What is inverse oflaplace ((s+3))/((s^2)) ?

The answer to inverse oflaplace ((s+3))/((s^2)) is 1+3t

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inverse oflaplace ((s+3))/((s^2))

inverse laplace

\frac{( s + 3 )}{( s ^{2} )}

1 + 3 t

Solution steps

L^{- 1} {\frac{( s + 3 )}{( s ^{2} )}}

Expand

= L^{- 1} {\frac{s}{s ^{2}} + \frac{3}{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}}} + 3 L^{- 1} {\frac{1}{s ^{2}}}

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

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

= 1 + 3 t

\int cos (x) tan (x) d x

\int_{0}^{\frac{π}{3}} x tan^{2} (x) d x

\frac{d}{d x} (ln (\frac{1 + x}{1 - x}))

5 x + 4 cos (x)

y = 5 e^{4 x} cos (5 x)

What is inverse oflaplace ((s+3))/((s^2)) ?
The answer to inverse oflaplace ((s+3))/((s^2)) is 1+3t