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

The answer to inverse oflaplace (3s+3)/((s+4)^2) is 3e^{-4t}-e^{-4t}*9t

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

inverse laplace

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

3 e^{- 4 t} - e^{- 4 t} \cdot 9 t

Solution steps

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

Take the partial fraction of

\frac{3 s + 3}{( s + 4 ) ^{2}} : \frac{3}{s + 4} - \frac{9}{( s + 4 ) ^{2}}

= L^{- 1} {\frac{3}{s + 4} - \frac{9}{( s + 4 ) ^{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{3}{s + 4}} - L^{- 1} {\frac{9}{( s + 4 ) ^{2}}}

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

L^{- 1} {\frac{9}{( s + 4 ) ^{2}}} : e^{- 4 t} \cdot 9 t

= 3 e^{- 4 t} - e^{- 4 t} \cdot 9 t

\int \frac{1}{441 - x ^{2}} d x

\frac{d}{d x} (cos (x) + x)

\frac{d}{d x} (3 x^{2} - 12 x)

\int ln (12 x) d x

2 x y y^{^{'}} = 4 x^{2} + 3 y^{2}

What is inverse oflaplace (3s+3)/((s+4)^2) ?
The answer to inverse oflaplace (3s+3)/((s+4)^2) is 3e^{-4t}-e^{-4t}*9t