What is inverse oflaplace 1/(s^2+2*pi*s+pi^2) ?

The answer to inverse oflaplace 1/(s^2+2*pi*s+pi^2) is e^{-pit}t

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inverse oflaplace 1/(s^2+2pis+pi^2)

inverse laplace

\frac{1}{s ^{2} + 2 \cdot π \cdot s + π ^{2}}

e^{- π t} t

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 2 π s + π ^{2}}}

\frac{1}{s ^{2} + 2 π s + π ^{2}} = \frac{1}{( s + π ) ^{2}}

= L^{- 1} {\frac{1}{( s + π ) ^{2}}}

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

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

L^{- 1} {\frac{1}{s ^{2}}} : t

= e^{- π t} t

n = 3 \sum \infty e^{5 - 6 n}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 0

tan (x) - 1 sec (x)

\int y^{2} + 1 d y

y = sin (sin (x)), at x = π

What is inverse oflaplace 1/(s^2+2*pi*s+pi^2) ?
The answer to inverse oflaplace 1/(s^2+2*pi*s+pi^2) is e^{-pit}t