What is inverse oflaplace 1/((s-2)^2+9) ?

The answer to inverse oflaplace 1/((s-2)^2+9) is e^{2t} 1/3 sin(3t)

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inverse oflaplace 1/((s-2)^2+9)

inverse laplace

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

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

Solution steps

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 j} (6 i + 10 j)

(x + y) d x - (x + y - 2) d y = 0

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

sin^{2} (2 x)

\int \frac{1 - x}{e ^{x}} d x

What is inverse oflaplace 1/((s-2)^2+9) ?
The answer to inverse oflaplace 1/((s-2)^2+9) is e^{2t} 1/3 sin(3t)