So, is the answer i posted correct?

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- Nov 26th 2010, 08:23 PMBelowzero78
So, is the answer i posted correct?

- Nov 27th 2010, 12:46 AMCaptainBlack
- Nov 27th 2010, 12:53 AMCaptainBlack
You can check this yourself. There are two requirements that the solution must full fill, it must satisfy the differential equation tofether with the initial condition and be continuous at the knot (the point where the forcing function changes its' form).

You solution appears to meet these requirements.

CB - Nov 27th 2010, 07:57 AMBelowzero78
$\displaystyle f(t)=\left\{\begin{array}{cc}3 - 8e^{-t}&\mbox{ for }

0\leq t <7\\-8e^{-t} + 6e^{7-t} - 3 & \mbox{ for } t\geq7\end{array}\right.$

Why doesn't this satisfy the initial condition? I'll show my work below.

$\displaystyle \displaystyle{Y(s) = \frac{-5}{s+1} -\frac{6e^{-7s}}{s(s+1)} + \frac{3}{s(s+1)}} $

$\displaystyle \displaystyle{Y(s) = \frac{-5}{s+1} +(\frac{-6}{s} + \frac{6}{s+1})e^{-7s} + \frac{3}{s} - \frac{3}{s+1}} $

$\displaystyle \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = \mathfrak{L^{-1}}\{\frac{-5}{s+1}\}(t) +\mathfrak{L^{-1}}\{(\frac{-6}{s} + \frac{6}{s+1})e^{-7s}\}(t) + \mathfrak{L^{-1}}\{\frac{3}{s}\}(t) - \mathfrak{L^{-1}}\{\frac{3}{s+1})\}(t)} $

$\displaystyle \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = -5e^{-t} + \mathfrak{L^{-1}}\{\frac{-6}{s} + \frac{6}{s+1}\}(t - 7)\cdot U_7(t) + 3 - 3e^{-t}} $

$\displaystyle \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{-t})|_{t = t - 7} \cdot U_7(t) $

$\displaystyle \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{7 - t}) \cdot U_7(t) $

So this is what gives me the answer. I still am not sure why it doesn't satisfy the initial conditions given the fact I used it right at the start when $\displaystyle Y(0) = -5. $ - Nov 27th 2010, 08:15 AMCaptainBlack
- Nov 27th 2010, 08:16 AMBelowzero78
So, the function f(t) i posted is correct given the intervals of t?

- Nov 28th 2010, 10:26 AMCaptainBlack