What is inverse oflaplace 1/((s+3)^2+25) ?

The answer to inverse oflaplace 1/((s+3)^2+25) is e^{-3t} 1/5 sin(5t)

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inverse oflaplace 1/((s+3)^2+25)

inverse laplace

\frac{1}{( s + 3 ) ^{2} + 25}

e^{- 3 t} \frac{1}{5} sin (5 t)

Solution steps

L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 25}}

Apply inverse transform rule:

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

then

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

For

\frac{1}{( s + 3 ) ^{2} + 25} : a = - 3, F (s) = \frac{1}{s ^{2} + 25}

= e^{- 3 t} L^{- 1} {\frac{1}{s ^{2} + 25}}

L^{- 1} {\frac{1}{s ^{2} + 25}} : \frac{1}{5} sin (5 t)

= e^{- 3 t} \frac{1}{5} sin (5 t)

g (x) = (1 - x^{2})^{- x}

x \to 0 lim (2 x ln (x^{2}))

7 x y^{^{'}} - 2 y = - \frac{x ^{2}}{y ^{6}}

\int cos (t^{2}) d t

\frac{1}{x - 1}, 8

What is inverse oflaplace 1/((s+3)^2+25) ?
The answer to inverse oflaplace 1/((s+3)^2+25) is e^{-3t} 1/5 sin(5t)