Anonymous

# 這樣的微分方程有沒有特殊的解法呢?

Update:

hi, 謝謝您

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

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

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• Anonymous

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~~~