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I don't fully understand how the output of a system can be derived from the system's differential equation and a given input.

For example:

$$y(0-) = 1 $$ $$y'(0-) = -2$$ $$ u(t) : \text{Heaviside function} $$ $$ S: y''(t) +2y'(t) +y(t) = u''(t) -2u'(t) + u(t)$$

Can someone show me a good way to find $y(t)$, given $u(t)$ and $S$?

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What you want to do is take the Laplace transform of both sides of S. The first and second derivative terms will be replaced by expressions of $Y(s)$ and the conditions at $t=0-$, so you can solve the equation for $Y(s)$ and then do the inverse Laplace transform, or more commonly, use partial fraction expansion to go back to the time domain.

See numbers 35 and 36 in the table of Laplace transforms here.

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  • $\begingroup$ Thnx a lot! Working this out for my example, I got the following: $$Y(s) = \frac{2s^2 - 3s + 1}{s(s^2 + 2s + 1)}$$ which gives me: $$y(t) = 1 - e^{-t}(6t - 1)$$ Does this seems correct to you? $\endgroup$ – NIKOOOOO Aug 4 '15 at 19:40
  • $\begingroup$ I get the same thing for $Y(s)$ except the second coefficient in the numerator is $-2$... when you plug in the initial conditions for the Heaviside function, you need to use $u(0-)$ and $u'(0-)$, not the initial conditions on $y(t)$... I think both initial conditions for a step are zero. $\endgroup$ – CMDoolittle Aug 4 '15 at 19:58
  • $\begingroup$ Indeed, both initial conditions for a step are zero, but I made a little mistake somewhere else, the second coefficient in the numerator is indeed $ -2$. This leads to the following: $ y(t) = 1 - e^{-t}(5t-1) $ I assume this is correct now? Thanks a lot for the explanation! $\endgroup$ – NIKOOOOO Aug 4 '15 at 20:10

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