What is laplacetransform 6te^{-5t}+e^{-3t} ?

The answer to laplacetransform 6te^{-5t}+e^{-3t} is 6/((s+5)^2)+1/(s+3)

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laplacetransform 6te^{-5t}+e^{-3t}

laplace transform

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

\frac{6}{( s + 5 ) ^{2}} + \frac{1}{s + 3}

Solution steps

L {6 e^{- 5 t} t + e^{- 3 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)}

= 6 L {t e^{- 5 t}} + L {e^{- 3 t}}

L {t e^{- 5 t}} : \frac{1}{( s + 5 ) ^{2}}

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

= 6 \cdot \frac{1}{( s + 5 ) ^{2}} + \frac{1}{s + 3}

Refine

6 \frac{1}{( s + 5 ) ^{2}} + \frac{1}{s + 3} : \frac{6}{( s + 5 ) ^{2}} + \frac{1}{s + 3}

= \frac{6}{( s + 5 ) ^{2}} + \frac{1}{s + 3}

\frac{d}{d x} (5 (2 - 7 x)^{4})

x^{3} + \frac{1}{x ^{2}} + 3

x + \frac{1}{x}

\int 6 e^{- t} d t

\int_{0}^{\frac{π}{6}} 9 sec^{2} (x) d x

What is laplacetransform 6te^{-5t}+e^{-3t} ?
The answer to laplacetransform 6te^{-5t}+e^{-3t} is 6/((s+5)^2)+1/(s+3)