What is inverse oflaplace s/(s^2+6s+10) ?

The answer to inverse oflaplace s/(s^2+6s+10) is e^{-3t}cos(t)-3e^{-3t}sin(t)

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inverse oflaplace s/(s^2+6s+10)

inverse laplace

\frac{s}{s ^{2} + 6 s + 10}

e^{- 3 t} cos (t) - 3 e^{- 3 t} sin (t)

Solution steps

L^{- 1} {\frac{s}{s ^{2} + 6 s + 10}}

Expand

\frac{s}{s ^{2} + 6 s + 10} : \frac{s + 3}{( s + 3 ) ^{2} + 1} - 3 \cdot \frac{1}{( s + 3 ) ^{2} + 1}

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

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

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

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

= e^{- 3 t} cos (t) - 3 e^{- 3 t} sin (t)

y^{^{''}} + 7 y^{^{'}} = 0, y (0) = - 5, y^{^{'}} (0) = - 21

\int 3 x e^{x} d x

\int \frac{1}{16 e ^{- 2 x} + e ^{2 x}} d x

t \to 0 lim (\frac{sin ( k ) t}{t})

\int arccot (15 x) d x

What is inverse oflaplace s/(s^2+6s+10) ?
The answer to inverse oflaplace s/(s^2+6s+10) is e^{-3t}cos(t)-3e^{-3t}sin(t)