What is laplacetransform 3(1+8e^{-t/2}-9e^{-t}) ?

The answer to laplacetransform 3(1+8e^{-t/2}-9e^{-t}) is 3/s+(48)/(2s+1)-(27)/(s+1)

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laplacetransform 3(1+8e^{-t/2}-9e^{-t})

laplace transform

3 (1 + 8 e^{- \frac{t}{2}} - 9 e^{- t})

\frac{3}{s} + \frac{48}{2 s + 1} - \frac{27}{s + 1}

Solution steps

L {3 (8 e^{- \frac{t}{2}} - 9 e^{- t} + 1)}

Expand

3 (1 + 8 e^{- \frac{t}{2}} - 9 e^{- t}) : 3 + 24 e^{- \frac{t}{2}} - 27 e^{- t}

= L {3 + 24 e^{- \frac{t}{2}} - 27 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)}

= L {3} + 24 L {e^{- \frac{1}{2} t}} - 27 L {e^{- t}}

L {3} : \frac{3}{s}

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

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

= \frac{3}{s} + 24 \cdot \frac{2}{2 s + 1} - 27 \cdot \frac{1}{s + 1}

Refine

\frac{3}{s} + 24 \frac{2}{2 s + 1} - 27 \frac{1}{s + 1} : \frac{3}{s} + \frac{48}{2 s + 1} - \frac{27}{s + 1}

= \frac{3}{s} + \frac{48}{2 s + 1} - \frac{27}{s + 1}

\frac{d y}{d x} = \frac{7 x}{6 y}

\frac{\partial}{\partial x} (4 x^{2} - 2 x y + y^{2} z - y^{3})

\int \frac{3 x ^{2}}{3 + x ^{2}} d x

\frac{3}{2 x + 3 y}

f (x) = 9 x - 2 x^{2}

What is laplacetransform 3(1+8e^{-t/2}-9e^{-t}) ?
The answer to laplacetransform 3(1+8e^{-t/2}-9e^{-t}) is 3/s+(48)/(2s+1)-(27)/(s+1)