hanley 發問時間: 科學數學 · 5 年前

想問關於微分方程的問題 dy/dx = (x+y+2)/(x-y+1) , y(0)=0 會碰到xy無法分離的狀況, 想問各位大大有沒有其他解法?

1 個解答

評分
  • 麻辣
    Lv 7
    5 年前
    最佳解答

    dy/dx = (x+y+2)/(x-y+1), y(0)=0 ==> x=y=0

    1.Transform coordinate

    (x + y + 2)dx + (-x + y -1)dy = 0

    => x + y + 2 = 0; -x + y - 1=0

    => x =-3/2; y = -1/2

    Parallel moving coordinate to (-3/2,-1/2):

    => x = X - 3/2, y = Y - 1/2

    => dx = dX, dy = dY

    => 0 = (X + Y)dX + (-X + Y)dY

    2.Set Y = X*T to delete Y

    => dY = XdT + TdX

    => T = Y/X

    0 = (X + XT)dX + (-X + XT)(XdT + TdX)

    = (1 + T)dX + (-1 + T)XdT + (-1 + T)TdX

    = (1 + T^2)dX + (-1 + T)XdT

    = dX/X + (T-1)dT/(T^2+1)

    lnC = ∫dX/X + ∫TdT/(T^2+1) - ∫dT/(T^2+1)

    = lnX + ln(T^2+1) - atanT

    ln[X(T^2+1)/C] = atanT

    X(T^2+1) = C*exp(atanT)

    (x+3/2)[(Y/X)^2+1] = C*exp[atan(Y/X)]

    C*exp{atan[(y+1/2)/(x+3/2)]}

    = (x+3/2)[(y+1/2)^2/(x+3/2)^2 + 1]

    = [(y+1/2)^2+(x+3/2)^2]/(x+3/2)

    3.Set x = y = 0 to find C = ?

    => C*exp(atan(1/3)) = 2/9*3

    => C = 2/[27*exp(atan(1/3))]

    Ans: [(y+1/2)^2+(x+3/2)^2]/(x+3/2) = (2/27)*exp{atan[(3y-x)/(3x+9/2)]}

    4.Note: atanA - atanB = atan((A-B)/(1+AB))

    atan(y+1/2)/(x+3/2) - atan(1/3)

    = atan{[(y+1/2)/(x+3/2)-(1/3)]/[1+(y+1/2)/3(x+3/2)]}

    = atan((3y-x)/(3x+9/2))

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