What is inverse oflaplace 1/(s(5s+3)) ?

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

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inverse oflaplace 1/(s(5s+3))

inverse laplace

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

\frac{1}{3} H (t) - \frac{1}{3} e^{- \frac{3 t}{5}}

Solution steps

L^{- 1} {\frac{1}{s ( 5 s + 3 )}}

Take the partial fraction of

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

= L^{- 1} {\frac{1}{3 s} - \frac{5}{3 ( 5 s + 3 )}}

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}{3 s}} - L^{- 1} {\frac{5}{3 ( 5 s + 3 )}}

L^{- 1} {\frac{1}{3 s}} : \frac{1}{3} H (t)

L^{- 1} {\frac{5}{3 ( 5 s + 3 )}} : \frac{1}{3} e^{- \frac{3 t}{5}}

= \frac{1}{3} H (t) - \frac{1}{3} e^{- \frac{3 t}{5}}

\frac{d y}{d x} + x y = x^{3}

\int x^{\frac{1}{2}} (x - 3)^{2} d x

y^{^{''''}} + 10 y^{^{''}} + 9 y = 0

t \to 0 lim (3 t + 3)

\frac{x - 2}{x - 4}

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