What is inverse oflaplace (10)/((5s+1)(s+2)^2) ?

The answer to inverse oflaplace (10)/((5s+1)(s+2)^2) is 50/81 e^{-t/5}-50/81 e^{-2t}-(10e^{-2t}t)/9

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inverse oflaplace (10)/((5s+1)(s+2)^2)

inverse laplace

\frac{10}{( 5 s + 1 ) ( s + 2 ) ^{2}}

\frac{50}{81} e^{- \frac{t}{5}} - \frac{50}{81} e^{- 2 t} - \frac{10 e ^{- 2 t} t}{9}

Solution steps

L^{- 1} {\frac{10}{( 5 s + 1 ) ( s + 2 ) ^{2}}}

Take the partial fraction of

\frac{10}{( 5 s + 1 ) ( s + 2 ) ^{2}} : \frac{250}{81 ( 5 s + 1 )} - \frac{50}{81 ( s + 2 )} - \frac{10}{9 ( s + 2 ) ^{2}}

= L^{- 1} {\frac{250}{81 ( 5 s + 1 )} - \frac{50}{81 ( s + 2 )} - \frac{10}{9 ( s + 2 ) ^{2}}}

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{250}{81 ( 5 s + 1 )}} - L^{- 1} {\frac{50}{81 ( s + 2 )}} - L^{- 1} {\frac{10}{9 ( s + 2 ) ^{2}}}

L^{- 1} {\frac{250}{81 ( 5 s + 1 )}} : \frac{50}{81} e^{- \frac{t}{5}}

L^{- 1} {\frac{50}{81 ( s + 2 )}} : \frac{50}{81} e^{- 2 t}

L^{- 1} {\frac{10}{9 ( s + 2 ) ^{2}}} : \frac{10 e ^{- 2 t} t}{9}

= \frac{50}{81} e^{- \frac{t}{5}} - \frac{50}{81} e^{- 2 t} - \frac{10 e ^{- 2 t} t}{9}

4 - t^{2}

x \to 2 - lim (3 x^{2} - 12)

\frac{d y}{d x} = \frac{2 x - y}{x + 4 y}

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

\int \frac{x ^{2}}{64 - x ^{2}} d x

What is inverse oflaplace (10)/((5s+1)(s+2)^2) ?
The answer to inverse oflaplace (10)/((5s+1)(s+2)^2) is 50/81 e^{-t/5}-50/81 e^{-2t}-(10e^{-2t}t)/9