What is inverse oflaplace 10*(1-3s)/(1+6s) ?

The answer to inverse oflaplace 10*(1-3s)/(1+6s) is 10(-1/2 \delta(t)+1/4 e^{-t/6})

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inverse oflaplace 10*(1-3s)/(1+6s)

inverse laplace

10 \cdot \frac{1 - 3 s}{1 + 6 s}

10 (- \frac{1}{2} δ (t) + \frac{1}{4} e^{- \frac{t}{6}})

Solution steps

L^{- 1} {10 \cdot \frac{- 3 s + 1}{6 s + 1}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= 10 L^{- 1} {\frac{1 - 3 s}{1 + 6 s}}

L^{- 1} {\frac{1 - 3 s}{1 + 6 s}} : - \frac{1}{2} δ (t) + \frac{1}{4} e^{- \frac{t}{6}}

= 10 (- \frac{1}{2} δ (t) + \frac{1}{4} e^{- \frac{t}{6}})

y = 4 x^{3} - 2 x, (- 1, - 2)

x^{2} y^{^{'}} = - 2 x y + (y)^{2}

f (x) = 3 x^{2}, at x = 10

\int_{1}^{10} \frac{37}{x ^{3}} d x

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

What is inverse oflaplace 10*(1-3s)/(1+6s) ?
The answer to inverse oflaplace 10*(1-3s)/(1+6s) is 10(-1/2 \delta(t)+1/4 e^{-t/6})