Suppose we have an equation y(t):
y(t) = c1e^((α−1)t) + c2e^(−2t)

Determine the values of α,if any,for which all solutions tend to zero as t → ∞, and also, determine the values of α,if any,for which all (nonzero) solutions become unbounded as t → ∞.

Question

Suppose we have an equation y(t):
y(t) = c1e^((α−1)t) + c2e^(−2t)

Determine the values of α,if any,for which all solutions tend to zero as t → ∞, and also, determine the values of α,if any,for which all (nonzero) solutions become unbounded as t → ∞.

sh3lsh · Answer

This isn't supposed to be complex, but I'm having a disagreement with the solution. 
I have alpha should be unbounded if it's greater than 1 and bounded if alpha <= 1.

mathmale · Answer

Hint:  the second term always decreases towards zero as t increases to infinity.  So you can focus on the first term alone.  
While not absolutely necessary, graphing the first term for a few values of alpha (one neg, one zero, one one, one positive) would tell you a lot in a hurry.

sh3lsh · Answer

The solution states: 
If α < 1, then limt→∞ y(t) = 0 and is bounded. 
There is no α such that y(t) is unbounded.

phi · Answer

if by "all solutions" they mean the value of y(t)
I don't understand the second part of the answer:There is no α such that y(t) is unbounded.

michele_laino · Answer

hint:
when $t$ goes to $+\infty$, then ${e^{ - 2t}}$ goes to $0$, so we can hope that the solution goes to zero, if and only if $\alpha-1<0$

michele_laino · Answer

when we say all non zero solutions, we think about the solutions for which $c_2 \neq 0$ and $c_1=0$. Now such solutions, as we can see, can not be unbounded

michele_laino · Answer

oops... I meant ...we $also$ think about to the solutions for which $c_2 \neq 0$ and $c_1=0$...