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

• 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))