What is inverse oflaplace (10)/((s+1)^2(s+3)) ?

The answer to inverse oflaplace (10)/((s+1)^2(s+3)) is -5/2 e^{-t}+e^{-t}*5t+5/2 e^{-3t}

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inverse oflaplace (10)/((s+1)^2(s+3))

inverse laplace

\frac{10}{( s + 1 ) ^{2} ( s + 3 )}

- \frac{5}{2} e^{- t} + e^{- t} \cdot 5 t + \frac{5}{2} e^{- 3 t}

Solution steps

L^{- 1} {\frac{10}{( s + 1 ) ^{2} ( s + 3 )}}

Take the partial fraction of

\frac{10}{( s + 1 ) ^{2} ( s + 3 )} : - \frac{5}{2 ( s + 1 )} + \frac{5}{( s + 1 ) ^{2}} + \frac{5}{2 ( s + 3 )}

= L^{- 1} {- \frac{5}{2 ( s + 1 )} + \frac{5}{( s + 1 ) ^{2}} + \frac{5}{2 ( s + 3 )}}

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{5}{2 ( s + 1 )}} + L^{- 1} {\frac{5}{( s + 1 ) ^{2}}} + L^{- 1} {\frac{5}{2 ( s + 3 )}}

L^{- 1} {\frac{5}{2 ( s + 1 )}} : \frac{5}{2} e^{- t}

L^{- 1} {\frac{5}{( s + 1 ) ^{2}}} : e^{- t} \cdot 5 t

L^{- 1} {\frac{5}{2 ( s + 3 )}} : \frac{5}{2} e^{- 3 t}

= - \frac{5}{2} e^{- t} + e^{- t} \cdot 5 t + \frac{5}{2} e^{- 3 t}

\int \frac{1}{169 - x ^{2}} d x

\int cos (x) ln (2) 2^{x} + 2^{x} sin (x) d x

\frac{d}{d x} ((3 x - 2)^{2} (5 x + 4))

\int (8 x + 3) (7 - x) d x

x \to - \infty lim (- 3 x + x)

What is inverse oflaplace (10)/((s+1)^2(s+3)) ?
The answer to inverse oflaplace (10)/((s+1)^2(s+3)) is -5/2 e^{-t}+e^{-t}*5t+5/2 e^{-3t}