What is inverse oflaplace 1/(s^2+4s+13) ?

The answer to inverse oflaplace 1/(s^2+4s+13) is e^{-2t} 1/3 sin(3t)

Popular >

inverse oflaplace 1/(s^2+4s+13)

inverse laplace

\frac{1}{s ^{2} + 4 s + 13}

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

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 4 s + 13}}

\frac{1}{s ^{2} + 4 s + 13} = \frac{1}{( s + 2 ) ^{2} + 9}

= L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 9}}

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 + 2 ) ^{2} + 9} : a = - 2, F (s) = \frac{1}{s ^{2} + 9}

= e^{- 2 t} L^{- 1} {\frac{1}{s ^{2} + 9}}

L^{- 1} {\frac{1}{s ^{2} + 9}} : \frac{1}{3} sin (3 t)

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

\frac{d}{d x} (e^{12 x})

\int \frac{1}{2 x ^{2} + 5 x - 7} d x

\frac{\partial}{\partial y} (z e^{x yz})

x \to 4 + lim (\frac{1}{( x + 4 ) ^{3}})

\int (\frac{1}{cos ^{2} ( x )}) d x

What is inverse oflaplace 1/(s^2+4s+13) ?
The answer to inverse oflaplace 1/(s^2+4s+13) is e^{-2t} 1/3 sin(3t)