What is inverse oflaplace 5/((s+1)(s^2+4s+3)) ?

The answer to inverse oflaplace 5/((s+1)(s^2+4s+3)) is -5/4 e^{-t}+(5e^{-t}t)/2+5/4 e^{-3t}

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inverse oflaplace 5/((s+1)(s^2+4s+3))

inverse laplace

\frac{5}{( s + 1 ) ( s ^{2} + 4 s + 3 )}

- \frac{5}{4} e^{- t} + \frac{5 e ^{- t} t}{2} + \frac{5}{4} e^{- 3 t}

Solution steps

L^{- 1} {\frac{5}{( s + 1 ) ( s ^{2} + 4 s + 3 )}}

Take the partial fraction of

\frac{5}{( s + 1 ) ( s ^{2} + 4 s + 3 )} : - \frac{5}{4 ( s + 1 )} + \frac{5}{2 ( s + 1 ) ^{2}} + \frac{5}{4 ( s + 3 )}

= L^{- 1} {- \frac{5}{4 ( s + 1 )} + \frac{5}{2 ( s + 1 ) ^{2}} + \frac{5}{4 ( 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}{4 ( s + 1 )}} + L^{- 1} {\frac{5}{2 ( s + 1 ) ^{2}}} + L^{- 1} {\frac{5}{4 ( s + 3 )}}

L^{- 1} {\frac{5}{4 ( s + 1 )}} : \frac{5}{4} e^{- t}

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

L^{- 1} {\frac{5}{4 ( s + 3 )}} : \frac{5}{4} e^{- 3 t}

= - \frac{5}{4} e^{- t} + \frac{5 e ^{- t} t}{2} + \frac{5}{4} e^{- 3 t}

\frac{d h}{d t} = - h

\int 2 x^{4} d x

x, 0, 4

6 - 5 cos (2 x)

f (x) = x x^{2} + 1

What is inverse oflaplace 5/((s+1)(s^2+4s+3)) ?
The answer to inverse oflaplace 5/((s+1)(s^2+4s+3)) is -5/4 e^{-t}+(5e^{-t}t)/2+5/4 e^{-3t}