What is laplacetransform 6(t)e^{-2t} ?

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

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laplacetransform 6(t)e^{-2t}

laplace transform

6 (t) e^{- 2 t}

\frac{6}{( s + 2 ) ^{2}}

Solution steps

L {6 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)}

= 6 L {t e^{- 2 t}}

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

= 6 \cdot \frac{1}{( s + 2 ) ^{2}}

Refine

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

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

\frac{\partial}{\partial x} ((1 + 2 x y^{2})^{3})

y = \frac{t}{( t - 5 ) ^{2}}

\int 5 cos^{3} (6 x) d x

\frac{\partial}{\partial x} (3 x^{3} + x^{2} y - 3 y^{3})

a x^{2}

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