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Popular >

inverse oflaplace 1/((s^2-2s+5))

Solution

inverse laplace

\frac{1}{( s ^{2} - 2 s + 5 )}

Solution

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

Solution steps

L^{- 1} {\frac{1}{( s ^{2} - 2 s + 5 )}}

\frac{1}{s ^{2} - 2 s + 5} = \frac{1}{( s - 1 ) ^{2} + 4}

= L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 4}}

Apply inverse transform rule:

if

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s - 1 ) ^{2} + 4} : a = 1, F (s) = \frac{1}{s ^{2} + 4}

= e^{t} L^{- 1} {\frac{1}{s ^{2} + 4}}

L^{- 1} {\frac{1}{s ^{2} + 4}} : \frac{1}{2} sin (2 t)

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

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\int (3^{- x}) d x

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\frac{d y}{d x}, y = x^{2} + 3

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\frac{d}{d x} (x sin (\frac{π}{x}))

f(x)=tan(ln(x))

f (x) = tan (ln (x))

Frequently Asked Questions (FAQ)

What is inverse oflaplace 1/((s^2-2s+5)) ?
The answer to inverse oflaplace 1/((s^2-2s+5)) is e^t 1/2 sin(2t)