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

The answer to inverse oflaplace (5s)/(12s^2+5s-2) is 5/44 e^{t/4}+10/33 e^{-(2t)/3}

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inverse oflaplace (5s)/(12s^2+5s-2)

inverse laplace

\frac{5 s}{12 s ^{2} + 5 s - 2}

\frac{5}{44} e^{\frac{t}{4}} + \frac{10}{33} e^{- \frac{2 t}{3}}

Solution steps

L^{- 1} {\frac{5 s}{12 s ^{2} + 5 s - 2}}

Take the partial fraction of

\frac{5 s}{12 s ^{2} + 5 s - 2} : \frac{5}{11 ( 4 s - 1 )} + \frac{10}{11 ( 3 s + 2 )}

= L^{- 1} {\frac{5}{11 ( 4 s - 1 )} + \frac{10}{11 ( 3 s + 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{5}{11 ( 4 s - 1 )}} + L^{- 1} {\frac{10}{11 ( 3 s + 2 )}}

L^{- 1} {\frac{5}{11 ( 4 s - 1 )}} : \frac{5}{44} e^{\frac{t}{4}}

L^{- 1} {\frac{10}{11 ( 3 s + 2 )}} : \frac{10}{33} e^{- \frac{2 t}{3}}

= \frac{5}{44} e^{\frac{t}{4}} + \frac{10}{33} e^{- \frac{2 t}{3}}

\frac{\partial}{\partial z} (x sin (y) + z e^{x})

\int 2 e^{- 0.1 x} d x

\frac{1}{( s ^{2} + 2 s + 2 ) ( s + 1 )}

\int 5 sin (6 π t) d t

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

What is inverse oflaplace (5s)/(12s^2+5s-2) ?
The answer to inverse oflaplace (5s)/(12s^2+5s-2) is 5/44 e^{t/4}+10/33 e^{-(2t)/3}