x+1) →  y = 1/c1 *ln[c1x+1] +c2


: 答案為 y=(1/c1) ln|c1x+1|+c2
: 但我算出來為 y=(-1/c1) ln|1-c1x|+c2
: 這答案也是正確的嗎= =? 還是她答案錯了!?
: 請高手解答搂!!感恩!!

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推 smallprawn:為什麼你的yp直接就知道是1/2= =..
推 squallting:因為等號右邊是常數 假設yp=A帶入 求得A=1/2