What is inverse oflaplace 1/((s^2-6s+10)) ?

The answer to inverse oflaplace 1/((s^2-6s+10)) is e^{3t}sin(t)

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inverse oflaplace 1/((s^2-6s+10))

inverse laplace

\frac{1}{( s ^{2} - 6 s + 10 )}

e^{3 t} sin (t)

Solution steps

L^{- 1} {\frac{1}{( s ^{2} - 6 s + 10 )}}

\frac{1}{s ^{2} - 6 s + 10} = \frac{1}{( s - 3 ) ^{2} + 1}

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

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

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

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

= e^{3 t} sin (t)

\frac{\partial ^{2}}{\partial x ^{2}} (e^{3 x + 4 y} cos (5 z))

\frac{2}{5 ( s + 3 ) ^{2} + 125}

\int \frac{lo g _{10} ( lo g _{10} ( x ) )}{x} d x

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

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

What is inverse oflaplace 1/((s^2-6s+10)) ?
The answer to inverse oflaplace 1/((s^2-6s+10)) is e^{3t}sin(t)