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)

f (x) = 2 cos (x^{3}) - \frac{2}{x}

\frac{d}{d x} (6 x^{5} y)

x^{3} + 1

f (p) = 325 - 3 p

\int 8 x + 4 d 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)