What is inverse oflaplace (e^{(-pi)/2})/(s^2+9) ?

The answer to inverse oflaplace (e^{(-pi)/2})/(s^2+9) is 1/(3e^{pi/2)}sin(3t)

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inverse oflaplace (e^{(-pi)/2})/(s^2+9)

inverse laplace

\frac{e ^{\frac{- π}{2}}}{s ^{2} + 9}

\frac{1}{3 e ^{\frac{π}{2}}} sin (3 t)

Solution steps

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

= L^{- 1} {\frac{1}{3 e ^{\frac{π}{2}}} \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{1}{3 e ^{\frac{π}{2}}} 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{1}{3 e ^{\frac{π}{2}}} sin (3 t)

\frac{\partial}{\partial x} (x^{3} + x^{2} y - 4 y^{3})

x (\frac{d y}{d x}) + y = 3 x y^{2}

\int 7 tan^{3} (x) sec (x) d x

\frac{\partial}{\partial y} (1 + x^{3} + y^{4})

What is inverse oflaplace (e^{(-pi)/2})/(s^2+9) ?
The answer to inverse oflaplace (e^{(-pi)/2})/(s^2+9) is 1/(3e^{pi/2)}sin(3t)