The Laplace Transform of sinh(bt) is definitively calculated as L{sinh(bt)} = b/(s^2 - b^2). This result is derived from the definition of the hyperbolic sine function, sinh(bt) = (e^(bt) - e^(-bt))/2, and applying the linearity of the Laplace Transform. The discussion highlights the importance of understanding the transformation of exponential functions, specifically L(e^(bt)) and L(e^(-bt)), to arrive at the correct solution without relying on transformation tables.
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first sinh(bt)=(e^bt-e^-bt))/2myusernameis said:Homework Statement
L{sinh(bt)}
Homework Equations
sinh(bt)=(e^bt-e^-bt))
not sure what you are doing here, I just went to the Tables & seemyusernameis said:The Attempt at a Solution
so the answer says it's b/(s^2-b^2)
but get get that far.. here's what i have
...
1/2L{e^bt}+1/2L{e^-bt}
= 1/2[e^((s-b)t)-e^((s+b)t)]
lanedance said:first sinh(bt)=(e^bt-e^-bt))/2
not sure what you are doing here, I just went to the Tables & see
L(e^{-\alpha t}}) = \frac{1}{s+\alpha}
can you show this? and use it?