What is laplacetransform 4te^{-2t} ?

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

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laplacetransform 4te^{-2t}

laplace transform

4 t e^{- 2 t}

\frac{4}{( s + 2 ) ^{2}}

Solution steps

L {4 e^{- 2 t} t}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 4 L {t e^{- 2 t}}

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

= 4 \cdot \frac{1}{( s + 2 ) ^{2}}

Refine

4 \frac{1}{( s + 2 ) ^{2}} : \frac{4}{( s + 2 ) ^{2}}

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

\frac{d}{d x} (ln (\frac{x ^{2} - 5 ^{2}}{x ^{2} + 5 ^{2}}))

\int 8 x^{13} e^{- x^{7}} d x

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

\frac{10 s}{s ^{2} + 16}

\int (sec^{2} (x) + 8) d x

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