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

The answer to inverse oflaplace (10)/(s^2+s+10) is (20}{sqrt(39)}e^{-t/2}sin(\frac{sqrt(39)t)/2)

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

inverse laplace

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

\frac{20}{39} e^{- \frac{t}{2}} sin (\frac{39 t}{2})

Solution steps

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

Expand

\frac{10}{s ^{2} + s + 10} : 10 \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}}

= L^{- 1} {10 \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{39}{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 + \frac{1}{2} ) ^{2} + \frac{39}{4}}}

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}}} : e^{- \frac{t}{2}} \frac{2}{39} sin (\frac{39 t}{2})

= 10 e^{- \frac{t}{2}} \frac{2}{39} sin (\frac{39 t}{2})

Refine

10 e^{- \frac{t}{2}} \frac{2}{39} sin (\frac{39 t}{2}) : \frac{20}{39} e^{- \frac{t}{2}} sin (\frac{39 t}{2})

= \frac{20}{39} e^{- \frac{t}{2}} sin (\frac{39 t}{2})

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

f (x) = \frac{3}{x + 1}, at x = 7

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 60 cos (8 t)

\frac{\partial}{\partial x} (e^{x + y})

n = 0 \sum \infty (x)^{n}

What is inverse oflaplace (10)/(s^2+s+10) ?
The answer to inverse oflaplace (10)/(s^2+s+10) is (20}{sqrt(39)}e^{-t/2}sin(\frac{sqrt(39)t)/2)