What is inverse oflaplace 1/(2s^2) ?

The answer to inverse oflaplace 1/(2s^2) is t/2

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inverse oflaplace 1/(2s^2)

inverse laplace

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

\frac{t}{2}

Solution steps

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

= L^{- 1} {\frac{1}{2} \cdot \frac{1}{s ^{2}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

Use Inverse Laplace Transform table:

L^{- 1} {\frac{( n ) !}{s ^{n + 1}}} = t^{n}

L^{- 1} {\frac{1}{s ^{2}}} = t

= \frac{1}{2} t

= \frac{t}{2}

\frac{d}{d x} (\frac{1}{3} x^{6} (x^{2} - 3))

\frac{d y}{d x} + x y = e^{x} y

\frac{d}{d x} (\frac{( x ^{4} - 4 x + 3 )}{x + 1})

f (x) = x^{2} + 4, at x = 1

f (x) = x^{5} - 3 (x)^{3} + 6

What is inverse oflaplace 1/(2s^2) ?
The answer to inverse oflaplace 1/(2s^2) is t/2