What is inverse oflaplace 1/((s-4)^2) ?

The answer to inverse oflaplace 1/((s-4)^2) is e^{4t}t

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inverse oflaplace 1/((s-4)^2)

inverse laplace

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

e^{4 t} t

Solution steps

L^{- 1} {\frac{1}{( 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{1}{( s - 4 ) ^{2}} : a = 4, F (s) = \frac{1}{s ^{2}}

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

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

= e^{4 t} t

\int \frac{4}{25 - x} d x

f (x) = 294 x^{5} - 120 x^{4} + 420 x^{2}

f (x) = 3 x^{2} + 4 x, at x = - 3

\int \frac{7}{2 y ^{4}} d y

y^{^{'}} = \frac{x ^{4} y ^{4} + 4 x ^{4}}{y ^{3}}

What is inverse oflaplace 1/((s-4)^2) ?
The answer to inverse oflaplace 1/((s-4)^2) is e^{4t}t