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

7 x - x^{2}, 2 x, (5, 10)

\int \frac{5 x + 3}{x ^{2} + 2 x - 3} d x

\frac{d}{d x} (2 x^{3} + 5 x)

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

(- 3) (2.1)

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