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

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

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inverse oflaplace [(3+s)/((s+1)^2)]

inverse laplace

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

e^{- t} + e^{- t} \cdot 2 t

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{1}{s + 1}} : e^{- t}

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

= e^{- t} + e^{- t} \cdot 2 t

1000

x \to 0 lim (\frac{8 ^{x} - 1 0 ^{x}}{x})

y = \frac{4 x}{3}

n = 1 \sum \infty (\frac{1}{2})^{n}

\frac{d y}{d x} - 4 x y = 0

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