What is inverse oflaplace 4/(s^2+9) ?

The answer to inverse oflaplace 4/(s^2+9) is 4/3 sin(3t)

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inverse oflaplace 4/(s^2+9)

inverse laplace

\frac{4}{s ^{2} + 9}

\frac{4}{3} sin (3 t)

Solution steps

L^{- 1} {\frac{4}{s ^{2} + 9}}

= L^{- 1} {\frac{4}{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{4}{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{4}{3} sin (3 t)

\frac{d y}{d x} = 1 - 2 y, y (0) = 0

\int 11 sin^{5} (x) d x

\frac{d y}{d x} = (\frac{( 2 y + 7 )}{( 4 x + 9 )})^{2}

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

y = 3 e^{- x} + e^{4 x}

What is inverse oflaplace 4/(s^2+9) ?
The answer to inverse oflaplace 4/(s^2+9) is 4/3 sin(3t)