What is inverse oflaplace s/((s+3)^2+1) ?

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

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inverse oflaplace s/((s+3)^2+1)

inverse laplace

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

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

Solution steps

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

Expand

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

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

L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 1}} : e^{- 3 t} cos (t)

L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 1}} : e^{- 3 t} sin (t)

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

\int \frac{10}{r ^{2}} d r

6 t \frac{d y}{d t} + y = t^{3}

x^{2} + 4 y^{2}

y^{^{'}} = \frac{( 3 t ^{2} - e ^{t} )}{2 y - 5}

x \to 5 lim (2 e^{x} - 2 ln (x))

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