What is inverse oflaplace 1/(s^2+4s+20) ?

The answer to inverse oflaplace 1/(s^2+4s+20) is e^{-2t} 1/4 sin(4t)

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inverse oflaplace 1/(s^2+4s+20)

inverse laplace

\frac{1}{s ^{2} + 4 s + 20}

e^{- 2 t} \frac{1}{4} sin (4 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 4 s + 20}}

\frac{1}{s ^{2} + 4 s + 20} = \frac{1}{( s + 2 ) ^{2} + 16}

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

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

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

= e^{- 2 t} \frac{1}{4} sin (4 t)

y^{^{'}} = 1 0^{- 21} \cdot y^{3}

\int_{- 1}^{0} \frac{x ^{3}}{x ^{2} - 2 x + 1} d x

\int \frac{1}{( 6 - y ^{2} ) ^{\frac{3}{2}}} d y

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

\int \frac{1}{x ^{2} + \frac{3}{4}} d x

What is inverse oflaplace 1/(s^2+4s+20) ?
The answer to inverse oflaplace 1/(s^2+4s+20) is e^{-2t} 1/4 sin(4t)