What is laplacetransform 1/2 e^{-4t}-1/2 e^{-6t} ?

The answer to laplacetransform 1/2 e^{-4t}-1/2 e^{-6t} is 1/((s+4)(s+6))

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laplacetransform 1/2 e^{-4t}-1/2 e^{-6t}

laplace transform

\frac{1}{2} e^{- 4 t} - \frac{1}{2} e^{- 6 t}

\frac{1}{( s + 4 ) ( s + 6 )}

Solution steps

L {\frac{1}{2} e^{- 4 t} - \frac{1}{2} 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}{2} L {e^{- 4 t}} - \frac{1}{2} L {e^{- 6 t}}

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

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

= \frac{1}{2} \cdot \frac{1}{s + 4} - \frac{1}{2} \cdot \frac{1}{s + 6}

Simplify

\frac{1}{2} \frac{1}{s + 4} - \frac{1}{2} \frac{1}{s + 6} : \frac{1}{( s + 4 ) ( s + 6 )}

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

x \to 0 - lim (sin (x))

x^{2} y^{^{''}} + 2 x y^{^{'}} - 30 y = 0, y (1) = - 5, y^{^{'}} (1) = - 6

y^{^{''}} + 3 y = x

\int t ln (t + 3) d t

\int \frac{1}{b} sin (b x) d x

What is laplacetransform 1/2 e^{-4t}-1/2 e^{-6t} ?
The answer to laplacetransform 1/2 e^{-4t}-1/2 e^{-6t} is 1/((s+4)(s+6))