What is inverse oflaplace ((3s+1))/(s(s+2)) ?

The answer to inverse oflaplace ((3s+1))/(s(s+2)) is 1/2 \H(t)+5/2 e^{-2t}

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inverse oflaplace ((3s+1))/(s(s+2))

inverse laplace

\frac{( 3 s + 1 )}{s ( s + 2 )}

\frac{1}{2} H (t) + \frac{5}{2} e^{- 2 t}

Solution steps

L^{- 1} {\frac{( 3 s + 1 )}{s ( s + 2 )}}

Take the partial fraction of

\frac{3 s + 1}{s ( s + 2 )} : \frac{1}{2 s} + \frac{5}{2 ( s + 2 )}

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

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

L^{- 1} {\frac{1}{2 s}} : \frac{1}{2} H (t)

L^{- 1} {\frac{5}{2 ( s + 2 )}} : \frac{5}{2} e^{- 2 t}

= \frac{1}{2} H (t) + \frac{5}{2} e^{- 2 t}

y^{^{'}} - \frac{6}{x} y = \frac{( y ^{5} )}{x ^{18}}, y (1) = 1

g (x) = 3 (5 - 7 x)^{5}

\frac{d x}{d y} = 7 sin (x) + 3 cos (x)

\frac{d}{d x} (sin (\frac{π}{4}))

\int \frac{7 e ^{2 t}}{1 + e ^{4 t}} d t

What is inverse oflaplace ((3s+1))/(s(s+2)) ?
The answer to inverse oflaplace ((3s+1))/(s(s+2)) is 1/2 \H(t)+5/2 e^{-2t}