What is laplacetransform 5+e^{-3t}+te^{-4t} ?

The answer to laplacetransform 5+e^{-3t}+te^{-4t} is 5/s+1/(s+3)+1/((s+4)^2)

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laplacetransform 5+e^{-3t}+te^{-4t}

laplace transform

5 + e^{- 3 t} + t e^{- 4 t}

\frac{5}{s} + \frac{1}{s + 3} + \frac{1}{( s + 4 ) ^{2}}

Solution steps

L {5 + e^{- 3 t} + e^{- 4 t} 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 {5} + L {e^{- 3 t}} + L {t e^{- 4 t}}

L {5} : \frac{5}{s}

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

L {t e^{- 4 t}} : \frac{1}{( s + 4 ) ^{2}}

= \frac{5}{s} + \frac{1}{s + 3} + \frac{1}{( s + 4 ) ^{2}}

\int \frac{1}{( e ^{- y} + 1 )} d y

\frac{d}{d x} (e^{x^{2}} + 2 e^{x^{2}} x^{2})

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

y = \frac{1}{x ^{5}}

\int 14 d x

What is laplacetransform 5+e^{-3t}+te^{-4t} ?
The answer to laplacetransform 5+e^{-3t}+te^{-4t} is 5/s+1/(s+3)+1/((s+4)^2)