What is laplacetransform xe^{-x} ?

The answer to laplacetransform xe^{-x} is 1/((s+1)^2)

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laplacetransform xe^{-x}

laplace transform

x e^{- x}

\frac{1}{( s + 1 ) ^{2}}

Solution steps

L {e^{- x} x}

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

x e^{- x} : f (x) = e^{- x}, k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- x}})

L {e^{- x}} : \frac{1}{s + 1}

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

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

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

\int_{0}^{\frac{π}{21}} cos (7 x) d x

d er i v a t i v e \frac{141}{7 - 2}

t an g e n t f (x) = 4 x^{2} + 2 x, at x = - 2

in v erse l a pl a ce \frac{1}{s ^{2} + 6 s + 8}

in v erse l a pl a ce \frac{s}{( s ^{2} + 2 s + 2 ) ( s ^{2} + 4 )}

What is laplacetransform xe^{-x} ?
The answer to laplacetransform xe^{-x} is 1/((s+1)^2)