What is laplacetransform 5+4e^{-2t} ?

The answer to laplacetransform 5+4e^{-2t} is 5/s+4/(s+2)

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laplacetransform 5+4e^{-2t}

laplace transform

5 + 4 e^{- 2 t}

\frac{5}{s} + \frac{4}{s + 2}

Solution steps

L {5 + 4 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)}

= L {5} + 4 L {e^{- 2 t}}

L {5} : \frac{5}{s}

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

= \frac{5}{s} + 4 \cdot \frac{1}{s + 2}

Refine

\frac{5}{s} + 4 \frac{1}{s + 2} : \frac{5}{s} + \frac{4}{s + 2}

= \frac{5}{s} + \frac{4}{s + 2}

\int (2 x) d x

\int x + y d x

\int y cos (\frac{y ^{2}}{x ^{2}}) d y

x^{2} - 43, 5 - 2 x

\int \frac{x ^{2} - 8 x - 13}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

What is laplacetransform 5+4e^{-2t} ?
The answer to laplacetransform 5+4e^{-2t} is 5/s+4/(s+2)