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

The answer to laplacetransform (e^{4t}-e^{-3t})^2 is (98)/((s-8)(s-1)(s+6))

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

laplace transform

(e^{4 t} - e^{- 3 t})^{2}

\frac{98}{( s - 8 ) ( s - 1 ) ( s + 6 )}

Solution steps

L {(e^{4 t} - e^{- 3 t})^{2}}

Expand

(e^{4 t} - e^{- 3 t})^{2} : e^{8 t} - 2 e^{t} + e^{- 6 t}

= L {e^{8 t} - 2 e^{t} + 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)}

= L {e^{8 t}} - 2 L {e^{t}} + L {e^{- 6 t}}

L {e^{8 t}} : \frac{1}{s - 8}

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

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

= \frac{1}{s - 8} - 2 \cdot \frac{1}{s - 1} + \frac{1}{s + 6}

Simplify

\frac{1}{s - 8} - 2 \frac{1}{s - 1} + \frac{1}{s + 6} : \frac{98}{( s - 8 ) ( s - 1 ) ( s + 6 )}

= \frac{98}{( s - 8 ) ( s - 1 ) ( s + 6 )}

\frac{d}{d x} ((x^{2} - 4)^{5} (3 x + 5)^{4})

\int \frac{5 x - 3}{( x + 1 ) ( x - 3 )} d x

9 x^{2}

(y^{2} - 2 x y + 6 x) d x - (x^{2} - 2 x y + 2) d y = 0

\int_{1}^{e^{2}} \frac{ln ^{2} ( x ^{2} )}{x} d x

What is laplacetransform (e^{4t}-e^{-3t})^2 ?
The answer to laplacetransform (e^{4t}-e^{-3t})^2 is (98)/((s-8)(s-1)(s+6))