What is inverse oflaplace (20)/(s^2+10) ?

The answer to inverse oflaplace (20)/(s^2+10) is 2sqrt(10)sin(sqrt(10)t)

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inverse oflaplace (20)/(s^2+10)

inverse laplace

\frac{20}{s ^{2} + 10}

210 sin (10 t)

Solution steps

L^{- 1} {\frac{20}{s ^{2} + 10}}

= L^{- 1} {210 \frac{10}{s ^{2} + ( 10 ) ^{2}}}

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)}

= 210 L^{- 1} {\frac{10}{s ^{2} + ( 10 ) ^{2}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{a}{s ^{2} + a ^{2}}} = sin (a t)

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

= 210 sin (10 t)

\int_{0}^{1} \frac{1}{1 + t ^{2}} d t

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x \to 4 lim (\frac{2}{x - 4})

\frac{\partial}{\partial x} (tan (6 + 2 x^{2} y^{4} z^{2}))

\frac{d}{d x} ((x^{3} + 1) (2 x^{2} + 8 x - 5))

What is inverse oflaplace (20)/(s^2+10) ?
The answer to inverse oflaplace (20)/(s^2+10) is 2sqrt(10)sin(sqrt(10)t)