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)

f (x) = \frac{7 x}{e ^{x}}

\int \frac{6}{y ^{2} - 2 y - 8} d y

\int (cos (3 x) - sin (x)) d x

\int_{- \infty}^{0} \frac{1}{2 - 5 x} d x

y = x^{3} - 2 x + 1

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)