What is inverse oflaplace (12s+5)/((s+3)^2(s+7)) ?

The answer to inverse oflaplace (12s+5)/((s+3)^2(s+7)) is 79/16 e^{-3t}-(31e^{-3t}t)/4-79/16 e^{-7t}

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inverse oflaplace (12s+5)/((s+3)^2(s+7))

inverse laplace

\frac{12 s + 5}{( s + 3 ) ^{2} ( s + 7 )}

\frac{79}{16} e^{- 3 t} - \frac{31 e ^{- 3 t} t}{4} - \frac{79}{16} e^{- 7 t}

Solution steps

L^{- 1} {\frac{12 s + 5}{( s + 3 ) ^{2} ( s + 7 )}}

Take the partial fraction of

\frac{12 s + 5}{( s + 3 ) ^{2} ( s + 7 )} : \frac{79}{16 ( s + 3 )} - \frac{31}{4 ( s + 3 ) ^{2}} - \frac{79}{16 ( s + 7 )}

= L^{- 1} {\frac{79}{16 ( s + 3 )} - \frac{31}{4 ( s + 3 ) ^{2}} - \frac{79}{16 ( s + 7 )}}

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{79}{16 ( s + 3 )}} - L^{- 1} {\frac{31}{4 ( s + 3 ) ^{2}}} - L^{- 1} {\frac{79}{16 ( s + 7 )}}

L^{- 1} {\frac{79}{16 ( s + 3 )}} : \frac{79}{16} e^{- 3 t}

L^{- 1} {\frac{31}{4 ( s + 3 ) ^{2}}} : \frac{31 e ^{- 3 t} t}{4}

L^{- 1} {\frac{79}{16 ( s + 7 )}} : \frac{79}{16} e^{- 7 t}

= \frac{79}{16} e^{- 3 t} - \frac{31 e ^{- 3 t} t}{4} - \frac{79}{16} e^{- 7 t}

\int 2 + 2 x d x

\int_{2}^{3} \frac{- 5}{x ^{2} - 1} d x

\frac{d}{d x} (\frac{x + 6}{7 x ^{2} e ^{x}})

\int e^{- \frac{x}{5}} d x

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

What is inverse oflaplace (12s+5)/((s+3)^2(s+7)) ?
The answer to inverse oflaplace (12s+5)/((s+3)^2(s+7)) is 79/16 e^{-3t}-(31e^{-3t}t)/4-79/16 e^{-7t}