Re: [理工] [工數]-ODE
※ 引述《imnothing11 (小嘉)》之銘言：
: (1+x^2)y"+1+(y')^2=0
: 這題算了好久
: 算不太出來
: 有人可以幫解一下嗎
2 2
(1 + x )y" + 1 + y' = 0
Let p = y'
2 dp 2
(1 + x )---- + 1 + p =0
dx
dp dx
-------- + -------- = 0
2 2
1 + p 1 + x
-1 -1
tan p + tan x = c1
Let tanα = p tanβ = x
tanα + tanβ p + x
tan(α+β) = -------------- = ------- = tan c1 = c2
1-tanαtanβ 1-px
p + x = c2 ( 1-px )
p + c2 px - c2 + x = 0
( 1+ c2 x )p - c2 +x = 0
2
1 1 + c2
----(1 + c2 x) - --------
x - c2 c2 c2
dy + ---------- dx = dy + --------------------------- = 0
1 + c2 x 1 + c2 x
2
1 1 + c2
dy + ( ---- - --------------- ) dx = 0
c2 c2 (1 + c2 x)
2
x 1 + c2
y + ---- - -------- ln| c2 (1 + c2 x) | = c3
c2 2
c2
1
稍微整理 Let ---- = c4 並把ln內常數提出
c2
2
y = -c4 x + ( 1+c4 ) ln | x+c4 | + c5
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08/12 20:32, , 1F
08/12 20:32, 1F
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