What is inverse oflaplace (100)/(s(s+10)^2) ?

The answer to inverse oflaplace (100)/(s(s+10)^2) is \H(t)-e^{-10t}-e^{-10t}*10t

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inverse oflaplace (100)/(s(s+10)^2)

inverse laplace

\frac{100}{s ( s + 10 ) ^{2}}

H (t) - e^{- 10 t} - e^{- 10 t} \cdot 10 t

Solution steps

L^{- 1} {\frac{100}{s ( s + 10 ) ^{2}}}

Take the partial fraction of

\frac{100}{s ( s + 10 ) ^{2}} : \frac{1}{s} - \frac{1}{s + 10} - \frac{10}{( s + 10 ) ^{2}}

= L^{- 1} {\frac{1}{s} - \frac{1}{s + 10} - \frac{10}{( s + 10 ) ^{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{1}{s}} - L^{- 1} {\frac{1}{s + 10}} - L^{- 1} {\frac{10}{( s + 10 ) ^{2}}}

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

L^{- 1} {\frac{1}{s + 10}} : e^{- 10 t}

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

= H (t) - e^{- 10 t} - e^{- 10 t} \cdot 10 t

y = \frac{4}{x}

x \to 2 π lim (f (x))

x \to - 4 lim (+ (x^{2} + 10 x + 8))

\frac{d}{d y} (1 - e^{y})

\int (10 - (14 - 2 16 - 2 x)) d x

What is inverse oflaplace (100)/(s(s+10)^2) ?
The answer to inverse oflaplace (100)/(s(s+10)^2) is \H(t)-e^{-10t}-e^{-10t}*10t