What is laplacetransform f(t)= 3/5 e^{-t}-22/5 e^{4t} ?

The answer to laplacetransform f(t)= 3/5 e^{-t}-22/5 e^{4t} is 3/(5(s+1))-(22)/(5(s-4))

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laplacetransform f(t)= 3/5 e^{-t}-22/5 e^{4t}

laplace transform

f (t) = \frac{3}{5} e^{- t} - \frac{22}{5} e^{4 t}

\frac{3}{5 ( s + 1 )} - \frac{22}{5 ( s - 4 )}

Solution steps

L {\frac{3}{5} e^{- t} - \frac{22}{5} e^{4 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{3}{5} L {e^{- t}} - \frac{22}{5} L {e^{4 t}}

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

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

= \frac{3}{5} \cdot \frac{1}{s + 1} - \frac{22}{5} \cdot \frac{1}{s - 4}

Refine

\frac{3}{5} \frac{1}{s + 1} - \frac{22}{5} \frac{1}{s - 4} : \frac{3}{5 ( s + 1 )} - \frac{22}{5 ( s - 4 )}

= \frac{3}{5 ( s + 1 )} - \frac{22}{5 ( s - 4 )}

\int (\frac{3}{x ^{2}} - 1) d x

\frac{d y}{d x} = 2 - y

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = 2

\int \frac{1}{9 x ^{2} + 5} d x

144 y^{^{'''}} + 120 y^{^{''}} + 25 y^{^{'}} = 0

What is laplacetransform f(t)= 3/5 e^{-t}-22/5 e^{4t} ?
The answer to laplacetransform f(t)= 3/5 e^{-t}-22/5 e^{4t} is 3/(5(s+1))-(22)/(5(s-4))