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

x y^{^{'}} + y = x y^{3}

n = 0 \sum \infty (\frac{3}{8})^{\frac{n}{2}}

\frac{d v}{d x} = x

y^{^{'''}} + y^{^{''}} = 0

\frac{d y}{d x} = \frac{x}{y - 2}

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