What is inverse oflaplace 2/((s-4)^2) ?

The answer to inverse oflaplace 2/((s-4)^2) is e^{4t}*2t

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inverse oflaplace 2/((s-4)^2)

inverse laplace

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

e^{4 t} \cdot 2 t

Solution steps

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

Apply inverse transform rule:

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

then

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

For

\frac{2}{( s - 4 ) ^{2}} : a = 4, F (s) = \frac{2}{s ^{2}}

= e^{4 t} L^{- 1} {\frac{2}{s ^{2}}}

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

= e^{4 t} \cdot 2 t

f (x) = \frac{7}{x}, 7, at x = 5

\int_{0}^{4} 2 π x x d x

\frac{d}{d x} (\frac{1}{1 + e ^{\frac{1}{x}}})

\int \frac{50}{( x + 1 ) ( x ^{2} + 9 ) ^{2}} d x

\frac{d}{d x} (\frac{d y}{d x}) + k^{2} y = a e^{- b x}

What is inverse oflaplace 2/((s-4)^2) ?
The answer to inverse oflaplace 2/((s-4)^2) is e^{4t}*2t