What is laplacetransform 1/10 e^{4t}+9/10 e^{-6t} ?

The answer to laplacetransform 1/10 e^{4t}+9/10 e^{-6t} is 1/(10(s-4))+9/(10(s+6))

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laplacetransform 1/10 e^{4t}+9/10 e^{-6t}

laplace transform

\frac{1}{10} e^{4 t} + \frac{9}{10} e^{- 6 t}

\frac{1}{10 ( s - 4 )} + \frac{9}{10 ( s + 6 )}

Solution steps

L {\frac{1}{10} e^{4 t} + \frac{9}{10} e^{- 6 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)}

= \frac{1}{10} L {e^{4 t}} + \frac{9}{10} L {e^{- 6 t}}

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

L {e^{- 6 t}} : \frac{1}{s + 6}

= \frac{1}{10} \cdot \frac{1}{s - 4} + \frac{9}{10} \cdot \frac{1}{s + 6}

Refine

\frac{1}{10} \frac{1}{s - 4} + \frac{9}{10} \frac{1}{s + 6} : \frac{1}{10 ( s - 4 )} + \frac{9}{10 ( s + 6 )}

= \frac{1}{10 ( s - 4 )} + \frac{9}{10 ( s + 6 )}

f (x) = 2 x^{2} - 16 x + 35

\int (- 3 x + 4 x) d x

d y - (y - 7)^{2} d x = 0

\int e^{4 x} cos (2 x) d x

\frac{\partial}{\partial y} ((sin (m x + n y))^{2})

What is laplacetransform 1/10 e^{4t}+9/10 e^{-6t} ?
The answer to laplacetransform 1/10 e^{4t}+9/10 e^{-6t} is 1/(10(s-4))+9/(10(s+6))