What is inverse oflaplace (10)/(s*(s+5)^2) ?

The answer to inverse oflaplace (10)/(s*(s+5)^2) is 2/5 \H(t)-2/5 e^{-5t}-e^{-5t}*2t

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inverse oflaplace (10)/(s*(s+5)^2)

inverse laplace

\frac{10}{s \cdot ( s + 5 ) ^{2}}

\frac{2}{5} H (t) - \frac{2}{5} e^{- 5 t} - e^{- 5 t} \cdot 2 t

Solution steps

L^{- 1} {\frac{10}{s ( s + 5 ) ^{2}}}

Take the partial fraction of

\frac{10}{s ( s + 5 ) ^{2}} : \frac{2}{5 s} - \frac{2}{5 ( s + 5 )} - \frac{2}{( s + 5 ) ^{2}}

= L^{- 1} {\frac{2}{5 s} - \frac{2}{5 ( s + 5 )} - \frac{2}{( s + 5 ) ^{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{2}{5 s}} - L^{- 1} {\frac{2}{5 ( s + 5 )}} - L^{- 1} {\frac{2}{( s + 5 ) ^{2}}}

L^{- 1} {\frac{2}{5 s}} : \frac{2}{5} H (t)

L^{- 1} {\frac{2}{5 ( s + 5 )}} : \frac{2}{5} e^{- 5 t}

L^{- 1} {\frac{2}{( s + 5 ) ^{2}}} : e^{- 5 t} \cdot 2 t

= \frac{2}{5} H (t) - \frac{2}{5} e^{- 5 t} - e^{- 5 t} \cdot 2 t

\int_{0}^{4} 3 2 x d x

\frac{x ^{2} - 4 x + 3}{x ^{2}}

7 y^{^{''}} + 7 y^{^{'}} - 14 y = 0

7 (3 x^{2} ln (5 x) + x^{2})

x \to 2 + lim (8 - x^{3})

What is inverse oflaplace (10)/(s*(s+5)^2) ?
The answer to inverse oflaplace (10)/(s*(s+5)^2) is 2/5 \H(t)-2/5 e^{-5t}-e^{-5t}*2t