applied maths diff. eq唔明(高手請進)

This is Mastering Applied Mathematics Vol. 1, P.104, example 46, right?

Your question is wrong, according to the book, it should be (2x + 3y + 2)dy/dx + 2x + 3y - 1 = 0

With this corrected question, you will get the reduced equation.


(2x + 3y + 2)dy/dx + (2x + 3y - 1) = 0

Let u = 2x + 3y

du/dy = 2dx/dy + 3

[du/dy - 3]/2 = dx/dy

dy/dx = 2/[du/dy - 3]

Therefore, the differential equation becomes:

2(u + 2)/(du/dy - 3) + (u - 1) = 0

2(u + 2) = (3 - du/dy)(u - 1)

2u + 4 = 3u - 3 - (u - 1)du/dy

(u - 1)du/dy = u - 7

∫[(u - 1)/(u - 7)]du = ∫dy

∫[1 + 6/(u - 7)]du = y

u + 6ln(u - 7) = y + c

6ln(u - 7) = y - u + c

6ln(2x + 3y - 7) = -2x - 2y + c

x + y = -3ln(2x + 3y - 7) + c'

2009-03-18 20:11:40 補充：
幸好check一check。如果是+1就做不到了。

難怪做不到