[大學]工數積分因子~~

[大學]工數積分因子~~

小皓 於 星期二 二月 28, 2012 6:55 pm


(2xy^2-y)dx+(2x-x^2y)dy=0
小弟一直找不到適合應對的積分因子工式麻煩一下各位  謝謝

小皓
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Re: [大學]工數積分因子~~

lskuo 於 星期三 二月 29, 2012 5:06 pm


小皓 寫到:(2xy^2-y)dx+(2x-x^2y)dy=0
小弟一直找不到適合應對的積分因子工式麻煩一下各位  謝謝


First, try to make the expression as the symmetric form in the left-hand side:

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

Recall d(xy) = xdy + ydx
LHS = (2xy - 1)(ydx - xdy) = (2xy - 1)[d(xy) - 2xdy]
RHS = x(xy - 1) dy

So one can get
(2xy - 1)d(xy) = x(5xy - 3)dy

Let z = xy, then x = z/y, one can re-write the above equation by using (y,z) instead  (x,y):

(2z - 1)dz = z/y (5z - 3) dy

Now the variables (y,z) are separable:

(2z-1)/(5z-3)/z dz = dy/y = d(lny)

[1/z - (3/10)/z/(z-1/5)]dz = 5/2 d(lny)
d(lnz) -3/2[1/(z-1/5) - 1/z]dz = 5/2 d(lny)

So, one can do integration term by term,

lnz - 3/2[ln(z-1/5) - lnz] = 5/2 lny + C
5/2ln(z/y) - 3/2ln(z-1/5) = C

Back to (x,y) and multiply 2 to each side:
5lnx - 3ln(xy-1/5) = 2C .

One can also re-write as:

5lnx - 3 ln(5xy-1) = D

where D is an integration constant.

lskuo
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