Question

find inverse Laplace transform for s/(s^2-1)^2​

Answer 1

The inverse Laplace transformation of the function is given by $\frac{e^{-t}t}{4} +\frac{e^{t}t}{4}$

The integral transformation of any given derivative function with real number variable "t" into a complex function with variable " s" is known as the Laplace transform.

If a function has only a finite number of breaks and does not reach up to infinity anywhere, it is said to be a piecewise continuous function. Assuming that f(t) is a piecewise continuous function, f(t) is defined using the Laplace transform.

A function's Laplace transform is represented by Lf(t) or F. (s). Laplace transform aids in the solution of differential equations by reducing the differential equation to an algebraic problem.

The given function is of the form: $\frac{2}{(s^2-1)^2}$

let us consider the function defined by 'f' on 's' as

$f(s)=\frac{2}{(s^2-1)^2}$

We can reduce the fraction in the partial form.

$f(s)=\frac{2}{(s^2-1)^2}$

$or, f(s)= -\frac{1}{4(s+1)^2} + \frac{1}{4(s-1)^2}$

Now we will put the inverse Laplace transform on the function.

$L^{-1}f(s)=L^{-1}[ -\frac{1}{4(s+1)^2} + \frac{1}{4(s-1)^2}]$

Now we know that:

$L^{-1} \{ -\frac{1}{4(s+1)^2} \} = \frac{e^{-t}t}{4}$

and:

$L^{-1} \{ \frac{1}{4(s-1)^2} \} = \frac{e^{t}t}{4}$

Therefore the required solution of the inverse Laplace transform is given by : $\frac{e^{-t}t}{4} +\frac{e^{t}t}{4}$ .

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