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)

\int \frac{t - 3}{t + 1} d t

y = x^{3} - 2 x^{2} + 2, y = 3 x^{2} + 5 x - 23

\frac{( x ^{2} + 1 ) d y}{d x} + 6 x (y - 1) = 0, y (0) = 3

t \frac{d y}{d t} = (t^{3} + 22 t^{3} \cdot y)

\int \frac{x ^{8}}{1 + x ^{18}} d x

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)