What is laplacetransform t+29 ?

The answer to laplacetransform t+29 is 1/(s^2)+(29)/s

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laplacetransform t+29

laplace transform

t + 29

\frac{1}{s ^{2}} + \frac{29}{s}

Solution steps

L {t + 29}

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 {29}

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

L {29} : \frac{29}{s}

= \frac{1}{s ^{2}} + \frac{29}{s}

d er i v a t i v e y - 7 y

\int_{1.5}^{2} cos (\frac{π x}{3}) d x

t a y l or sin (9 x)

d er i v a t i v e x sin (x^{2})

y^{^{''}} - 4 y^{^{'}} + 3 y = - e^{- s}, y (0) = 0, y^{^{'}} (0) = 0

What is laplacetransform t+29 ?
The answer to laplacetransform t+29 is 1/(s^2)+(29)/s