Question

Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2

Answer 1

Hello there, hope I can help!

I assume you mean
`L\left\{(ekt+e-kt)/(2)\right\}`
With that, let's begin


`(ekt+e-kt)/(2)=(ekt)/(2)+(e)/(2)-(kt)/(2) \ \textgreater \ L\left\{(ekt)/(2)-(kt)/(2)+(e)/(2)\right\}`


`\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}`

`\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b`

`L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}`

`(ek)/(2)L\left\{t\right\}+L\left\{(e)/(2)\right\}-(k)/(2)L\left\{t\right\}`


`L\left\{t\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=(1)/(s^2) \ \textgreater \ L\left\{t\right\}=(1)/(s^2)`


`L\left\{(e)/(2)\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=(a)/(s) \ \textgreater \ L\left\{(e)/(2)\right\}=((e)/(2))/(s) \ \textgreater \ (e)/(2s)`


`(ek)/(2)\cdot (1)/(s^2)+(e)/(2s)-(k)/(2)\cdot (1)/(s^2)`


`(ek)/(2)\cdot (1)/(s^2) \ \textgreater \ \mathrm{Multiply\:fractions}: (a)/(b)\cdot (c)/(d)=(a\:\cdot \:c)/(b\:\cdot \:d) \ \textgreater \ (ek\cdot \:1)/(2s^2) \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a`

`(ek)/(2s^2)`


`(k)/(2)\cdot (1)/(s^2) \ \textgreater \ \mathrm{Multiply\:fractions}: (a)/(b)\cdot (c)/(d)=(a\:\cdot \:c)/(b\:\cdot \:d) \ \textgreater \ (k\cdot \:1)/(2s^2) \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a`

`(k)/(2s^2)`


`(ek)/(2s^2)+(e)/(2s)-(k)/(2s^2)`

Hope this helps!