What is inverse oflaplace 5/(s(4s^2+4s+1)) ?

The answer to inverse oflaplace 5/(s(4s^2+4s+1)) is 5\H(t)-5e^{-t/2}-(5e^{-t/2}t)/2

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inverse oflaplace 5/(s(4s^2+4s+1))

inverse laplace

\frac{5}{s ( 4 s ^{2} + 4 s + 1 )}

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

Solution steps

L^{- 1} {\frac{5}{s ( 4 s ^{2} + 4 s + 1 )}}

Take the partial fraction of

\frac{5}{s ( 4 s ^{2} + 4 s + 1 )} : \frac{5}{s} - \frac{10}{2 s + 1} - \frac{10}{( 2 s + 1 ) ^{2}}

= L^{- 1} {\frac{5}{s} - \frac{10}{2 s + 1} - \frac{10}{( 2 s + 1 ) ^{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{5}{s}} - L^{- 1} {\frac{10}{2 s + 1}} - L^{- 1} {\frac{10}{( 2 s + 1 ) ^{2}}}

L^{- 1} {\frac{5}{s}} : 5 H (t)

L^{- 1} {\frac{10}{2 s + 1}} : 5 e^{- \frac{t}{2}}

L^{- 1} {\frac{10}{( 2 s + 1 ) ^{2}}} : \frac{5 e ^{- \frac{t}{2}} t}{2}

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

x \to 3 lim (- x 6 x - 8)

\int_{- \infty}^{1} 3^{x} d x

\frac{d}{d x} (3 π^{8})

f (x) = - 4 x^{\frac{1}{2}} - 10, at x = 9

8 x + 7 cos (x)

What is inverse oflaplace 5/(s(4s^2+4s+1)) ?
The answer to inverse oflaplace 5/(s(4s^2+4s+1)) is 5\H(t)-5e^{-t/2}-(5e^{-t/2}t)/2