What is inverse oflaplace ((6s+3))/(s^2) ?

The answer to inverse oflaplace ((6s+3))/(s^2) is 6+3t

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inverse oflaplace ((6s+3))/(s^2)

inverse laplace

\frac{( 6 s + 3 )}{s ^{2}}

6 + 3 t

Solution steps

L^{- 1} {\frac{( 6 s + 3 )}{s ^{2}}}

Expand

= L^{- 1} {\frac{6 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)}

= 6 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

= 6 \cdot 1 + 3 t

Refine

61 + 3 t : 6 + 3 t

= 6 + 3 t

\frac{d}{d x} ((\frac{1 + x}{1 - x})^{2})

\frac{d ^{4}}{d x ^{4}} (\frac{1}{3 x + 1})

\int 2 x^{3} + 4 x + 5 d x

x \to π lim (sin (\frac{11}{6}))

\frac{1}{8} x^{4} - \frac{32}{x ^{3}}

What is inverse oflaplace ((6s+3))/(s^2) ?
The answer to inverse oflaplace ((6s+3))/(s^2) is 6+3t