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

\int_{0}^{4} e^{2 x} d x

(a x^{2} + b x + c) \cdot e^{- \frac{x}{2}}

\int x^{5} e^{- x} d x

f (x) = \frac{5}{x}, at x = 2

X

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