Eigenvalue of LTI System I'm sure I am overlooking something simple, but I can't figure it out.
For an LTI system with impulse response $h(t)$, we have $y(t) = x(t) * h(t)$. So 
$$ y(t) = \int_{-\infty}^{\infty}{x(t-\tau)h(\tau)d\tau}. $$
If we consider the case of $x(t) = e^{st}$, then
$$ y(t) = \int_{-\infty}^{\infty}{e^{s(t-\tau)}h(\tau)d\tau} $$
$$ = e^{st}\int_{-\infty}^{\infty}{e^{-s\tau}h(\tau)d\tau} $$
$$ = x(t) \mathcal{L}\lbrace{h(t)}\rbrace $$
$$ = H(s) x(t). $$
To my understanding, the above states that the transfer function $H(s)$ is the eigenvalue of the system corresponding to the eigenfunction $x(t) = e^{s t}$.
In an attempt to verify this in a more concrete case, I attempted to work through an example using the system
$$ y'(t) + y(t) = x(t)$$
with input $x(t) = e^{5 t}$ and initial value $y(0) = 0$.
Using the Laplace transform, I quickly was able to compute 
$$H(s) = \frac{1}{1+s}.$$
So for this input, $H(5) = \frac{1}{6}$.
Therefore I expect $y(t) = H(5)e^{5 t} = \frac{1}{6}e^{5 t}$, which is indeed a component of the output, but somewhere I have lost the transient response. Using normal methods of solving the equation, I calculate $y(t) = \frac{1}{6}e^{5t} - \frac{1}{6}e^{-t}$. Choosing the initial value carefully I can eliminate the other component of the response, but I was under the impression that the initial result held for null initial conditions.
Where have I gone wrong? I've got a feeling it is a missed assumption or some such, but I haven't been able to find my mistake.
Thanks.
EDIT: Some more information.
The initial value problem $y' + y = 0; y(0) = 0$ has solution $y(t) = 0$.
The impulse response of the system described by $y' + y = x$ is
$h(t) = e^{-t}u(t)$. Taking the convolution integral with $x(t) = e^{5t}$, I get $y(t) = \frac{1}{6}e^{5t}$, consistent with the result above. But $Y(s) = H(s)X(s) = \frac{1}{(s-5)(s+1)} = \frac{1}{6(s-5)} -\frac{1}{6(s+1)}$, which gives $y(t) = \frac{1}{6}e^{5t} - \frac{1}{6}e^{-t}$. These methods should give the same result for all time, so I must be misapplying them or otherwise missing something. I can't find the source of this inconsistency.
 A: You are missing a $u(t)$ in all your input and output relationships in the second approach. $e^{5t}$ is not equal to $e^{5t}u(t)$ (well, after the transient response is damped you may assume there has not been a $u(t)$ and use the eigen approach). Notice that $\frac{1}{s+a}$ is the LT of $e^{-at}u(t)$.
"I was under the impression that the initial result held for null initial conditions"
This is a valid observation and is because:

A system described by ordinary differential equations with constant coefficients is only linear if the initial conditions are zero.

So any other initial condition except zero will make the system nonlinear, in which case the eigen-analysis is not valid anymore.
We can do something to make the system look like a linear system for $t\ge0$ by adjusting $y(0)$ such that the transient response is canceled.
Since the LT is $$sY(s)-y(0)+Y(s)=\frac{1}{s-5}$$ we have $y(t)=(y(0)-\frac{1}{6})e^{-t}u(t)+\frac{1}{6}e^{5t}u(t)$. Hence, with $y(0)=\frac{1}{6}$ the transient response is canceled for $t>0$. Note that $t=0$ is not an "initial" condition in this context (but $t=-\infty$ is).
