The Laplace transform of Sinh(at) is a/(s^{2}-a^{2}). Note sinh at is the hyperbolic sine function, where t is real. In this post, we will learn how to find the Laplace of sinh(at).

The Laplace transform formula of sinh at is given below:

L{sinh at} = a/(s^{2}-a^{2}).

Table of Contents

## Proof of Laplace Transform of sinh(at)

The Laplace of sinh at is a/(s^{2}-a^{2}). |

By the definition of sine hyperbolic function, we know that

sinh(at) = $\dfrac{e^{at}-e^{-at}}{2}$

Taking Laplace transforms on both sides, we get that

L{sinh (at)} = $L\big( \dfrac{e^{at}-e^{-at}}{2} \big)$

= $\dfrac{1}{2} L(e^{at}-e^{-at})$

= $\dfrac{1}{2} \big[ L( e^{at}) – L(e^{-at}) \big]$ by the linearity property of Laplace transforms

= $\dfrac{1}{2} \big[ \dfrac{1}{s-a} – \dfrac{1}{s+a} \big]$ by the Laplace formula of exponential functions: L{e^{at}} =1/(s-a).

= $\dfrac{1}{2} \dfrac{s+a-s+a}{(s-a)(s+a)}$

= $\dfrac{1}{2} \dfrac{2a}{s^2-a^2}$

= $\dfrac{a}{s^2-a^2}$

So the Laplace transform of sinh(at) is equal to a/(s^{2}-a^{2}).

Remark:

Putting a=1, the Laplace Transform of sinh t is equal to 1/(s^{2}-1). In other words,

L{sinh t} = 1/(s^{2}-1).

More Laplace Transforms:

## FAQs

**Q1: What is the Laplace transform of sinh(at)?**

**Answer:** The Laplace transform of sinh(at) is equal to a/(s^{2}-a^{2})

**Q2: What is the Laplace transform of sinh t?**

**Answer:** The Laplace transform of sinh t is equal to 1/(s^{2}-1)