What is inverse oflaplace (40)/(1+25p+100p^2) ?

The answer to inverse oflaplace (40)/(1+25p+100p^2) is 8/3 e^{-t/(20)}-8/3 e^{-t/5}

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inverse oflaplace (40)/(1+25p+100p^2)

inverse laplace

\frac{40}{1 + 25 p + 100 p ^{2}}

\frac{8}{3} e^{- \frac{t}{20}} - \frac{8}{3} e^{- \frac{t}{5}}

Solution steps

L^{- 1} {\frac{40}{1 + 25 p + 100 p ^{2}}}

Take the partial fraction of

\frac{40}{1 + 25 p + 100 p ^{2}} : \frac{160}{3 ( 20 p + 1 )} - \frac{40}{3 ( 5 p + 1 )}

= L^{- 1} {\frac{160}{3 ( 20 p + 1 )} - \frac{40}{3 ( 5 p + 1 )}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{160}{3 ( 20 p + 1 )}} - L^{- 1} {\frac{40}{3 ( 5 p + 1 )}}

L^{- 1} {\frac{160}{3 ( 20 p + 1 )}} : \frac{8}{3} e^{- \frac{t}{20}}

L^{- 1} {\frac{40}{3 ( 5 p + 1 )}} : \frac{8}{3} e^{- \frac{t}{5}}

= \frac{8}{3} e^{- \frac{t}{20}} - \frac{8}{3} e^{- \frac{t}{5}}

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\int 8 cot^{4} (x) d x

\frac{d}{d x} (x^{\frac{3}{2}} - 1)

\frac{d}{d x} (4 a sec (x) tan (x))

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

What is inverse oflaplace (40)/(1+25p+100p^2) ?
The answer to inverse oflaplace (40)/(1+25p+100p^2) is 8/3 e^{-t/(20)}-8/3 e^{-t/5}