What is inverse oflaplace (5s)/((s^2+4)*(s+7)) ?

The answer to inverse oflaplace (5s)/((s^2+4)*(s+7)) is 35/53 cos(2t)+(10sin(2t))/(53)-35/53 e^{-7t}

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inverse oflaplace (5s)/((s^2+4)*(s+7))

inverse laplace

\frac{5 s}{( s ^{2} + 4 ) \cdot ( s + 7 )}

\frac{35}{53} cos (2 t) + \frac{10 sin ( 2 t )}{53} - \frac{35}{53} e^{- 7 t}

Solution steps

L^{- 1} {\frac{5 s}{( s ^{2} + 4 ) ( s + 7 )}}

Take the partial fraction of

\frac{5 s}{( s ^{2} + 4 ) ( s + 7 )} : \frac{35 s + 20}{53 ( s ^{2} + 4 )} - \frac{35}{53 ( s + 7 )}

= L^{- 1} {\frac{35 s + 20}{53 ( s ^{2} + 4 )} - \frac{35}{53 ( s + 7 )}}

Expand

= L^{- 1} {\frac{35 s}{53 ( s ^{2} + 4 )} + \frac{20}{53 ( s ^{2} + 4 )} - \frac{35}{53 ( 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)}

= \frac{35}{53} L^{- 1} {\frac{s}{s ^{2} + 4}} + \frac{20}{53} L^{- 1} {\frac{1}{s ^{2} + 4}} - \frac{35}{53} L^{- 1} {\frac{1}{s + 7}}

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

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

L^{- 1} {\frac{1}{s + 7}} : e^{- 7 t}

= \frac{35}{53} cos (2 t) + \frac{20}{53} \cdot \frac{1}{2} sin (2 t) - \frac{35}{53} e^{- 7 t}

Refine

\frac{35}{53} cos (2 t) + \frac{20}{53} \frac{1}{2} sin (2 t) - \frac{35}{53} e^{- 7 t} : \frac{35}{53} cos (2 t) + \frac{10 sin ( 2 t )}{53} - \frac{35}{53} e^{- 7 t}

= \frac{35}{53} cos (2 t) + \frac{10 sin ( 2 t )}{53} - \frac{35}{53} e^{- 7 t}

f (x) = (\frac{5 x ^{2} + 6 x + 1}{3 x - 5})^{\frac{3}{2}}

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

\int \frac{6 x + 4}{3 x ^{2} + 4 x + 2} d x

\frac{\partial}{\partial x} (x \cdot ln (x + 2 y))

y^{^{'}} = \frac{4 t + y}{4 y - t}

What is inverse oflaplace (5s)/((s^2+4)*(s+7)) ?
The answer to inverse oflaplace (5s)/((s^2+4)*(s+7)) is 35/53 cos(2t)+(10sin(2t))/(53)-35/53 e^{-7t}