What is laplacetransform t^2+1 ?

The answer to laplacetransform t^2+1 is 2/(s^3)+1/s

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laplacetransform t^2+1

laplace transform

t^{2} + 1

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

Solution steps

L {t^{2} + 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^{2}} + L {1}

L {t^{2}} : \frac{2}{s ^{3}}

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

= \frac{2}{s ^{3}} + \frac{1}{s}

y = sec (x) - 6 cos (x)

y = x^{8} + 7 x^{6} - 5 x + 4

\frac{d}{d x} (sin^{2} (x - 1))

\int (2 x^{2} - 1) (2 x^{3} - 3 x + 9)^{\frac{1}{3}} d x

\frac{( x ^{2} )}{y ^{2} - 5} y^{^{'}} = \frac{1}{2 y}

What is laplacetransform t^2+1 ?
The answer to laplacetransform t^2+1 is 2/(s^3)+1/s