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

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

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inverse oflaplace 5/(s+2)

inverse laplace

\frac{5}{s + 2}

5 e^{- 2 t}

Solution steps

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

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

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)}

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

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s + 2}} = e^{- 2 t}

= 5 e^{- 2 t}

x^{2 x}

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

x y^{^{'}} + y^{2} + y = 0

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

f (x) = sin^{2} (x) + cos^{2} (x)

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