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

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

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

inverse laplace

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

\frac{10}{3} e^{- t} sin (3 t)

Solution steps

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

Expand

\frac{10}{s ^{2} + 2 s + 10} : 10 \cdot \frac{1}{( s + 1 ) ^{2} + 9}

= L^{- 1} {10 \cdot \frac{1}{( s + 1 ) ^{2} + 9}}

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} + 9}}

L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 9}} : e^{- t} \frac{1}{3} sin (3 t)

= 10 e^{- t} \frac{1}{3} sin (3 t)

Refine

10 e^{- t} \frac{1}{3} sin (3 t) : \frac{10}{3} e^{- t} sin (3 t)

= \frac{10}{3} e^{- t} sin (3 t)

\int \frac{x ^{2}}{x ^{3} - 1} d x

\frac{d}{d x} (ln (x + 7))

\frac{d}{d x} (3^{t a n (x)})

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

\int \frac{1 - 2 x}{x ( 1 - x )} d x

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