How Do You Solve a Second Order ODE Using Reduction of Order?

[praecox]

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? :/

 

[Some Pig]

Let y=tⁿ
n(n-1)tⁿ-4ntⁿ+6tⁿ=0
(n-2)(n-3)tⁿ=0
n=2 or 3

 

[matematikawan]

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.

 

[praecox]

That makes so much sense now! Thank you so much! :D