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The DE is

t

^{2}y'' - 4ty' + 6y = 0, with y

_{1}= t

^{2}.

So I set up my y = t

^{2}v(t); y' = t

^{2}v' + 2tv; y'' = t

^{2}v'' + 4tv' + 2v.

I then substituted back in and got:

t

^{4}v'' = 0

This is where I'm getting stuck. The book says that the answer I'm looking for is y

_{2}= t

^{3}, 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'':

t

^{4}dw/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 t

^{3}.

Any ideas? Anybody? :/