Question

if f(t)= cosh at then laplace transform of

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Answer 1

Answer:

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Answer 2

Laplace Transform of $f(t) = cosh \ at$ is $L[cosh \ at] = \frac{s}{s^{2} - a^{2}}$

Step-by-step explanation:

Given: The function $f(t) = cosh \ at$

To Find: Laplace Transform of $f(t) = cosh \ at$

Solution:

• Laplace Transform of the function $f(t)$

The Laplace Transform $L[f(t)]$ of the function $f(t)$ is given by,

$L[f(t)] = \int\limits^{\infty}_0 {f(t)e^{-st}} \, dt$

• Laplace Transform of $f(t) = cosh \ at$

The Laplace Transform $L[cosh \ at]$ of the function $cosh \ at$ is,

$L[cosh \ at] = L \Big[ \frac{e^{at} + e^{-at}}{2} \Big]$

$\Rightarrow L[cosh \ at] = \frac{1}{2} \Big( L[e^{at}] + L[e^{-at}] \Big) \ \cdots \cdots (1)$

Now, considering $L[e^{at}]$ such that

$\Rightarrow L[e^{at}] = \int\limits^{\infty}_0 {e^{at}e^{-st}} \, dt$

$\Rightarrow L[e^{at}] = \int\limits^{\infty}_0 {e^{-(s-a)t}} \, dt$

$\Rightarrow L[e^{at}] = \Big[ \frac{e^{-(s-a)t}}{-(s-a)} \Big]^{\infty}_0 = \frac{1}{(s-a)} \ \cdots \cdots (2)$

Similarly, considering $L[e^{-at}]$ such that

$\Rightarrow L[e^{-at}] = \Big[ \frac{e^{(s+a)t}}{(s+a)} \Big]^{\infty}_0 = \frac{1}{(s+a)} \ \cdots \cdots (3)$

Using (2) and (3) in (1), we get;

$\Rightarrow L[cosh \ at] = \frac{1}{2} \Big( \frac{1}{s-a} + \frac{1}{s+a} \Big) = \frac{1}{2} \Big( \frac{s+a+s-a}{s^2-a^2} \Big)$

$\Rightarrow L[cosh \ at] = \Big( \frac{s}{s^2-a^2} \Big)$

Hence, Laplace Transform of $f(t) = cosh \ at$ is $L[cosh \ at] = \frac{s}{s^{2} - a^{2}}$