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

The answer to inverse oflaplace (2s+3)/(s^2+2s+5) is 2e^{-t}cos(2t)+1/2 e^{-t}sin(2t)

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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)}

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

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

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

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

Refine

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

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

csc^{7} (2 t^{π} + π^{3})

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

\frac{\partial}{\partial x} (x^{\frac{1}{y}})

\int \frac{1}{5 x lo g _{7} ( x )} d x

x \to 6 - lim (f (x))

What is inverse oflaplace (2s+3)/(s^2+2s+5) ?
The answer to inverse oflaplace (2s+3)/(s^2+2s+5) is 2e^{-t}cos(2t)+1/2 e^{-t}sin(2t)