What is inverse oflaplace (s+4)/(s^2+6s+25) ?

The answer to inverse oflaplace (s+4)/(s^2+6s+25) is e^{-3t}cos(4t)+1/4 e^{-3t}sin(4t)

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inverse oflaplace (s+4)/(s^2+6s+25)

inverse laplace

\frac{s + 4}{s ^{2} + 6 s + 25}

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

Solution steps

L^{- 1} {\frac{s + 4}{s ^{2} + 6 s + 25}}

Expand

\frac{s + 4}{s ^{2} + 6 s + 25} : \frac{s + 3}{( s + 3 ) ^{2} + 16} + 1 \cdot \frac{1}{( s + 3 ) ^{2} + 16}

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

Refine

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

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

(ln (3) + e^{x^{e}})^{e}

x \to - 1 + lim (1 - x^{2})

\int \frac{1}{( 1 + x ^{2} ) ( 1 + arctan ( x ) )} d x

x \to 7 lim (e^{x} - e^{- x} - x)

9 x^{2} + 2 x, at x = 3

What is inverse oflaplace (s+4)/(s^2+6s+25) ?
The answer to inverse oflaplace (s+4)/(s^2+6s+25) is e^{-3t}cos(4t)+1/4 e^{-3t}sin(4t)