Question

Find the Laplace transform of the given function; a and b are real constants. f(t) = eat sinh bt Your answer should be an expression in terms of a, b and s. L{f(t)}(s) = F(s) =

Answer 1

Laplace transform of f(t) is  

$L[f(t)] = \frac{1}{s-(a+b)}-\frac{1}{s-(a-b)}$

Step-by-step explanation:

Given that $f(t) = e^{at} sinh bt$

Also, $sinhx=\frac{e^{x}-e^{-x} }{2}$

Simplyfing,

$f(t) = e^{at} \frac{e^{bt}-e^{-bt} }{2}$

$f(t) = \frac{e^{bt+at}-e^{-bt+at} }{2}$

$f(t) = \frac{e^{(a+b)t}-e^{(a-b)t} }{2}$

Now, we know

$L(e^{at}) = \frac{1}{s-a}$

Therefore,

$L[f(t)] =L[ \frac{e^{(a+b)t}}{2}- \frac{e^{(a-b)t}}{2} ]$

$L[f(t)] = \frac{1}{s-(a+b)}-\frac{1}{s-(a-b)}$