What is inverse oflaplace 8/((s+2)^2) ?

The answer to inverse oflaplace 8/((s+2)^2) is e^{-2t}*8t

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inverse oflaplace 8/((s+2)^2)

inverse laplace

\frac{8}{( s + 2 ) ^{2}}

e^{- 2 t} \cdot 8 t

Solution steps

L^{- 1} {\frac{8}{( s + 2 ) ^{2}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{8}{( s + 2 ) ^{2}} : a = - 2, F (s) = \frac{8}{s ^{2}}

= e^{- 2 t} L^{- 1} {\frac{8}{s ^{2}}}

L^{- 1} {\frac{8}{s ^{2}}} : 8 t

= e^{- 2 t} \cdot 8 t

9 y^{^{''}} + 60 y^{^{'}} + 100 y = 0

8 (x + 2)^{\frac{3}{2}}

x^{3} e^{x} - 3 x^{2} e^{x} + 6 x e^{x} - 6 e^{x}

2 ln (+ ln (x))

\frac{\partial}{\partial x} (9 - 2 x + 4 y - x^{2} - 4 y^{2})

What is inverse oflaplace 8/((s+2)^2) ?
The answer to inverse oflaplace 8/((s+2)^2) is e^{-2t}*8t