- #1
praecox
- 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? :/
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? :/