Answer to Question #221321 in Chemical Engineering for Mohsin Al Mamun

Question

Find the Laplace Transform of the following functions:

i)ft=e-tsinh 3t

                                                       ii)ft=t9e5t

Answer 1

i

$\mathrm{Apply\:transform\:rule:\quad if\:}L\left\{f\left(t\right)\right\}=F\left(s\right)\mathrm{\:then\:}\:L\left\{e^{at}f\left(t\right)\right\}=F\left(s-a\right)\\ \mathrm{For\:}e^{-t}\sinh \left(3t\right):\quad f\left(t\right)=\sinh \left(3t\right),\:\quad a=-1\\ =L\left\{\sinh \left(3t\right)\right\}\left(s+1\right)\\ =\frac{3}{s^2-9}\\ =\frac{3}{\left(s+1\right)^2-9}$

ii

$\mathrm{Use\:Laplace\:Transform\:table}:\quad \:L\left\{t^kf\left(t\right)\right\}=\left(-1\right)^k\frac{d^k}{ds^k}\left(L\left\{f\left(t\right)\right\}\right)\\ \mathrm{For\:}t^9e^{5t}:\quad f\left(t\right)=e^{5t},\:\quad \:k=9\\ =\left(-1\right)^9\frac{d^9}{ds^9}\left(L\left\{e^{5t}\right\}\right)\\ L({e^{5t}})=\frac{1}{s-5}\\ \frac{d^9}{ds^9}(\frac{1}{s-5})=-\frac{362880}{\left(s-5\right)^{10}}\\ =\left(-1\right)^9\left(-\frac{362880}{\left(s-5\right)^{10}}\right)\\ =\frac{362880}{\left(s-5\right)^{10}}$