Reduction of Order Problem

  • Thread starter praecox
  • Start date
17
0
Hey, guys. I'm having a hard time with a 2nd ODE reduction of order problem.

The DE is
t2y'' - 4ty' + 6y = 0, with y1 = t2.

So I set up my y = t2v(t); y' = t2v' + 2tv; y'' = t2v'' + 4tv' + 2v.
I then substituted back in and got:
t4v'' = 0

This is where I'm getting stuck. The book says that the answer I'm looking for is y2 = t3, but I can't seem to get there from here.

I don't think I can use an integrating factor here (if I'm wrong there let me know), so I tried separating the variables, but that got me no where. I think I did it wrong though.

Here was my separation of variables attempt:
letting w = v', so w' = v'':
t4dw/dt = 0
dw = t-4 dt
Integrating both sides (and taking constants of integration to be zero):
w = -1/3 t-3
since w = v':
dv = -1/3 t-3 dt
Integrating again to find v and still no where near t3.

Any ideas? Anybody? :/
 
38
0
Let y=tn
n(n-1)tn-4ntn+6tn=0
(n-2)(n-3)tn=0
n=2 or 3
 
Hey, guys. I'm having a hard time with a 2nd ODE reduction of order problem.

The DE is
t2y'' - 4ty' + 6y = 0, with y1 = t2.

So I set up my y = t2v(t); y' = t2v' + 2tv; y'' = t2v'' + 4tv' + 2v.
I then substituted back in and got:
t4v'' = 0


Any ideas? Anybody? :/
v"(t)=0 hence v(t) is linear. Choose v(t)=t so that y = t2v(t)=t3.
 
17
0
That makes so much sense now! Thank you so much! :D
 

Related Threads for: Reduction of Order Problem

  • Posted
Replies
2
Views
2K
  • Posted
Replies
2
Views
3K
  • Posted
Replies
4
Views
5K
Replies
4
Views
1K
Replies
3
Views
1K
  • Posted
Replies
11
Views
3K
Replies
1
Views
2K
  • Posted
Replies
7
Views
2K
Top