What is inverse oflaplace 1/(s+42)-1/((s+3)^4) ?

The answer to inverse oflaplace 1/(s+42)-1/((s+3)^4) is e^{-42t}-(e^{-3t}t^3)/6

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inverse oflaplace 1/(s+42)-1/((s+3)^4)

inverse laplace

\frac{1}{s + 42} - \frac{1}{( s + 3 ) ^{4}}

e^{- 42 t} - \frac{e ^{- 3 t} t ^{3}}{6}

Solution steps

L^{- 1} {\frac{1}{s + 42} - \frac{1}{( s + 3 ) ^{4}}}

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 + 42}} - L^{- 1} {\frac{1}{( s + 3 ) ^{4}}}

L^{- 1} {\frac{1}{s + 42}} : e^{- 42 t}

L^{- 1} {\frac{1}{( s + 3 ) ^{4}}} : \frac{e ^{- 3 t} t ^{3}}{6}

= e^{- 42 t} - \frac{e ^{- 3 t} t ^{3}}{6}

\int - π cos (1 π) t d t

\frac{\partial}{\partial x} (- 2 e^{3 x})

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

11 - x^{2}

\int_{0}^{2} \frac{1}{1 - x ^{2}} d x

What is inverse oflaplace 1/(s+42)-1/((s+3)^4) ?
The answer to inverse oflaplace 1/(s+42)-1/((s+3)^4) is e^{-42t}-(e^{-3t}t^3)/6