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

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

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inverse oflaplace 1/(s^2+s^2)

inverse laplace

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

\frac{t}{2}

Solution steps

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

Factor

s^{2} + s^{2} : 2 s^{2}

= 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}

x \to 6 lim (cos (x \cdot π))

\frac{\partial}{\partial x} (\frac{I}{2 x})

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

\frac{d}{d x} ((x^{2} + 4)^{- 2})

\frac{d y}{d t} = b (1 - y) y

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