Laplace transform of cosh(at) | L{cosh t}

The Laplace transform of cosh(at) is s/(s²-a²). Here we focus on how to find the Laplace of cosh at, the hyperbolic cosine function.

The Laplace transform of cosh at is denoted by L{cosh at} and its formula is given by

L{cosh at} = s/(s²-a²).

Find Laplace Transform of cosh(at)

Answer: The Laplace transform of cosh at is s/(s2-a2).

Proof:

By the definition of cosine hyperbolic functions, we have that

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

Now, we take Laplace transforms on both sides. By doing so, we get that

L{cosh (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 which is L{e^at} =1/(s-a).

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

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

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

So the Laplace transform of cosh(at) is equal to s/(s²-a²).

Laplace Transform of cosh t

From above, L{cosh at} = s/(s²-a²).

Put a=1 in order to get the Laplace Transform of cosh t. Therefore,

L{cosh t} = s/(s²-1).

Hence, the Laplace of cosh t is equal to s/(s²-1).

More Laplace Transforms:

L{1}	L{tn}
L{sin at}	L{cos at}
L{eat}	L{tet}
L{sinh at}	L{sin2t}

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