What is laplacetransform e^{2t}(t-1)^2 ?

The answer to laplacetransform e^{2t}(t-1)^2 is (s^2-6s+10)/((s-2)^3)

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laplacetransform e^{2t}(t-1)^2

laplace transform

e^{2 t} (t - 1)^{2}

\frac{s ^{2} - 6 s + 10}{( s - 2 ) ^{3}}

Solution steps

L {e^{2 t} (t - 1)^{2}}

Apply transform rule:

L {f (t)} = F (s)

then

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

For

e^{2 t} (t - 1)^{2} : f (t) = (t - 1)^{2}, a = 2

= L {(t - 1)^{2}} (s - 2)

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

= \frac{2}{( s - 2 ) ^{3}} - \frac{2}{( s - 2 ) ^{2}} + \frac{1}{( s - 2 )}

= \frac{2}{( s - 2 ) ^{3}} - \frac{2}{( s - 2 ) ^{2}} + \frac{1}{s - 2}

Simplify

\frac{2}{( s - 2 ) ^{3}} - \frac{2}{( s - 2 ) ^{2}} + \frac{1}{s - 2} : \frac{s ^{2} - 6 s + 10}{( s - 2 ) ^{3}}

= \frac{s ^{2} - 6 s + 10}{( s - 2 ) ^{3}}

\frac{d}{d x} ((1 - 3 x) arccoth (3 x))

y^{^{'}} + 4 x y^{3} = 0

\frac{d}{d x} (\frac{4 x ^{3}}{3} + cos (x) + 1)

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

\frac{\partial}{\partial y} (ln (yz))

What is laplacetransform e^{2t}(t-1)^2 ?
The answer to laplacetransform e^{2t}(t-1)^2 is (s^2-6s+10)/((s-2)^3)