What is inverse oflaplace ((s+6))/((s^2+6s+25)) ?

The answer to inverse oflaplace ((s+6))/((s^2+6s+25)) is e^{-3t}cos(4t)+3/4 e^{-3t}sin(4t)

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inverse oflaplace ((s+6))/((s^2+6s+25))

inverse laplace

\frac{( s + 6 )}{( s ^{2} + 6 s + 25 )}

e^{- 3 t} cos (4 t) + \frac{3}{4} e^{- 3 t} sin (4 t)

Solution steps

L^{- 1} {\frac{( s + 6 )}{( s ^{2} + 6 s + 25 )}}

Expand

\frac{s + 6}{s ^{2} + 6 s + 25} : \frac{s + 3}{( s + 3 ) ^{2} + 16} + 3 \cdot \frac{1}{( s + 3 ) ^{2} + 16}

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

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

L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 16}} : e^{- 3 t} cos (4 t)

L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 16}} : e^{- 3 t} \frac{1}{4} sin (4 t)

= e^{- 3 t} cos (4 t) + 3 e^{- 3 t} \frac{1}{4} sin (4 t)

Refine

e^{- 3 t} cos (4 t) + 3 e^{- 3 t} \frac{1}{4} sin (4 t) : e^{- 3 t} cos (4 t) + \frac{3}{4} e^{- 3 t} sin (4 t)

= e^{- 3 t} cos (4 t) + \frac{3}{4} e^{- 3 t} sin (4 t)

What is inverse oflaplace ((s+6))/((s^2+6s+25)) ?
The answer to inverse oflaplace ((s+6))/((s^2+6s+25)) is e^{-3t}cos(4t)+3/4 e^{-3t}sin(4t)