What is laplacetransform y=(2-7/5)e^{3t}+7/5 e^{-2t} ?

The answer to laplacetransform y=(2-7/5)e^{3t}+7/5 e^{-2t} is 3/(5(s-3))+7/(5(s+2))

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laplacetransform y=(2-7/5)e^{3t}+7/5 e^{-2t}

laplace transform

y = (2 - \frac{7}{5}) e^{3 t} + \frac{7}{5} e^{- 2 t}

\frac{3}{5 ( s - 3 )} + \frac{7}{5 ( s + 2 )}

Solution steps

L {(- \frac{7}{5} + 2) e^{3 t} + \frac{7}{5} e^{- 2 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)}

= (2 - \frac{7}{5}) L {e^{3 t}} + \frac{7}{5} L {e^{- 2 t}}

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

L {e^{- 2 t}} : \frac{1}{s + 2}

= (2 - \frac{7}{5}) \frac{1}{s - 3} + \frac{7}{5} \cdot \frac{1}{s + 2}

Refine

(2 - \frac{7}{5}) \frac{1}{s - 3} + \frac{7}{5} \frac{1}{s + 2} : \frac{3}{5 ( s - 3 )} + \frac{7}{5 ( s + 2 )}

= \frac{3}{5 ( s - 3 )} + \frac{7}{5 ( s + 2 )}

3 sin (3 t)

\frac{d}{d x} ((x - 2)^{3} (6 - x)^{4})

\frac{d}{d x} (2^{x} sin (x))

\int cos (b + a x) d x

2 x^{2} y y^{^{'}} + 2 x y^{2} + 1 = 0

What is laplacetransform y=(2-7/5)e^{3t}+7/5 e^{-2t} ?
The answer to laplacetransform y=(2-7/5)e^{3t}+7/5 e^{-2t} is 3/(5(s-3))+7/(5(s+2))