What is inverse oflaplace (2s+1)/(s^2+2s+10) ?

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

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inverse oflaplace (2s+1)/(s^2+2s+10)

inverse laplace

\frac{2 s + 1}{s ^{2} + 2 s + 10}

2 e^{- t} cos (3 t) - \frac{1}{3} e^{- t} sin (3 t)

Solution steps

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

Expand

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

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

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

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

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

= 2 e^{- t} cos (3 t) - 1 \cdot e^{- t} \frac{1}{3} sin (3 t)

Refine

2 e^{- t} cos (3 t) - 1 e^{- t} \frac{1}{3} sin (3 t) : 2 e^{- t} cos (3 t) - \frac{1}{3} e^{- t} sin (3 t)

= 2 e^{- t} cos (3 t) - \frac{1}{3} e^{- t} sin (3 t)

\int \frac{8}{5 t 25 t ^{2} - 25} d t

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

\int sin^{2} (v) d v

x \to 0 lim ((\frac{5}{6})^{\frac{1}{6}})

y^{^{'}} - x yy = 2 x y

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