What is inverse oflaplace (5s+31)/(s^2+6s+9) ?

The answer to inverse oflaplace (5s+31)/(s^2+6s+9) is 5e^{-3t}+e^{-3t}*16t

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inverse oflaplace (5s+31)/(s^2+6s+9)

inverse laplace

\frac{5 s + 31}{s ^{2} + 6 s + 9}

5 e^{- 3 t} + e^{- 3 t} \cdot 16 t

Solution steps

L^{- 1} {\frac{5 s + 31}{s ^{2} + 6 s + 9}}

Take the partial fraction of

\frac{5 s + 31}{s ^{2} + 6 s + 9} : \frac{5}{s + 3} + \frac{16}{( s + 3 ) ^{2}}

= L^{- 1} {\frac{5}{s + 3} + \frac{16}{( s + 3 ) ^{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 + 3}} + L^{- 1} {\frac{16}{( s + 3 ) ^{2}}}

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

L^{- 1} {\frac{16}{( s + 3 ) ^{2}}} : e^{- 3 t} \cdot 16 t

= 5 e^{- 3 t} + e^{- 3 t} \cdot 16 t

\int_{1}^{3} x^{- 2} d x

x \to 2 + lim (\frac{5}{x - 2})

x \to \infty lim (- 2)

\int y cos (x y) - 3 y^{2} + e^{x} d x

x = - 2, x = 3, y = x^{2} - 3, y = 2 x

What is inverse oflaplace (5s+31)/(s^2+6s+9) ?
The answer to inverse oflaplace (5s+31)/(s^2+6s+9) is 5e^{-3t}+e^{-3t}*16t