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

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

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

inverse laplace

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

e^{- t} + e^{- t} \cdot 3 t

Solution steps

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

Take the partial fraction of

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

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

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

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

= e^{- t} + e^{- t} \cdot 3 t

y = (1 + 3 x)^{9}, (0, 1)

\frac{6}{x - 1}

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

\frac{d}{d x} (x^{2} - x y + y^{2})

\int_{0}^{1} \frac{63}{4 y - 1} d y

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