Question

$L\left\{\cos 2tU\right(t-\pi \left)\right\}$

Short answer

The Laplace transform is$\frac{s}{s^2+4}{e}^{-\pi s}$

Step-by-step solution

The Second Translation Theorem

If$F\left(s\right)=L\left\{f\right(t\left)\right\}$and$a>0$, then$L\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)$

Apply Laplace transform

We will use an alternative form of the theorem, 7.3.2. Written below

$L\left\{g\right(t\left)U\right(t-a\left)\right\}=e^{-as}L\left\{g\right(t+a\left)\right\}$

With, $g\left(t\right)=\cos \left(2t\right),a=\pi$then

$$\begin{gathered} g(t+\pi )=\cos \left(2\right(t+\pi \left)\right) \\ =\cos (2t+2\pi ) \\ =\cos \left(2t\right)\cos \left(2\pi \right)-\sin \left(2t\right)\sin \left(2\pi \right) \\ =\cos \left(2t\right) \end{gathered}$$

Hence, by the above formula,

$$\begin{gathered} L\left\{\cos \right(2t\left)U\right(t-\pi \left)\right\}=e^{-\pi s}L\left(\cos \right(2t\left)\right\} \\ =\frac{s}{s^2+4}{e}^{-\pi s} \end{gathered}$$

Since $L\left\{\cos \right(kt\left)\right\}=\frac{s}{s^2+k^2}$

Hence, the solution is $Y\left(s\right)=\frac{s}{s^2+4}{e}^{-\pi s}$