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{d}{d x} ((x^{2} - 8 x + 8) e^{\frac{9 x}{8}})

(1 - x)^{- 1}

y = (x^{2} + 6 x + 2)^{3}

\int \frac{1}{v ^{2} + v} d v

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