Laplace Transform of Hyperbolic Functions 📂Odinary Differential Equations

Laplace Transform of Hyperbolic Functions

Formulas¹

The Laplace transforms of hyperbolic sine and hyperbolic cosine functions are as follows.

$$ \mathcal{L} \left\{ \sinh (at) \right\} = \dfrac{a}{s^2-a^2},\quad s>|a| \\ \mathcal{L} \left\{ \cosh (at) \right\} = \dfrac{s}{s^2-a^2},\quad s>|a| $$

Description

The definition of hyperbolic functions is as follows.

$$ \sinh (ax) = \dfrac{ e^{ax} - e^{-ax} }{ 2 } \\ \cosh (ax) = \dfrac{ e^{ax} + e^{-ax} }{ 2 } $$

Derivation

Use the results of the Laplace transform of the exponential function.

$\sinh (at)$

$$ \begin{align*} \mathcal{ L } \left\{ \sinh (at) \right\} &= \int_{0}^\infty e^{-st} \sinh (at) dt \\ &= \int _{0} ^\infty e^{-st} \left( \dfrac{ e^{at} - e^{-at} }{ 2 } \right) dt \\ &= \dfrac{1}{2}\int _{0}^\infty e^{-st}e^{at} dt -\dfrac{1}{2}\int _{0}^\infty e^{-st}e^{-at} dt \\ &= \dfrac{1}{2}\int _{0}^\infty e^{-(s-a)t} dt -\dfrac{1}{2}\int _{0}^\infty e^{-(s+a)t} dt \\ &= \dfrac{1}{2}\int _{0}^\infty e^{-(s-a)t} dt -\dfrac{1}{2}\int _{0}^\infty e^{-(s+a)t} dt \\ &= \dfrac{1}{2} \dfrac{1}{s-a} - \dfrac{1}{2} \dfrac{1}{s+a} \\ &= \dfrac{1}{2} \left( \dfrac{1}{s-a} -\dfrac{1}{s+a} \right) \\ &= \dfrac{1}{2}\dfrac{2a}{s^2-a^2} \\ &= \dfrac{a}{s^2-a^2} \end{align*} $$

However, as $\lim \limits_{A \to \infty} e^{-(s-a)A}$ and $\lim \limits_{A \to \infty} e^{-(s+a)A}$ converge to $0$,

$$ s>a \quad \text{and} \quad s>-a $$

Consequently, the condition $s>|a|$ applies.

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$\cosh (at)$

Using the already calculated Laplace transform result of $\sinh (at)$,

$$ \begin{align*} &&\mathcal {L} \left\{ e^{at} \right\} &= \mathcal{ L} \left\{ \cosh (at) \right\} + \mathcal{L} \left\{ \sinh (at) \right\} \\ \implies&& \mathcal{L} \left\{ \cosh (at) \right\} &= \dfrac{1}{s-a} -\dfrac{a}{s^2-a^2} \\ && &=\dfrac{s+a}{s^2-a^2}-\dfrac{a}{s^2-a^2} \\ && &=\dfrac{s}{s^2-a^2},\quad s>|a| \end{align*} $$

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