What is inverse oflaplace (5s)/((s-2)^2) ?

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

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inverse oflaplace (5s)/((s-2)^2)

inverse laplace

\frac{5 s}{( s - 2 ) ^{2}}

5 e^{2 t} + e^{2 t} \cdot 10 t

Solution steps

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

Take the partial fraction of

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

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

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

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

= 5 e^{2 t} + e^{2 t} \cdot 10 t

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