What is laplacetransform 200e^{-2t}+100e^{-t}+10 ?

The answer to laplacetransform 200e^{-2t}+100e^{-t}+10 is (200}{s+2}+(100)/(s+1)+\frac{10)/s

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laplacetransform 200e^{-2t}+100e^{-t}+10

laplace transform

200 e^{- 2 t} + 100 e^{- t} + 10

\frac{200}{s + 2} + \frac{100}{s + 1} + \frac{10}{s}

Solution steps

L {200 e^{- 2 t} + 100 e^{- t} + 10}

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)}

= 200 L {e^{- 2 t}} + 100 L {e^{- t}} + L {10}

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

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

L {10} : \frac{10}{s}

= 200 \cdot \frac{1}{s + 2} + 100 \cdot \frac{1}{s + 1} + \frac{10}{s}

Refine

200 \frac{1}{s + 2} + 100 \frac{1}{s + 1} + \frac{10}{s} : \frac{200}{s + 2} + \frac{100}{s + 1} + \frac{10}{s}

= \frac{200}{s + 2} + \frac{100}{s + 1} + \frac{10}{s}

\frac{d y}{d x} + 2 x y = x

x \to 1 lim (x - 2)

\frac{d}{d y} (e^{x} cos (x y))

(- 3, - 8), (4, 6)

\int \frac{1}{x ^{2} - 2} d x

What is laplacetransform 200e^{-2t}+100e^{-t}+10 ?
The answer to laplacetransform 200e^{-2t}+100e^{-t}+10 is (200}{s+2}+(100)/(s+1)+\frac{10)/s