Laplace transform of f(t) is
[tex]L[f(t)] = \frac{1}{s-(a+b)}-\frac{1}{s-(a-b)}[/tex]
Step-by-step explanation:
Given that [tex]f(t) = e^{at} sinh bt[/tex]
Also, [tex]sinhx=\frac{e^{x}-e^{-x} }{2}[/tex]
Simplyfing,
[tex]f(t) = e^{at} \frac{e^{bt}-e^{-bt} }{2} [/tex]
[tex]f(t) = \frac{e^{bt+at}-e^{-bt+at} }{2} [/tex]
[tex]f(t) = \frac{e^{(a+b)t}-e^{(a-b)t} }{2} [/tex]
Now, we know
[tex]L(e^{at}) = \frac{1}{s-a}[/tex]
Therefore,
[tex]L[f(t)] =L[ \frac{e^{(a+b)t}}{2}- \frac{e^{(a-b)t}}{2} ] [/tex]
[tex]L[f(t)] = \frac{1}{s-(a+b)}-\frac{1}{s-(a-b)}[/tex]
