What is inverse oflaplace 1/(4s+1) ?

The answer to inverse oflaplace 1/(4s+1) is 1/4 e^{-t/4}

inverse oflaplace 1/(4s+1)

inverse laplace

\frac{1}{4 s + 1}

\frac{1}{4} e^{- \frac{t}{4}}

Solution steps

L^{- 1} {\frac{1}{4 s + 1}}

= L^{- 1} {\frac{1}{4} \cdot \frac{1}{s + \frac{1}{4}}}

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)}

= \frac{1}{4} L^{- 1} {\frac{1}{s + \frac{1}{4}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s + \frac{1}{4}}} = e^{- \frac{t}{4}}

= \frac{1}{4} e^{- \frac{t}{4}}

y^{^{'}} + 2 y = 6, y (0) = 0

s l o p e (2.5) (1.1)

\int \frac{( x ^{2} + x + 1 )}{( x ^{2} + 1 ) ^{2}} d x

\int_{a}^{x} \frac{1}{b - a} d x

\frac{d}{d x} (\frac{cos ( 2 x ) + x ^{3}}{tan ( 2 x ) - 3 x})

What is inverse oflaplace 1/(4s+1) ?
The answer to inverse oflaplace 1/(4s+1) is 1/4 e^{-t/4}