What is inverse oflaplace (se^{-3s})/((s-4)^3) ?

The answer to inverse oflaplace (se^{-3s})/((s-4)^3) is \H(t-3)(e^{4(t-3)}(t-3)+e^{4(t-3)}*2(t-3)^2)

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inverse oflaplace (se^{-3s})/((s-4)^3)

inverse laplace

\frac{s e ^{- 3 s}}{( s - 4 ) ^{3}}

H (t - 3) (e^{4 (t - 3)} (t - 3) + e^{4 (t - 3)} \cdot 2 (t - 3)^{2})

Solution steps

L^{- 1} {\frac{e ^{- 3 s} s}{( s - 4 ) ^{3}}}

Apply inverse transform rule:

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

then

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

Where

H (t)

is Heaviside step function

= H (t - 3) L^{- 1} {\frac{s}{( s - 4 ) ^{3}}} (t - 3)

L^{- 1} {\frac{s}{( s - 4 ) ^{3}}} : e^{4 t} t + e^{4 t} \cdot 2 t^{2}

= H (t - 3) (e^{4 (t - 3)} (t - 3) + e^{4 (t - 3)} \cdot 2 (t - 3)^{2})

\frac{d}{d x} (sin (13 x))

\frac{\partial}{\partial h} (\frac{1}{3} π r^{2} h)

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

f (t) = t^{2} + 4 t

\int \frac{1}{( 1 + y ) ^{2}} d y

What is inverse oflaplace (se^{-3s})/((s-4)^3) ?
The answer to inverse oflaplace (se^{-3s})/((s-4)^3) is \H(t-3)(e^{4(t-3)}(t-3)+e^{4(t-3)}*2(t-3)^2)