What is inverse oflaplace s/(s^2+a^2) ?

The answer to inverse oflaplace s/(s^2+a^2) is cos(tsqrt(a^2))

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inverse oflaplace s/(s^2+a^2)

inverse laplace

\frac{s}{s ^{2} + a ^{2}}

cos (t a^{2})

Solution steps

L^{- 1} {\frac{s}{s ^{2} + a ^{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} + a ^{2}} : F (s) = \frac{1}{s ^{2} + a ^{2}}

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

L^{- 1} {\frac{1}{s ^{2} + a ^{2}}} : \frac{sin ( t a ^{2} )}{a ^{2}}

\frac{d}{d t} \frac{sin ( t a ^{2} )}{a ^{2}} = cos (t a^{2})

Evaluate

\frac{sin ( t a ^{2} )}{a ^{2}}

at point

0 = 0

= cos (t a^{2})

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What is inverse oflaplace s/(s^2+a^2) ?
The answer to inverse oflaplace s/(s^2+a^2) is cos(tsqrt(a^2))