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

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

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

inverse laplace

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

2 + t

Solution steps

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

Expand

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

= 2 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

= 2 \cdot 1 + t

Refine

21 + t : 2 + t

= 2 + t

n = 0 \sum \infty n \cdot 2^{n - 1}

\frac{\partial}{\partial x} (x^{4} + y^{4} - 4 x y)

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

x \to - \infty lim (- 2)

\frac{\partial}{\partial y} (3 y cos (x))

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