Laplace Transform of t cosht | Find L{tcosh at}

The Laplace Transform of t cosht is equal to (s²+1)/(s²-1)². The Laplace of tcosh t is denoted by L{t cosht}, and it is given as follows.

$\boxed{L\{t \cosh t\} = \dfrac{s^2+1}{(s^2-1)^2}}$

More generally, L{t coshat} = (s²+a²)/(s²-a²)². Let us now find the Laplace transform of tcosh(at).

Find the Laplace of t cosht

To find the Laplace transform of t cosht, we will use the multiplication by t Laplace transform formula. It says that if L{f(t)}=F(s), then

$L\{t f(t)\} = – \dfrac{d}{ds}(F(s))$ …(∗)

Step 1: Put f(t) = cosht

∴ F(s) = L{f(t)} = L{cosht} = s/(s²-1) by the Laplace of cosh(at).

Step 2: So the Laplace transform of tcosh(t) using (∗) will be

$L\{t\cosh t\} = – \dfrac{d}{ds}\left(\dfrac{s}{s^2-1}\right)$

Step 3: Now using the quotient rule of derivatives, we have

$L\{t\cosh t \}$ $= – \dfrac{(s^2-1)\frac{d}{ds}(s)-s \cdot \frac{d}{ds}(s^2-1)}{(s^2-1)^2}$

$= – \dfrac{(s^2-1)\cdot 1- s\cdot 2s}{(s^2-1)^2}$

$= – \dfrac{s^2-1-2s^2}{(s^2-1)^2}$

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

So the Laplace transform of tcosh t is equal to (s²+1)/(s²-1)².

More Laplace Transforms:

Laplace transform of sinh(at)	a/(s2-a2)
Laplace transform of cosh(at)	s/(s2-a2)
Laplace transform of sin t:	1/(s2+1)
Laplace transform of sin(t)/t:	tan-1(1/s)
Laplace transform of cos t:	s/(s2+1)
Laplace transform of t cost:	(s2-1)/(s2+1)2
Laplace transform of cos(t)/t:	Does Not Exist

What is the Laplace Transform of t cosh(at)?

Answer: L{t coshat} = (s2+a2)/(s2-a2)2.

Proof:

As L{cosh at} = s/(s²+a²), the above formula (∗) implies that

$L\{t\cosh(at)\} = – \dfrac{d}{ds}\left(\dfrac{s}{s^2-a^2}\right)$

= $- \dfrac{(s^2-a^2)\frac{d}{ds}(s)-s \frac{d}{ds}(s^2-a^2)}{(s^2-a^2)^2}$

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

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

So the Laplace transform of tcosh(at) is (s²+a²)/(s²-a²)².

Also Read:

Laplace transform of (1-sint)/t

Laplace transform of (1-cost)/t

Laplace transform of (1-e^t)/t

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