What is inverse oflaplace s/((s-3)^2) ?

The answer to inverse oflaplace s/((s-3)^2) is e^{3t}+e^{3t}*3t

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inverse oflaplace s/((s-3)^2)

inverse laplace

\frac{s}{( s - 3 ) ^{2}}

e^{3 t} + e^{3 t} \cdot 3 t

Solution steps

L^{- 1} {\frac{s}{( s - 3 ) ^{2}}}

Take the partial fraction of

\frac{s}{( s - 3 ) ^{2}} : \frac{1}{s - 3} + \frac{3}{( s - 3 ) ^{2}}

= L^{- 1} {\frac{1}{s - 3} + \frac{3}{( s - 3 ) ^{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 - 3}} + L^{- 1} {\frac{3}{( s - 3 ) ^{2}}}

L^{- 1} {\frac{1}{s - 3}} : e^{3 t}

L^{- 1} {\frac{3}{( s - 3 ) ^{2}}} : e^{3 t} \cdot 3 t

= e^{3 t} + e^{3 t} \cdot 3 t

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What is inverse oflaplace s/((s-3)^2) ?
The answer to inverse oflaplace s/((s-3)^2) is e^{3t}+e^{3t}*3t