What is inverse oflaplace 1/((s+2)^2) ?

The answer to inverse oflaplace 1/((s+2)^2) is e^{-2t}t

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inverse oflaplace 1/((s+2)^2)

inverse laplace

\frac{1}{( s + 2 ) ^{2}}

e^{- 2 t} t

Solution steps

L^{- 1} {\frac{1}{( 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{1}{( s + 2 ) ^{2}} : a = - 2, F (s) = \frac{1}{s ^{2}}

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

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

= e^{- 2 t} t

\int \frac{sin ( x )}{cos ( x ) + 1} d x

(y - y x^{2}) \frac{d y}{d x} = (y + 1)^{2}

f (x) = 24

\frac{\partial}{\partial y} (6 x + 5 y)

(- 3 sin (x) - y sin (x) + 5 cos (x)) d x + cos (x) d y = 0

What is inverse oflaplace 1/((s+2)^2) ?
The answer to inverse oflaplace 1/((s+2)^2) is e^{-2t}t