What is inverse oflaplace 1/((s^2+6s+25)) ?

The answer to inverse oflaplace 1/((s^2+6s+25)) is e^{-3t} 1/4 sin(4t)

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inverse oflaplace 1/((s^2+6s+25))

inverse laplace

\frac{1}{( s ^{2} + 6 s + 25 )}

e^{- 3 t} \frac{1}{4} sin (4 t)

Solution steps

L^{- 1} {\frac{1}{( s ^{2} + 6 s + 25 )}}

\frac{1}{s ^{2} + 6 s + 25} = \frac{1}{( s + 3 ) ^{2} + 16}

= L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 16}}

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} + 16} : a = - 3, F (s) = \frac{1}{s ^{2} + 16}

= e^{- 3 t} L^{- 1} {\frac{1}{s ^{2} + 16}}

L^{- 1} {\frac{1}{s ^{2} + 16}} : \frac{1}{4} sin (4 t)

= e^{- 3 t} \frac{1}{4} sin (4 t)

\int 4 sin (t) d t

\int \frac{2}{7 e ^{x} + 4} d x

y^{^{''}} + 9 y = t^{2} e^{3 t} + 6

\frac{\partial}{\partial y} (x y sin (z))

x \to 0 lim (\frac{e ^{- \frac{1}{x}}}{x})

What is inverse oflaplace 1/((s^2+6s+25)) ?
The answer to inverse oflaplace 1/((s^2+6s+25)) is e^{-3t} 1/4 sin(4t)