Possible 3rd Order Non-Linear Diffeq Solutions

[HerpaDerp]

Alright, so I have a general question on finding the general solution of the eq:

t^3y"'(t)-4t^2y''(t)+8ty'(t)-8y(t)=t

next, turn it into a homogenous equation:

t^3y"'(t)-4t^2y''(t)+8ty'(t)-8y(t)=0

The assumed solutions are and derivatives:
Y1 = 8t Y1'= 8 Y1'' = 0 Y1'''=0
Y2 = -4y^2 Y2' = -8t Y2'' = -8 Y2''' = 0
Y3 = t^3 Y3'=3t^2 Y3'' = 6t Y3'''=6

Y1 and Y2 are solutions, but Y3 is not, I get the equation being equal to 2.

So, I am wondering for a 3rd order non-linear Diffeq, is it possible for there to be only 2 solutions, I thought there had to be 3?

 

[JJacquelin]

The assumed solutions are and derivatives:
Y1 = 8t Y1'= 8 Y1'' = 0 Y1'''=0
Y2 = -4y^2 Y2' = -8t Y2'' = -8 Y2''' = 0
Y3 = t^3 Y3'=3t^2 Y3'' = 6t Y3'''=6

Why assuming thoses solutions ? There is no reason to that.
It's just by chance that the two first ones are convenient. And there is no surprise for the third to be not convenient.
Let y = t^k
y'=k*t^(k-1)
y'' = k(k-1)t^(k-2)
y''' = k(k-1)(k-2)t^(k-3)
bringing back into the homogeneous equation leads to :
(k^3)-7k²+14k-8 = 0
(k-1)(k-2)(k-4)=0
So your solution Y1 is convenient only because, by chance, k=1 is solution. If k=1 wasn't solution, it would have failed. The same for Y2.
The third solution is given by k=4.