What is inverse oflaplace (-5s+0.02)/(s^2) ?

The answer to inverse oflaplace (-5s+0.02)/(s^2) is -5+0.02t

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inverse oflaplace (-5s+0.02)/(s^2)

inverse laplace

\frac{- 5 s + 0.02}{s ^{2}}

- 5 + 0.02 t

Solution steps

L^{- 1} {\frac{- 5 s + 0.02}{s ^{2}}}

Expand

= L^{- 1} {\frac{0.02}{s ^{2}} - \frac{5 s}{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)}

= - 5 L^{- 1} {\frac{s}{s ^{2}}} + 0.02 L^{- 1} {\frac{1}{s ^{2}}}

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

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

= - 5 \cdot 1 + 0.02 t

Refine

- 51 + 0.02 t : - 5 + 0.02 t

= - 5 + 0.02 t

y^{^{'}} - 2 \cdot x \cdot y = e^{x^{2}}

n = 0 \sum \infty 4^{\frac{1}{2 ^{n}}}

y = \frac{1}{x}, y = \frac{1}{x ^{2}}, y = 4

x \to + \frac{π}{2} lim (sec^{2} (x))

3 x^{2} + \frac{16 x}{3} - 2

What is inverse oflaplace (-5s+0.02)/(s^2) ?
The answer to inverse oflaplace (-5s+0.02)/(s^2) is -5+0.02t