What is inverse oflaplace (s-3)/(s^2+10s+9) ?

The answer to inverse oflaplace (s-3)/(s^2+10s+9) is -1/2 e^{-t}+3/2 e^{-9t}

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inverse oflaplace (s-3)/(s^2+10s+9)

inverse laplace

\frac{s - 3}{s ^{2} + 10 s + 9}

- \frac{1}{2} e^{- t} + \frac{3}{2} e^{- 9 t}

Solution steps

L^{- 1} {\frac{s - 3}{s ^{2} + 10 s + 9}}

Take the partial fraction of

\frac{s - 3}{s ^{2} + 10 s + 9} : - \frac{1}{2 ( s + 1 )} + \frac{3}{2 ( s + 9 )}

= L^{- 1} {- \frac{1}{2 ( s + 1 )} + \frac{3}{2 ( s + 9 )}}

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 + 1 )}} + L^{- 1} {\frac{3}{2 ( s + 9 )}}

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

L^{- 1} {\frac{3}{2 ( s + 9 )}} : \frac{3}{2} e^{- 9 t}

= - \frac{1}{2} e^{- t} + \frac{3}{2} e^{- 9 t}

x \to 4 lim (2 x + 1)

\int \frac{1}{- u ^{2} + 1} d u

\frac{( x ^{8} )}{8 ( 1 - x ^{2} ) ^{4}}

\frac{d}{d x} (\frac{1}{( 9 + x ) ^{2}})

\frac{d}{d x} (x^{- 3} + 4 x^{- 5} - 3 x^{- 8})

What is inverse oflaplace (s-3)/(s^2+10s+9) ?
The answer to inverse oflaplace (s-3)/(s^2+10s+9) is -1/2 e^{-t}+3/2 e^{-9t}