What is inverse oflaplace (0.26s)/((s+1)) ?

The answer to inverse oflaplace (0.26s)/((s+1)) is 0.26\delta(t)-0.26e^{-t}

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inverse oflaplace (0.26s)/((s+1))

inverse laplace

\frac{0.26 s}{( s + 1 )}

0.26 δ (t) - 0.26 e^{- t}

Solution steps

L^{- 1} {\frac{0.26 s}{( s + 1 )}}

Take the partial fraction of

\frac{0.26 s}{s + 1} : 0.26 - \frac{0.26}{s + 1}

= L^{- 1} {0.26 - \frac{0.26}{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} {0.26} - L^{- 1} {\frac{0.26}{s + 1}}

L^{- 1} {0.26} : 0.26 δ (t)

L^{- 1} {\frac{0.26}{s + 1}} : 0.26 e^{- t}

= 0.26 δ (t) - 0.26 e^{- t}

16 - x, x + 4, 0, - 9

y = (sin (x)) (cos (x))

\int e^{- 2 x + 1} d x

\frac{\partial}{\partial x} (- 3 x y (- 9 - x - y))

\int 3 e^{- x} d x

What is inverse oflaplace (0.26s)/((s+1)) ?
The answer to inverse oflaplace (0.26s)/((s+1)) is 0.26\delta(t)-0.26e^{-t}