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

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

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

laplace transform

2 e^{- t} - \frac{3}{2} e^{- 2 t}

\frac{2}{s + 1} - \frac{3}{2 ( s + 2 )}

Solution steps

L {2 e^{- t} - \frac{3}{2} e^{- 2 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)}

= 2 L {e^{- t}} - \frac{3}{2} L {e^{- 2 t}}

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

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

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

Refine

2 \frac{1}{s + 1} - \frac{3}{2} \frac{1}{s + 2} : \frac{2}{s + 1} - \frac{3}{2 ( s + 2 )}

= \frac{2}{s + 1} - \frac{3}{2 ( s + 2 )}

\frac{d}{d x} (sec (x) - 2 cos (x))

\int 8 sin (t) d t

\frac{\partial}{\partial x} (3 x^{4} y^{5})

\frac{d}{d x} ((x^{2} + 1)^{5})

\int cos (x) - cos^{3} (x) d x

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