<s>[INST] Third order ODE initial value problem,solution obeys $y(x) \rightarrow 0 $ as $x \rightarrow \infty$ ??? $y''' + y'' -y' -y=0$
$y(0)=7,y'(0)=-3,y''(0)=\alpha$
Find all values of $\alpha$ for which the solution obeys $y(x) \rightarrow 0 $ as $x 
\rightarrow \infty$
Here is my work
I used the cubic characteristic equation $r^3 +r^2 -r -1 = 0$ and find the three roots to be
$-1,-1$and $1$
So my general solution is $y(x) = Ae^x+Be^{-x}+Cxe^{-x}$
and
$y'(x) = Ae^x-Be^{-x}+Ce^{-x}-Cxe^{-x}$
$y''(x) = Ae^x+Be^{-x}+-2Ce^{-x}+Cxe^{-x}$
Use the given initial values, I get
$y(0) = 7 = A+B$
$y'(0) = -3 = A-B+C$
$y''(0)=\alpha = A+B-2C$
What should I do next to determine all the values of $\alpha$ such that the solution obeys $y(x) \rightarrow 0 $ as $x 
\rightarrow \infty$ ???
Thanks in advance for the help
 [/INST] As $x \to \infty$, $e^x \to \infty$, so we must set $A = 0$ in order to guarantee that $y(x) \to 0$ as $x \to \infty$. Then from $y(0) = 7 = A + B$, we see that $A = 0$ implies that $B = 7$. 
From $y'(0) = -3 = A - B + C$, we get $-3 = -B + C = -7 + C$, so $C = 4$.
From $y''(0) = \alpha = B - 2C$, we get $\alpha = 7 - 8 = -1$. Then $\alpha = -1$. 
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