What is inverse oflaplace (7s+4)/(2s^2+16s+30) ?

The answer to inverse oflaplace (7s+4)/(2s^2+16s+30) is -17/4 e^{-3t}+31/4 e^{-5t}

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inverse oflaplace (7s+4)/(2s^2+16s+30)

inverse laplace

\frac{7 s + 4}{2 s ^{2} + 16 s + 30}

- \frac{17}{4} e^{- 3 t} + \frac{31}{4} e^{- 5 t}

Solution steps

L^{- 1} {\frac{7 s + 4}{2 s ^{2} + 16 s + 30}}

Take the partial fraction of

\frac{7 s + 4}{2 s ^{2} + 16 s + 30} : - \frac{17}{4 ( s + 3 )} + \frac{31}{4 ( s + 5 )}

= L^{- 1} {- \frac{17}{4 ( s + 3 )} + \frac{31}{4 ( s + 5 )}}

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{17}{4 ( s + 3 )}} + L^{- 1} {\frac{31}{4 ( s + 5 )}}

L^{- 1} {\frac{17}{4 ( s + 3 )}} : \frac{17}{4} e^{- 3 t}

L^{- 1} {\frac{31}{4 ( s + 5 )}} : \frac{31}{4} e^{- 5 t}

= - \frac{17}{4} e^{- 3 t} + \frac{31}{4} e^{- 5 t}

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

x \to 2 lim (- x + 2 x^{2} - 1)

\frac{10}{s ^{2} + 2 s + 10}

\frac{d}{d x} ((\frac{1}{x ^{2}})^{t a n (x)})

\frac{3}{( 4 x + 1 )}

What is inverse oflaplace (7s+4)/(2s^2+16s+30) ?
The answer to inverse oflaplace (7s+4)/(2s^2+16s+30) is -17/4 e^{-3t}+31/4 e^{-5t}