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Anonymous
Anonymous asked in 科學其他:科學 · 1 decade ago

這樣的微分方程有沒有特殊的解法呢?

題目1: y'=(3x-y-9)/(x+y+1)

正解: 3(x-2)^2 - 2(x-2)(y+3)-(y+3)^2 = c

題目2: y'=(2x-5y-9)/(- 4x+y+1)

正解: (2x+y-3)^2=c(y-x+3)

我一開始的解法是要座標平移的方式將題目先轉換成homogeneous,然後再

用解一階homogeneous differential equation的方式將的新變數的解,最後再將

變數轉回成原來的x,y變數。但我求出來的解總是跟正解不太一樣,我的比較

複雜一些,正解的答案感覺很順。請問這樣的題目是否有特殊的解法呢? 該

怎麼解出這樣的結果呢?

謝謝

Update:

hi, 謝謝您

但,請教一下

u + Xdu/dX = (2- 5u) / (u - 4)

Solving, X = c(u - 1) / (u + 2)^2

之間的過程是怎麼做的呢?

可以請你再指教一下嗎? 謝謝您

1 Answer

Rating
  • Anonymous
    1 decade ago
    Favorite Answer

    其實,所答出的解與正解有些微出入是沒關系的。只要你把答案微分,再作比較,若是對的便可以了。

    1. y' = (3x - y - 9) / (x + y + 1)

    -(3x - y - 9) + (x + y + 1)y' = 0

    { 3x - y - 9 = 0

    { x + y + 1 = 0

    Solving the intersection point, it is (2 , -3)

    Let x = X + 2, y = Y - 3

    Let Y = uX

    dY/dX = u + Xdu/dX

    Therefore, -(3X - Y) + (X + Y)dY/dX = 0

    u + Xdu/dX = (3X - Y) / (X + Y)

    Xdu/dX = (-u^2 - 2u + 3) / (1 + u)

    ∫1/X dX = ∫(1 + u)/(-u^2 - 2u + 3) du

    lnX = -1/2 ln(u^2 - 2u + 3) + C

    X^2 = c / (u^2 - 2u + 3), where c is a constant

    X^2 = cX^2 / (Y^2 + 2XY - 3X^2)

    Therefore, 3(x - 2)^2 - 2(x - 2)(y + 3) - (y + 3)^2 = c

    2. -(2x - 5y - 9) + (-4x + y + 1)y' = 0

    The intersection point is (-2/9 , -17/9)

    Let x = X - 2/9 , y = Y - 17/9

    Let Y = uX

    dY/dX = u + Xdu/dX

    So, -(2X - 5Y) + (-4X + Y)dY/dX = 0

    u + Xdu/dX = (2- 5u) / (u - 4)

    Solving, X = c(u - 1) / (u + 2)^2

    (Y + 2X)^2 = c(Y - X)

    (2x + y - 3)^2 = c(y - x + 3)

    2008-07-19 09:17:00 補充:

    Xdu/dX = (2 - 5u) / (u - 4) - u

    Xdu/dX = -(u + 2)(u - 1) / (u - 4)

    ∫1/X dX = -∫(u - 4) / (u + 2)(u - 1) du

    lnX = -∫2/(u + 2)du + ∫1/(u - 1)du

    lnX = -2ln(u + 2) + ln(u - 1) + c'

    X = c(u - 1) / (u + 2)^2

    Source(s): Myself~~~
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