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

The answer to laplacetransform 3e^{-2t}+e^{-t} is 3/(s+2)+1/(s+1)

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

laplace transform

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

\frac{3}{s + 2} + \frac{1}{s + 1}

Solution steps

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

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= 3 L {e^{- 2 t}} + L {e^{- t}}

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

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

= 3 \cdot \frac{1}{s + 2} + \frac{1}{s + 1}

Refine

3 \frac{1}{s + 2} + \frac{1}{s + 1} : \frac{3}{s + 2} + \frac{1}{s + 1}

= \frac{3}{s + 2} + \frac{1}{s + 1}

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

\int \frac{1}{x 1 + x} d x

\int e^{3 x + 9} d x

\frac{d}{d x} (e^{- 2 x^{3} + 2})

\frac{\partial}{\partial y} (e^{x} cos (x + y))

What is laplacetransform 3e^{-2t}+e^{-t} ?
The answer to laplacetransform 3e^{-2t}+e^{-t} is 3/(s+2)+1/(s+1)