crazy notation u say?

i found this at www.sosmath.com. it''s called integrating factor method
if diff eq. is something like:
a(x)dy/dx + b(x)y = c(x), we rewrite this as:

dy/dx + p(x)y = q(x), where p(x) = b(x)/a(x) and q(x) = c(x)/a(x)

then u "can" do integrating factor (called I,u(x), whatever)
u(x) = e^(INT[p(x)dx]), while the general solution is
y = (INT[u(x)q(x)dx] + C) / u(x)

as i see it the second method is called seperable variables, where u put all y-based variables on one side, and x''s on the other.

so from our equation: dy/dx(x-1) = y(x+1) ->
dy/y = (x+1)/(x-1)dx, now we integrate each part
INT[dy/y] = ln[y]
INT[(x+1)/(x-1)dx] = x + 2ln[x-1]
so,
ln[y] = x + 2ln[x-1]
here i get a bit confuesed... we have to "ditch" the log''s, right?
so, y = e^(x+2ln[x-1]) ?

now, both methods should yield same results or what? do they?

to test if y is correct, we have to do a derivation of it, right?
y'' = (e^(x+2ln[x-1]))''
(e^x)'' = e^x, since x in this case is (x+2ln[x-1]), we have to derivate this too? so, y'' = e^(x+2ln[x-1]) * (x+2ln[x-1])'' ->
x'' + (2ln[x-1])'' -> 1 + ???
how to derivate 2ln[x-1] ?

original*,

Thank you for the sosmath link. I''m adding it to the forum FAQ!

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

I''ve never used the integrating factor method before, but since there''s literally dozens upon dozens of integration methods and formulas out there, I''m not surprised It appears to be nifty, but if you can recognize the implicit differentation going on beforehand then it saves you a lot of work.

quote:

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how to derivate 2ln[x-1] ?

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Bah, just reverse the integration It''s what you had originally.

think i got it
the method i find best is the one found at sosmath (integrating factor), others (separable variables, constants variation?) are still not quite clear to me...
so, here we go:

example: solve DE at given condition

xy''-y''= xy-y condition P(2,2)

first we rewite this in the following form:
y'' + p(x)y = q(x)

xy''-y'' = xy-y
y''(x-1) = y(x+1) / x-1)
y'' = y(x+1)/(x-1)
y'' - y(x+1)/(x-1) = 0
p(x) = -( (x+1)/(x-1) )
q(x) = 0

second, find the integrating factor u(x):
u(x) = e^( INT[p(x)dx] )
INT[ p(x)dx ]
INT[ -(x+1)/(x-1)dx ] -> - INT[ (x+1)/(x-1) ]
t = x-1
dt = 1*dx
- INT[ (t+2)/t * dt ] -> - ( INT[t/t dt] + INT[2/t dt] ) -> - ( x + 2Log[x-1] )
u(x) = e^(-x - 2Log[x-1])

third:
y = (INT[ u(x)q(x) ] + C) / u(x)
since q(x) = 0, INT[ 0 ] -> 0
and we get
y = C / u(x) -> y = C / e^(- x -2Log[x-1])
question: this is general solution?

plug in the condition P(2,2):
2 = C / e^(-2)
C = 2e^(-2)
C = 0.27
y = 0.27 / e^(- x -2Log[x-1])
question: this is particular solution?

now, to give it a test, we try to solve starting equation:
xy'' - y'' = xy - y
we have x,y (2,2), we only need the y''
y'' = ( C / e^(- 2 - 2Log[x-1]) )''
this can be solved like
(f/g)'' = (f''g - fg'') / g^2
since f = C, f'' = 0, so first part f''g = 0
and we get:
y'' = - C * e^(-2-2Log[x-1] * (-1-(2/(x-1))) ) / (e^(-2-2log[x-1])^2

y''(x-1) = y(x+1)
plugin 2,2 and C

y'' = 2*3
y'' = 6
so, spinach again...
(- 0.27 * e^(-2) * (-3)) / (e^(-2))^2 = 6
0.81 * e^(-2) / e^(-4) = 6
0.81 * 0.135 / 0.0183 = 6
0.81 * 7.278 = 6 if my calc does not lie, this is true
the end...

would be nice if someone checked this, just to make sure
also, a short enlightment on other methods would be cool, till then peace, vsc out