What is inverse oflaplace (s+3)/(s^2+4s+5) ?

The answer to inverse oflaplace (s+3)/(s^2+4s+5) is e^{-2t}cos(t)+e^{-2t}sin(t)

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inverse oflaplace (s+3)/(s^2+4s+5)

inverse laplace

\frac{s + 3}{s ^{2} + 4 s + 5}

e^{- 2 t} cos (t) + e^{- 2 t} sin (t)

Solution steps

L^{- 1} {\frac{s + 3}{s ^{2} + 4 s + 5}}

Expand

\frac{s + 3}{s ^{2} + 4 s + 5} : \frac{s + 2}{( s + 2 ) ^{2} + 1} + 1 \cdot \frac{1}{( s + 2 ) ^{2} + 1}

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

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

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

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

= e^{- 2 t} cos (t) + 1 \cdot e^{- 2 t} sin (t)

Refine

e^{- 2 t} cos (t) + 1 e^{- 2 t} sin (t) : e^{- 2 t} cos (t) + e^{- 2 t} sin (t)

= e^{- 2 t} cos (t) + e^{- 2 t} sin (t)

\int \frac{11}{2 x 4 x ^{2} - 16} d x

\int \frac{( a x + b )}{( c x + d )} d x

\frac{7}{2 x ^{2}}

x \to \infty lim (arctan (x^{6} - x^{9}))

y = arctan (3 x)

What is inverse oflaplace (s+3)/(s^2+4s+5) ?
The answer to inverse oflaplace (s+3)/(s^2+4s+5) is e^{-2t}cos(t)+e^{-2t}sin(t)