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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\int_{0}^{\frac{π}{2}} 3 sin^{2} (2 x) d x

f (x) = 3 x^{2} - x

f (x) = 18

y^{^{'}} = \frac{x y ^{3}}{4}, y (0) = 6

\frac{s}{s ^{4} + 4}

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