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If we consider the following differential equation $$ a_0x + a_1\dot{x} + a_2\ddot{x} + \ldots + a_n x^{(n)} = b_0y+b_1\dot{y} b_2\ddot{y} + \ldots + b_m y^{(m)} \ , $$ my text first substitutes the ansatz $x(t) = e^{st}$, after which the solution for $y(t)$ seems to be given by $$ y(t) = \frac{a_0+a_1s+a_2s^2+\ldots+a_ns^n}{b_0+b_1s+b_2 s^2+\ldots+b_ms^m}e^{st} \ . $$

I'm having trouble seeing how we arrive at the given solution for $y(t)$. How would one show that this is indeed a solution for $y$ given $x(t) = e^{st}$?

After we substituted for $x(t)$, I'd first look for a homogeneous solution in $y$. Then I would add up a particular solution but, this doesn't seem to yield the required coefficients of our given solution that easily.

Plugging in our given solution for $y$, to see if it satisfies the ODE doesn't seem that instructive either, since the coefficients make it a huge mess.

Context

The given ODE describes a linear time-invariant system that transforms an input signal $x(t)$ in to an output signal $y(t)$. $H(s) = \frac{a_0+a_1s+a_2s^2+\ldots+a_ns^n}{b_0+b_1s+b_2 s^2+\ldots+b_ms^m}$ is called the transfer function of the system.

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    Plug $x(t) = e^{st}$ into the left hand side of your equation. Then the equation becomes $F(s)e^{st} = b_0y+\cdots$ which means $y(t) = G(s)e^{st}$. – 2017-02-15
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    @hkr Could you elaborate on how we arrive at $y(t)=G(s)e^{st}$? – 2017-02-15
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    Whatever $y(t)$ is, it must produce derivatives of the form $g(s,t)e^{st}$ because the equality would fail otherwise. If $g$ were dependent on $t$ then for at least some of the terms, they would have $t$ present not in the exponential. Hence $g(s,t) = G(s)$ for some function $G$. Since $H(s)$ (your transfer function) is such a function, you are done. – 2017-02-15
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    Nice. But, what gives you the assurance as to why our specific $H(s)$ is valid? – 2017-02-15
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    Uniqueness. Especially when doing physics, never overlook a solution you got by guessing, since physics is basically learning to cultivate and refine your guessing ability by collecting data. – 2017-02-15
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    Could you be more concrete on why it is valid, before we discuss if it is unique? What do you mean by *never to overlook a solution by guessing in physics*? – 2017-02-15
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    So sorry for not responding. In physics, many problems are solved by looking at the physical data, and guessing the form of the solution, then plugging it in and seeing if it works. Uniqueness guarantees that if it does, it is indeed a specific solution. We are not discussing whether or not it is unique, not for an ODE. If it solves the equation, then for that specific set of initial conditions at that point, it is unique. – 2017-04-03

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We can check the given solution by substituting it in the ODE, which, for the separate terms add up to the denominator of H(s).