What is inverse oflaplace 3/s+(16)/(s(s+1)) ?

The answer to inverse oflaplace 3/s+(16)/(s(s+1)) is 19\H(t)-16e^{-t}

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inverse oflaplace 3/s+(16)/(s(s+1))

inverse laplace

\frac{3}{s} + \frac{16}{s ( s + 1 )}

19 H (t) - 16 e^{- t}

Solution steps

L^{- 1} {\frac{3}{s} + \frac{16}{s ( s + 1 )}}

Take the partial fraction of

\frac{16}{s ( s + 1 )} : \frac{16}{s} - \frac{16}{s + 1}

= L^{- 1} {\frac{3}{s} + \frac{16}{s} - \frac{16}{s + 1}}

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}} + L^{- 1} {\frac{16}{s}} - L^{- 1} {\frac{16}{s + 1}}

L^{- 1} {\frac{3}{s}} : 3 H (t)

L^{- 1} {\frac{16}{s}} : 16 H (t)

L^{- 1} {\frac{16}{s + 1}} : 16 e^{- t}

= 3 H (t) + 16 H (t) - 16 e^{- t}

Refine

3 H (t) + 16 H (t) - 16 e^{- t} : 19 H (t) - 16 e^{- t}

= 19 H (t) - 16 e^{- t}

\int_{1}^{e^{2}} \frac{ln ( x )}{x ^{2}} d x

\frac{\partial}{\partial x} (\frac{a - b ^{2}}{x - c ^{2}})

\frac{d}{d x} (5 - \frac{4}{x})

\frac{- u ^{2} + 7}{( u ^{2} + 7 ) ^{2}}

\frac{d}{d x} (sec (x^{5} + x))

What is inverse oflaplace 3/s+(16)/(s(s+1)) ?
The answer to inverse oflaplace 3/s+(16)/(s(s+1)) is 19\H(t)-16e^{-t}