What is laplacetransform (t^3-3t+2)e^{-2t} ?

The answer to laplacetransform (t^3-3t+2)e^{-2t} is 6/((s+2)^4)-3/((s+2)^2)+2/(s+2)

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laplacetransform (t^3-3t+2)e^{-2t}

laplace transform

(t^{3} - 3 t + 2) e^{- 2 t}

\frac{6}{( s + 2 ) ^{4}} - \frac{3}{( s + 2 ) ^{2}} + \frac{2}{s + 2}

Solution steps

L {e^{- 2 t} (t^{3} - 3 t + 2)}

Apply transform rule:

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

then

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

For

(t^{3} - 3 t + 2) e^{- 2 t} : f (t) = (t^{3} - 3 t + 2), a = - 2

= L {(t^{3} - 3 t + 2)} (s + 2)

L {(t^{3} - 3 t + 2)} : \frac{6}{s ^{4}} - \frac{3}{s ^{2}} + \frac{2}{s}

= \frac{6}{( s + 2 ) ^{4}} - \frac{3}{( s + 2 ) ^{2}} + \frac{2}{( s + 2 )}

= \frac{6}{( s + 2 ) ^{4}} - \frac{3}{( s + 2 ) ^{2}} + \frac{2}{s + 2}

\int_{0}^{5} \frac{1}{1 + x ^{2}} d x

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

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}, y (1) = \frac{1}{4}

y^{^{''}} + y^{^{'}} = sec (x) tan (x)

What is laplacetransform (t^3-3t+2)e^{-2t} ?
The answer to laplacetransform (t^3-3t+2)e^{-2t} is 6/((s+2)^4)-3/((s+2)^2)+2/(s+2)