laplace

Dave

Joined Nov 17, 2003
6,960
hi does anyone know how to solve y' + y = e [-t] using laplace?? thanks!
Yes. y' = sY(s), y = Y(s) and e[-t] = 1/(s-a)

Looking at your basic Laplacian theory should give you the derivations.

Dave
 

kautilya

Joined Apr 26, 2007
39
y' = sY(s)
y = Y(s)
e[-t]

y' + y = e[-t]

Y(s){s + 1} = 1/(s+a)

Y(s) = 1/(s+a)^2
= te[-t]
 

BlackBox

Joined Apr 22, 2007
20
Sorry to contraddict, but the Laplace transform of a derivative is not that simple... remember that we are transforming between time and complex frequency, so the "memory" included in the derivative is to be found explicitly in the Laplace domain. Otherwise you are not gonna be able to solve the Cauchy problem deriving form your differential equation:

\[L(f'(t)) = s F(s) + f(t)|_{s=0^+}\]

so the correct solution of the problem (given you don't know any initial states) is:

\[sY(s) \ + \ y(0^+) \ + \ Y(s) = \frac{1}{s+1}\\
Y(s) (s +1) \ = \ \frac{1}{s+1} \ - \ y(0^+) \\
Y(s) \ = \ \frac{1}{(s+1)^2} \ -\ \frac{y(0^+)}{s+1}\]

if the y(t) function is continuous in the origin you could put \[y(0)\] instead of \[y(0^+)\]
 
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