Laplace transform of sinh(at): Formula, Proof

The Laplace transform of Sinh(at) is a/(s²-a²). 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/(s²-a²).

Proof of Laplace Transform of sinh(at)

The Laplace of sinh at is a/(s2-a2).

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/(s²-a²).

Remark:

Putting a=1, the Laplace Transform of sinh t is equal to 1/(s²-1). In other words,

L{sinh t} = 1/(s²-1).

More Laplace Transforms:

L{1}	L{tn}	L{t sint}
L{sin at}	L{cos at}	L{sin2t}
L{eat}	L{tet}	L{sint/t}

FAQs