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jeepey · Jan 7, 2008, 09:49 AM

Hello
Can anyone please help me and tell me if I have done the inverse laplace transform on this equation correctly?

See attached word document

galactus · Jan 7, 2008, 02:12 PM

$\frac{1}{5s}+\frac{s+2}{5(s^{2}+2s+5)}$

What you do is complete the square in the denominator.

$\frac{s+2}{5(s^{2}+2s+5)}$

$=\frac{s}{5(s^{2}+2s+5)}+\frac{2}{5(s^{2}+2s+5)}$

$s^{2}+2s+5=(s+1)^{2}+4$

Also, $\frac{1}{s}=1$ , so $\frac{1}{5s}=\frac{1}{5}$

It should be:

$L^{-1}(s)=\frac{1}{5}e^{-t}cos(2t)+\frac{1}{10}e^{-t}sin(2t)+\frac{1}{5}$

n.nadia · Dec 12, 2011, 04:15 PM

y' - y = te^2t