# The Laplace inverse of L^{-1}left[frac{s}{s^2+5^2}right] is a) cos(5t) b) sin h(5t) c) sin(5t) d) cos h(5t)

The Laplace inverse of ${L}^{-1}\left[\frac{s}{{s}^{2}+{5}^{2}}\right]$ is
$a\right)\mathrm{cos}\left(5t\right)$
$b\right)\mathrm{sin}h\left(5t\right)$
$c\right)\mathrm{sin}\left(5t\right)$
$d\right)\mathrm{cos}h\left(5t\right)$
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Step 1
To find the Laplace transformation of ${L}^{-1}\left[\frac{s}{{s}^{2}+{5}^{2}}\right]$
Solution:
let $\phi \left(s\right)$ be the function whose Laplace transformation will be $\frac{s}{{s}^{2}+{5}^{2}}$
So,
$L\left(\phi \left(s\right)\right)=\frac{s}{{s}^{2}+{5}^{2}}$
Applying inverse we get,
$\phi \left(s\right)={L}^{-1}\left[\frac{s}{{s}^{2}+{5}^{2}}\right]\dots \left(1\right)$
Step 2
since, we know that
$L\left(\mathrm{cos}as\right)=\frac{s}{{s}^{2}+{a}^{2}}$

From equation (1),
$\mathrm{cos}\left(5s\right)=\frac{s}{{s}^{2}+{5}^{2}}$
Hence, the inverse Laplace for given expression is cos5t.