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.
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:
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).