What is inverse oflaplace 2/((s^2+40s+400)) ?

The answer to inverse oflaplace 2/((s^2+40s+400)) is 2e^{-20t}t

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inverse oflaplace 2/((s^2+40s+400))

inverse laplace

\frac{2}{( s ^{2} + 40 s + 400 )}

2 e^{- 20 t} t

Solution steps

L^{- 1} {\frac{2}{( s ^{2} + 40 s + 400 )}}

Expand

\frac{2}{s ^{2} + 40 s + 400} : 2 \cdot \frac{1}{( s + 20 ) ^{2}}

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

L^{- 1} {\frac{1}{( s + 20 ) ^{2}}} : e^{- 20 t} t

= 2 e^{- 20 t} t

2 x + 5, x^{2} + 2 x + 1, [- 2, 2]

\int 4 x^{\frac{1}{3}} + 2 x^{- \frac{2}{3}} + 9 d x

y = x^{2} - 6, y = 5 x

\frac{d}{d x} (sin (tan (sin (x))))

y = 1 - 4 x

What is inverse oflaplace 2/((s^2+40s+400)) ?
The answer to inverse oflaplace 2/((s^2+40s+400)) is 2e^{-20t}t