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I have the following O.D.E. relating a system's input and output, where x(t) is the input and y(t) is the output:

$2x(t) = {d^2\over dt^2} [y(t)] + 6{d\over dt} [y(t)] + 8y(t)$

It is also known that the system is causal and stable.

I figure there are several ways to approach this to find the impulse response, but I was wondering if the most practical method would be to simply select the Dirac Delta signal as the input for x(t). I should be able to do this after I simply divide everything by 2, and then my output yield will be the impulse response h(t).

Is there another way that might yield the impulse response more quickly if my method is correct? If not, what would you suggest as an approach to handling the O.D.E. structure?

Thanks!

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Given the impulse location $a$, the quickest way is produce $y$ such that $$y''+6y'+8=2\delta_a \tag{1}$$ is to (a) Solve the homogeneous equation; (b) Choose constants in its general solution differently to the left and to the right of $a$, so that $y(a-)=y(a+)$ and $y'(a-)+2=y'(a+)$. The jump of $y'$ creates a delta function component in $y''$.

The homogeneous solution is $Ae^{-2t}+Be^{-4t}$; the aforementioned conditions are $$A_-e^{-2a}+B_-e^{-4a} = A_+e^{-2a}+B_+e^{-4a} $$ and $$-2A_-e^{-2a}-4B_-e^{-4a} +2 = -2A_+e^{-2a}-4B_+e^{-4a} $$ You can use this system to solve for $A_+, B_+$ thus obtaining a general solution of (1). If you need just one solution, set $A_-=B_-=0$; this corresponds to $y(t)=0$ for $t<a$.

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