What is inverse oflaplace 1/((s+3)^2+4) ?

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

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inverse oflaplace 1/((s+3)^2+4)

inverse laplace

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

e^{- 3 t} \frac{1}{2} sin (2 t)

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s + 3 ) ^{2} + 4} : a = - 3, F (s) = \frac{1}{s ^{2} + 4}

= e^{- 3 t} L^{- 1} {\frac{1}{s ^{2} + 4}}

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

= e^{- 3 t} \frac{1}{2} sin (2 t)

x^{2} y^{^{'}} - 2 x^{3} e^{- \frac{1}{x}} = y

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

e^{t^{2}} y^{^{'}} + (t + t y^{2}) = 0

(7 x^{2} + 2) (6 x + x)

y^{^{'}} = - 0.000631594 \cdot y

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