# Second order DE

• Aug 30th 2010, 05:39 PM
Second order DE
Which method can i use to solve u'' - (t-1)/t u' = t/(t+1)??

This is an equation I have formed by making it U absent and I let p = dy/dx and dp/du = d^y/dx^2 but then i'm stuck after that -.-"" any help would be appreciated I've attempted this question so many times...I tried using an integrating factor to solve but i couldnt do it..
• Aug 31st 2010, 02:16 AM
Ackbeet
Well, you can reduce the order immediately by setting v = u', as you have already (sort of) mentioned. Then I would probably use the integrating factor on the resulting DE:

v' - v (t-1)/t = t/(t+1).

What is the integrating factor here?
• Aug 31st 2010, 02:23 AM
The integarting factor I found was e^ integral of (1/t -1) which i worked out to be te^-t but then when i multiplied it with t/(t+1) I had no idea how to integrate it....
• Aug 31st 2010, 02:35 AM
Ackbeet
Right. Looking good so far. So now we must integrate the RHS which is, as you've pointed out, the integral

$\displaystyle{\int\frac{t^{2}e^{-t}}{t+1}\,dt.}$

This is not a very nice integral, I grant you. At this point, I would probably substitute $s=t+1,$ in order to get rid of that sum in the denominator. What does this get you?
• Aug 31st 2010, 02:39 AM
s = t + 1
t = s -1
dt = ds
[(s-1)^2e^-(s-1)]/s ?
• Aug 31st 2010, 02:49 AM
Ackbeet
That's right. Keep going... what would you do next?
• Aug 31st 2010, 03:00 AM
uuuummm I tried to split it up as in (s-1)^2/s x e^1-s/2 then expanded the (s-1)^2/2 to get (s - 2 - 1/s). e^(1-s)/s ...but that didnt really help haha
• Aug 31st 2010, 03:05 AM
Ackbeet
Hmm. Clean up your notation there and check your minus signs. The integral should break up into three pieces, each of which you handle differently.
• Aug 31st 2010, 03:10 AM