\( L \{ \cosh 5 t \} \)

To find the Laplace transform of \( \cosh(5t) \), we can use the definition of the Laplace transform: \[ L\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt \] For \( f(t) = \cosh(5t) \), we have: \[ L\{\cosh(5t)\} = \int_0^\infty e^{-st} \cosh(5t) \, dt \] Recall that the hyperbolic cosine function can be expressed in terms of exponential functions: \[ \cosh(5t) = \frac{e^{5t} + e^{-5t}}{2} \] Thus, we can rewrite the Laplace transform as: \[ L\{\cosh(5t)\} = \int_0^\infty e^{-st} \left( \frac{e^{5t} + e^{-5t}}{2} \right) dt \] This can be separated into two integrals: \[ L\{\cosh(5t)\} = \frac{1}{2} \left( \int_0^\infty e^{(5-s)t} \, dt + \int_0^\infty e^{(-5-s)t} \, dt \right) \] Now, we can evaluate each integral. The integral \( \int_0^\infty e^{\alpha t} \, dt \) converges for \( \alpha < 0 \) and is given by: \[ \int_0^\infty e^{\alpha t} \, dt = \frac{1}{-\alpha} \] 1. For the first integral, \( \alpha = 5 - s \): \[ \int_0^\infty e^{(5-s)t} \, dt = \frac{1}{s - 5} \quad \text{(valid for } s > 5\text{)} \] 2. For the second integral, \( \alpha = -5 - s \): \[ \int_0^\infty e^{(-5-s)t} \, dt = \frac{1}{5 + s} \quad \text{(valid for } s > -5\text{)} \] Putting it all together, we have: \[ L\{\cosh(5t)\} = \frac{1}{2} \left( \frac{1}{s - 5} + \frac{1}{5 + s} \right) \] To combine these fractions, we find a common denominator: \[ L\{\cosh(5t)\} = \frac{1}{2} \left( \frac{(5+s) + (s-5)}{(s-5)(5+s)} \right) = \frac{1}{2} \left( \frac{2s}{(s-5)(s+5)} \right) \] Thus, we simplify to: \[ L\{\cosh(5t)\} = \frac{s}{(s-5)(s+5)} \] Therefore, the final result is: \[ \boxed{\frac{s}{s^2 - 25}} \]