What is laplacetransform 7e^{3t}-8e^{-t}+3 ?

The answer to laplacetransform 7e^{3t}-8e^{-t}+3 is 7/(s-3)-8/(s+1)+3/s

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laplacetransform 7e^{3t}-8e^{-t}+3

laplace transform

7 e^{3 t} - 8 e^{- t} + 3

\frac{7}{s - 3} - \frac{8}{s + 1} + \frac{3}{s}

Solution steps

L {7 e^{3 t} - 8 e^{- t} + 3}

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)}

= 7 L {e^{3 t}} - 8 L {e^{- t}} + L {3}

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

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

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

= 7 \cdot \frac{1}{s - 3} - 8 \cdot \frac{1}{s + 1} + \frac{3}{s}

Refine

7 \frac{1}{s - 3} - 8 \frac{1}{s + 1} + \frac{3}{s} : \frac{7}{s - 3} - \frac{8}{s + 1} + \frac{3}{s}

= \frac{7}{s - 3} - \frac{8}{s + 1} + \frac{3}{s}

\int tan^{- 3} (x) sec^{4} (x) d x

\int_{- 10}^{x} t^{2} + 11 d t

f (x) = \frac{8}{x}, at x = 1

\frac{\partial}{\partial x} (sin (\frac{5 x}{y}))

\frac{\partial}{\partial x} ((3 x y - 1) e^{- x y})

What is laplacetransform 7e^{3t}-8e^{-t}+3 ?
The answer to laplacetransform 7e^{3t}-8e^{-t}+3 is 7/(s-3)-8/(s+1)+3/s