What is inverse oflaplace (-5)/(s(6(5s+1))) ?

The answer to inverse oflaplace (-5)/(s(6(5s+1))) is -5/6 \H(t)+5/6 e^{-t/5}

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inverse oflaplace (-5)/(s(6(5s+1)))

inverse laplace

\frac{- 5}{s ( 6 ( 5 s + 1 ) )}

- \frac{5}{6} H (t) + \frac{5}{6} e^{- \frac{t}{5}}

Solution steps

L^{- 1} {\frac{- 5}{s ( 6 ( 5 s + 1 ) )}}

Take the partial fraction of

\frac{- 5}{s 6 ( 5 s + 1 )} : - \frac{5}{6 s} + \frac{25}{6 ( 5 s + 1 )}

= L^{- 1} {- \frac{5}{6 s} + \frac{25}{6 ( 5 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{5}{6 s}} + L^{- 1} {\frac{25}{6 ( 5 s + 1 )}}

L^{- 1} {\frac{5}{6 s}} : \frac{5}{6} H (t)

L^{- 1} {\frac{25}{6 ( 5 s + 1 )}} : \frac{5}{6} e^{- \frac{t}{5}}

= - \frac{5}{6} H (t) + \frac{5}{6} e^{- \frac{t}{5}}

f (x) = (7 x^{6} + 4 x^{3})^{4}

\frac{\partial}{\partial x} (2 x - y^{2})

x \to 1 lim (x^{2} - 6 x + 2)

\int \frac{( e ^{- x} )}{x} d x

\int \frac{1}{( 64 + x ^{2} ) ^{\frac{3}{2}}} d x

What is inverse oflaplace (-5)/(s(6(5s+1))) ?
The answer to inverse oflaplace (-5)/(s(6(5s+1))) is -5/6 \H(t)+5/6 e^{-t/5}