# Output of system defined by differential equation

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 $S$?

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.
• 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? – NIKOOOOO Aug 4 '15 at 19:40
• 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. – CMDoolittle Aug 4 '15 at 19:58
• 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! – NIKOOOOO Aug 4 '15 at 20:10