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

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

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

inverse laplace

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

1 + t

Solution steps

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

Expand

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

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

= 1 + t

\frac{\partial}{\partial x} (8 e^{x - y})

\int 2 - x - x^{2} d x

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

\int 5 sin^{3} (x co) s^{2} x d x

\frac{d y}{d x} = \frac{y}{x} + csc (\frac{y}{x})

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