What is inverse oflaplace (6e^{-s})/(s(10s+1)) ?

The answer to inverse oflaplace (6e^{-s})/(s(10s+1)) is \H(t-1)(6-6e^{-(t-1)/(10)})

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inverse oflaplace (6e^{-s})/(s(10s+1))

inverse laplace

\frac{6 e ^{- s}}{s ( 10 s + 1 )}

H (t - 1) (6 - 6 e^{- \frac{t - 1}{10}})

Solution steps

L^{- 1} {\frac{6 e ^{- s}}{s ( 10 s + 1 )}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - 1) L^{- 1} {\frac{6}{s ( 10 s + 1 )}} (t - 1)

L^{- 1} {\frac{6}{s ( 10 s + 1 )}} : 6 H (t) - 6 e^{- \frac{t}{10}}

= H (t - 1) (6 H (t - 1) - 6 e^{- \frac{t - 1}{10}})

H (t - 1) = {01 t < 1 t \geq 1

= {0 6 - 6 e^{- \frac{t - 1}{10}} t < 1 t \geq 1

= H (t - 1) (6 - 6 e^{- \frac{t - 1}{10}})

\frac{\partial}{\partial x} (\frac{( 3 )}{( x + y ) ^{3}})

\int sec^{3} (x) tan (x) d x

x \to (- \frac{π}{2}) - lim (sec (x))

\frac{\partial}{\partial x} (6 x^{2} + 3 y^{2} + x y)

7 x + 1 (9.8)

What is inverse oflaplace (6e^{-s})/(s(10s+1)) ?
The answer to inverse oflaplace (6e^{-s})/(s(10s+1)) is \H(t-1)(6-6e^{-(t-1)/(10)})