What is laplacetransform {t+e^{2t}} ?

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

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laplacetransform {t+e^{2t}}

laplace transform

{t + e^{2 t}}

\frac{1}{s ^{2}} + \frac{1}{s - 2}

Solution steps

L {t + 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)}

= L {t} + L {e^{2 t}}

L {t} : \frac{1}{s ^{2}}

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

= \frac{1}{s ^{2}} + \frac{1}{s - 2}

x \cdot (y^{2} - 1) = y \cdot (4 - x^{2}) \cdot y^{^{'}}

\frac{\partial}{\partial y} (y^{2} e^{- 2 x})

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

x \to 0 lim (\frac{3 ^{s i n (x)} - 1}{4 x})

25 y^{^{''}} + 10 y = 0

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