What is laplacetransform 8te^{2t} ?

The answer to laplacetransform 8te^{2t} is 8/((s-2)^2)

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laplacetransform 8te^{2t}

laplace transform

8 t e^{2 t}

\frac{8}{( s - 2 ) ^{2}}

Solution steps

L {8 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)}

= 8 L {t e^{2 t}}

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

= 8 \cdot \frac{1}{( s - 2 ) ^{2}}

Refine

8 \frac{1}{( s - 2 ) ^{2}} : \frac{8}{( s - 2 ) ^{2}}

= \frac{8}{( s - 2 ) ^{2}}

f (x) = \frac{5 - x e ^{x}}{x + e ^{x}}

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

y = x e^{- 2 x}

\frac{1}{2} sin^{2} (x)

x \to - 1 lim ((1 + \frac{1}{x})^{x})

What is laplacetransform 8te^{2t} ?
The answer to laplacetransform 8te^{2t} is 8/((s-2)^2)