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

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

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

inverse laplace

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

2 e^{- 2 t} - 2 e^{- 3 t} - e^{- 3 t} \cdot 2 t

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{2}{s + 3}} : 2 e^{- 3 t}

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

= 2 e^{- 2 t} - 2 e^{- 3 t} - e^{- 3 t} \cdot 2 t

\int 5 x e^{x} d x

\int_{\frac{π}{6}}^{\frac{π}{4}} tan (x) d x

\frac{1}{s ^{4} ( s ^{2} - 4 )}

y^{2} - 4 x = 4, 4 x - y = 16

x \to 1 lim (7 - 2 x - x^{2})

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