What is inverse oflaplace (10)/(s^2+2s+5) ?

The answer to inverse oflaplace (10)/(s^2+2s+5) is 5e^{-t}sin(2t)

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inverse oflaplace (10)/(s^2+2s+5)

inverse laplace

\frac{10}{s ^{2} + 2 s + 5}

5 e^{- t} sin (2 t)

Solution steps

L^{- 1} {\frac{10}{s ^{2} + 2 s + 5}}

Expand

\frac{10}{s ^{2} + 2 s + 5} : 10 \cdot \frac{1}{( s + 1 ) ^{2} + 4}

= L^{- 1} {10 \cdot \frac{1}{( s + 1 ) ^{2} + 4}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= 10 L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 4}}

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

= 10 e^{- t} \frac{1}{2} sin (2 t)

Refine

10 e^{- t} \frac{1}{2} sin (2 t) : 5 e^{- t} sin (2 t)

= 5 e^{- t} sin (2 t)

\int \frac{1}{x ^{2} - 441} d x

\int - 16 sin (4 x - \frac{π}{2}) d x

n = 1 \sum \infty \frac{1}{n} - \frac{1}{n + 2}

\int \frac{21 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

y^{^{'}} = 3 y, y (2) = 4

What is inverse oflaplace (10)/(s^2+2s+5) ?
The answer to inverse oflaplace (10)/(s^2+2s+5) is 5e^{-t}sin(2t)