What is laplacetransform t+1 ?

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

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laplacetransform t+1

laplace transform

t + 1

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

Solution steps

L {t + 1}

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 {1}

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

L {1} : \frac{1}{s}

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

\int \frac{( 1 + cos ( 2 x ) )}{2} d x

y = \frac{2 x}{1 - cot ( x )}

\frac{\partial}{\partial x} (3 e^{x y + 9})

f (x) = \frac{3 - 2 x}{3 + 2 x}

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

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