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hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1do you know the identity \(\huge L^{1}[e^{as}F(s)]=f(ta)u(ta)\) ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1so, you first need laplace inverse of 1/(2s^2+s+2) right ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1ohh...so you were able to find laplace inverse of 1/(2s^2+s+2) ??

gorica
 2 years ago
Best ResponseYou've already chosen the best response.0yes. It's this \[\frac{ 2 }{ \sqrt{15} }e ^{t/4}\sin(\frac{ \sqrt{15} }{ 4 }t)\]

gorica
 2 years ago
Best ResponseYou've already chosen the best response.0and this is where I don't know what to do next

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1yes, thats correct, now just replace, t by 't5' and add u(t5) in the end. using that formula....

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1\(\huge \frac{ 2 }{ \sqrt{15} }e ^{(t+5)/4}\sin(\frac{ \sqrt{15} }{ 4 }(t5))u(t5)\)
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