What is inverse oflaplace s/(s^2+8s+25) ?

The answer to inverse oflaplace s/(s^2+8s+25) is e^{-4t}cos(3t)-4/3 e^{-4t}sin(3t)

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inverse oflaplace s/(s^2+8s+25)

inverse laplace

\frac{s}{s ^{2} + 8 s + 25}

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

Solution steps

L^{- 1} {\frac{s}{s ^{2} + 8 s + 25}}

Expand

\frac{s}{s ^{2} + 8 s + 25} : \frac{s + 4}{( s + 4 ) ^{2} + 9} - 4 \cdot \frac{1}{( s + 4 ) ^{2} + 9}

= L^{- 1} {\frac{s + 4}{( s + 4 ) ^{2} + 9} - 4 \cdot \frac{1}{( s + 4 ) ^{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{s + 4}{( s + 4 ) ^{2} + 9}} - 4 L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 9}}

L^{- 1} {\frac{s + 4}{( s + 4 ) ^{2} + 9}} : e^{- 4 t} cos (3 t)

L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 9}} : e^{- 4 t} \frac{1}{3} sin (3 t)

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

Refine

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

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

y = \frac{2 x + 3}{3 x - 2}, (1, 5)

f (x) = \frac{x - 3}{6 x - 1}, at x = 1

y^{^{''}} + 4 y^{^{'}} + 4 y = 12 t e^{- 2 t} + 4 t

(x + 2)^{2} + (y - 3)^{2} = 37, (- 1, 9)

\int tan (4 x) sec^{2} (4 x) d x

What is inverse oflaplace s/(s^2+8s+25) ?
The answer to inverse oflaplace s/(s^2+8s+25) is e^{-4t}cos(3t)-4/3 e^{-4t}sin(3t)