What is inverse oflaplace (10)/(s(s+50)) ?

The answer to inverse oflaplace (10)/(s(s+50)) is 1/5 \H(t)-1/5 e^{-50t}

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inverse oflaplace (10)/(s(s+50))

inverse laplace

\frac{10}{s ( s + 50 )}

\frac{1}{5} H (t) - \frac{1}{5} e^{- 50 t}

Solution steps

L^{- 1} {\frac{10}{s ( s + 50 )}}

Take the partial fraction of

\frac{10}{s ( s + 50 )} : \frac{1}{5 s} - \frac{1}{5 ( s + 50 )}

= L^{- 1} {\frac{1}{5 s} - \frac{1}{5 ( s + 50 )}}

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{1}{5 s}} - L^{- 1} {\frac{1}{5 ( s + 50 )}}

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

L^{- 1} {\frac{1}{5 ( s + 50 )}} : \frac{1}{5} e^{- 50 t}

= \frac{1}{5} H (t) - \frac{1}{5} e^{- 50 t}

y = 2 x, y = x

\frac{\partial}{\partial x} (sin (4 x^{2}))

\int - 3 x e^{x} d x

\int \frac{3}{x} + x d x

\int_{0}^{27} \frac{5}{x ^{\frac{2}{3}}} d x

What is inverse oflaplace (10)/(s(s+50)) ?
The answer to inverse oflaplace (10)/(s(s+50)) is 1/5 \H(t)-1/5 e^{-50t}