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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

Expand

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

= L^{- 1} {\frac{1}{s} - \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)}

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

L^{- 1} {\frac{1}{s}} : H (t)

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)

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

Refine

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

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

\frac{\partial}{\partial y} (x y + y^{2} + y)

f (x) = \frac{2}{x}, at (\frac{1}{9})

\int x arctan (4 x) d x

x^{2} - 4 x + 1

\int sin (x) cos^{6} (x) d x

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