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

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

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})

n = 0 \sum \infty 4^{n}

\int_{- \infty}^{- 4} \frac{3}{x ^{4}} d x

\int \frac{1}{8 + 2 x ^{2}} d x

\int 11 e^{10 x} d x

\frac{d y}{d x} = \frac{( y ^{2} - 2 x y )}{( x ^{2} - 2 x y )}

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