What is inverse oflaplace (10)/(s+10) ?

The answer to inverse oflaplace (10)/(s+10) is 10e^{-10t}

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inverse oflaplace (10)/(s+10)

inverse laplace

\frac{10}{s + 10}

10 e^{- 10 t}

Solution steps

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

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

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}{s + 10}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

L^{- 1} {\frac{1}{s + 10}} = e^{- 10 t}

= 10 e^{- 10 t}

\int \frac{5 x}{( 6 - x ^{2} ) ^{3}} d x

\int_{0}^{3} \frac{x}{x ^{2} + 16} d x

a (x \frac{d y}{d x} + 2 y) = x y \frac{d y}{d x}

3 x e^{x}

y^{^{'}} + \frac{y}{x} = \frac{1}{x y ^{2}}

What is inverse oflaplace (10)/(s+10) ?
The answer to inverse oflaplace (10)/(s+10) is 10e^{-10t}