What is laplacetransform F(t)=2+4e^{-4t}-3e^{3t} ?

The answer to laplacetransform F(t)=2+4e^{-4t}-3e^{3t} is 2/s+4/(s+4)-3/(s-3)

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laplacetransform F(t)=2+4e^{-4t}-3e^{3t}

laplace transform

F (t) = 2 + 4 e^{- 4 t} - 3 e^{3 t}

\frac{2}{s} + \frac{4}{s + 4} - \frac{3}{s - 3}

Solution steps

L {2 + 4 e^{- 4 t} - 3 e^{3 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 {2} + 4 L {e^{- 4 t}} - 3 L {e^{3 t}}

L {2} : \frac{2}{s}

L {e^{- 4 t}} : \frac{1}{s + 4}

L {e^{3 t}} : \frac{1}{s - 3}

= \frac{2}{s} + 4 \cdot \frac{1}{s + 4} - 3 \cdot \frac{1}{s - 3}

Refine

\frac{2}{s} + 4 \frac{1}{s + 4} - 3 \frac{1}{s - 3} : \frac{2}{s} + \frac{4}{s + 4} - \frac{3}{s - 3}

= \frac{2}{s} + \frac{4}{s + 4} - \frac{3}{s - 3}

\frac{\partial}{\partial x} (ln (x + \frac{y}{2 x}))

x \to \infty lim ((1 + \frac{7}{x})^{2 x})

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

x \to + 3 - lim (\frac{3}{x ^{2} - 9})

y^{3} - (14 x + 8) + 3 x y^{2} y^{^{'}} = 0

What is laplacetransform F(t)=2+4e^{-4t}-3e^{3t} ?
The answer to laplacetransform F(t)=2+4e^{-4t}-3e^{3t} is 2/s+4/(s+4)-3/(s-3)