Re: [微積] 微分方程
※ 引述《BeMgCaSrBaRa (鈹鎂鈣鍶鋇鐳)》之銘言：
: Solve the differential equation dy/dx = r(M-y)y ,0<y<M
用分離變數法
1
∫------ dy = ∫r dx + C
(M-y)y
1 1 1 1
------ = --- (----- + ---)
(M-y)y M M-y y
1
=> ---[ln y - ln(M-y)] = rx + C
M
M-y
=> ln(M-y) - ln y = ln(-----) = - M(rx + C)
y
M -M(rx+C)
=> --- = 1 + e
y
M
=> y = ------------- for some K
-Mrx
1 + Ke
: ∞ x^n
: Find the radius and interval of convergence of the power series Σ -----
: n=J n
n+1 n
By ratio test, if |(x / (n+1)) / (x / n) | = |nx/(n+1)| < 1 as n->∞,
the series converges.
=> |x| < 1 => -1 < x < 1
∞ 1
If x = 1, we have Σ -----, which diverges
n=J n
所以收斂半徑 = 1, 區間 (-1, 1)
: ln(x^2+y^2)
: Evaluate ∫∫ ------------- dxdy ,where D={(x,y):1≦x^2+y^2≦e}
: D √(x^2+y^2)
2 2 2
座標變換, 令 r = (x + y )
2
2π √e ln r
則原式 = ∫ ∫ ------ r dr dθ
0 1 r
2π √e
= ∫ ∫ 2 ln r dr dθ
0 1
2π √e
= 2(∫ dθ)(∫ ln r dr )
0 1
∫ln r dr = r ln r - ∫ r/r dr + C = ln r - r + C
原式 = 2(2π-0)[(√e ln √e -ln 1) - (√e -1)] = 4π(1 - √e/2) = 4π-2π√e
: ∞ 1
: Determine the convergence or divergence of the series Σ -----------
: n=2 (√n) ln(n)
: Give reasons for your answer
1 1
---------- > -------- for n > 2
√n ln n n ln n
∞ 1 ∞ 1 ∞
∫ ------- dx = ∫ --- du = ln u | = ∞
2 x ln x ln 2 u ln 2
∞ 1
By comparison test, Σ ----------- diverges
n=2 (√n) ln(n)
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 111.251.180.160
※ 編輯: yueayase 來自: 111.251.180.160 (06/03 23:42)
推
06/04 00:21, , 1F
06/04 00:21, 1F
※ 編輯: yueayase 來自: 111.251.180.160 (06/04 05:26)
※ 編輯: yueayase 來自: 111.251.180.160 (06/04 05:27)
※ 編輯: yueayase 來自: 111.251.180.160 (06/04 05:29)
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