What is inverse oflaplace s/((s^2-k^2)) ?

The answer to inverse oflaplace s/((s^2-k^2)) is cosh(tsqrt(k^2))

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inverse oflaplace s/((s^2-k^2))

inverse laplace

\frac{s}{( s ^{2} - k ^{2} )}

cosh (t k^{2})

Solution steps

L^{- 1} {\frac{s}{( s ^{2} - k ^{2} )}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{s}{s ^{2} - k ^{2}} : F (s) = \frac{1}{s ^{2} - k ^{2}}

= \frac{d}{d t} (L^{- 1} {\frac{1}{s ^{2} - k ^{2}}}) + L^{- 1} {\frac{1}{s ^{2} - k ^{2}}} (0)

L^{- 1} {\frac{1}{s ^{2} - k ^{2}}} : \frac{sinh ( t k ^{2} )}{k ^{2}}

\frac{d}{d t} \frac{sinh ( t k ^{2} )}{k ^{2}} = cosh (t k^{2})

Evaluate

\frac{sinh ( t k ^{2} )}{k ^{2}}

at point

0 = 0

= cosh (t k^{2})

\frac{\partial}{\partial x} ((x^{3}) x^{2} + 9)

x \to 1 - lim (\frac{x ^{2} - 1}{x - 1})

\int \frac{x}{( x + 1 ) ( x + 2 ) ^{2}} d x

\frac{d y}{d x} = 2 y + 1

n = 0 \sum \infty \frac{2}{1.1 5 ^{n}}

What is inverse oflaplace s/((s^2-k^2)) ?
The answer to inverse oflaplace s/((s^2-k^2)) is cosh(tsqrt(k^2))