What is inverse oflaplace (10)/(s^2)-1 ?

The answer to inverse oflaplace (10)/(s^2)-1 is 10t-\delta(t)

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inverse oflaplace (10)/(s^2)-1

inverse laplace

\frac{10}{s ^{2}} - 1

10 t - δ (t)

Solution steps

L^{- 1} {\frac{10}{s ^{2}} - 1}

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{10}{s ^{2}}} - L^{- 1} {1}

L^{- 1} {\frac{10}{s ^{2}}} : 10 t

L^{- 1} {1} : δ (t)

= 10 t - δ (t)

f (x) = (2 x^{2} + 4 x)^{3}

x y^{^{'}} = y + 8 x^{2} sin (x), y (π) = 0

3 x^{2} + 4 x y + (2 y + 2 x^{2}) \frac{d y}{d x} = 0

x^{2} \frac{d y}{d x} + 2 x y - y^{3} = 0

\frac{d}{d x} (\frac{10}{x ^{7}})

What is inverse oflaplace (10)/(s^2)-1 ?
The answer to inverse oflaplace (10)/(s^2)-1 is 10t-\delta(t)