What is inverse oflaplace (2.98)/(9+s^2) ?

The answer to inverse oflaplace (2.98)/(9+s^2) is (2.98)/3 sin(3t)

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inverse oflaplace (2.98)/(9+s^2)

inverse laplace

\frac{2.98}{9 + s ^{2}}

\frac{2.98}{3} sin (3 t)

Solution steps

L^{- 1} {\frac{2.98}{9 + s ^{2}}}

= L^{- 1} {\frac{2.98}{3} \cdot \frac{3}{s ^{2} + 3 ^{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)}

= \frac{2.98}{3} L^{- 1} {\frac{3}{s ^{2} + 3 ^{2}}}

Use Inverse Laplace Transform table:

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

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

= \frac{2.98}{3} sin (3 t)

\int \frac{28}{4 x + x x} d x

\int_{0}^{2} y^{2} 1 + y^{3} d y

x \to - \frac{1}{2} lim (\frac{6 x - 1}{x})

\int x^{2} 2 x - x^{2} d x

f (x) = x^{3} - 5 x^{2}, at x = 2

What is inverse oflaplace (2.98)/(9+s^2) ?
The answer to inverse oflaplace (2.98)/(9+s^2) is (2.98)/3 sin(3t)