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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

= L^{- 1} {3 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} - 2 \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)}

= 3 L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 1}} - 2 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)

= 3 e^{- 2 t} cos (t) - 2 e^{- 2 t} sin (t)

\frac{\partial}{\partial x} (\frac{x}{x ^{5} - y ^{6}})

\int \frac{cos ^{2} ( x )}{sin ( x )} d x

\frac{\partial}{\partial x} (\frac{( x + y )}{( x - y )})

e^{t} \cdot cos (t)

\frac{d}{d x} (\frac{2}{3} sin (\frac{x + 1}{x - 1}))

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