What is inverse oflaplace 1/((s-5)^2) ?

The answer to inverse oflaplace 1/((s-5)^2) is e^{5t}t

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inverse oflaplace 1/((s-5)^2)

inverse laplace

\frac{1}{( s - 5 ) ^{2}}

e^{5 t} t

Solution steps

L^{- 1} {\frac{1}{( s - 5 ) ^{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 - 5 ) ^{2}} : a = 5, F (s) = \frac{1}{s ^{2}}

= e^{5 t} L^{- 1} {\frac{1}{s ^{2}}}

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

= e^{5 t} t

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

y^{^{''}} + 4 y^{^{'}} + 13 y = 0

\frac{d ^{2}}{d x ^{2}} (e^{2 x})

f (x) = 2 x - 4, 3, 6

\int (3 x^{2} - 7 x)^{4} (6 x - 7) d x

What is inverse oflaplace 1/((s-5)^2) ?
The answer to inverse oflaplace 1/((s-5)^2) is e^{5t}t