What is inverse oflaplace 1/(s^2+40s+2000) ?

The answer to inverse oflaplace 1/(s^2+40s+2000) is e^{-20t} 1/40 sin(40t)

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inverse oflaplace 1/(s^2+40s+2000)

inverse laplace

\frac{1}{s ^{2} + 40 s + 2000}

e^{- 20 t} \frac{1}{40} sin (40 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 40 s + 2000}}

\frac{1}{s ^{2} + 40 s + 2000} = \frac{1}{( s + 20 ) ^{2} + 1600}

= L^{- 1} {\frac{1}{( s + 20 ) ^{2} + 1600}}

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 + 20 ) ^{2} + 1600} : a = - 20, F (s) = \frac{1}{s ^{2} + 1600}

= e^{- 20 t} L^{- 1} {\frac{1}{s ^{2} + 1600}}

L^{- 1} {\frac{1}{s ^{2} + 1600}} : \frac{1}{40} sin (40 t)

= e^{- 20 t} \frac{1}{40} sin (40 t)

8 t^{2} y^{^{''}} - 16 y = 11 t^{2} - 1

cot (3 x)

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 6

ln (1 + x^{3})

What is inverse oflaplace 1/(s^2+40s+2000) ?
The answer to inverse oflaplace 1/(s^2+40s+2000) is e^{-20t} 1/40 sin(40t)