What is inverse oflaplace (s^2+3)/(2s^2+5) ?

The answer to inverse oflaplace (s^2+3)/(2s^2+5) is 1/2 \delta(t)+1/(2sqrt(10))sin(sqrt(5/2)t)

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inverse oflaplace (s^2+3)/(2s^2+5)

inverse laplace

\frac{s ^{2} + 3}{2 s ^{2} + 5}

\frac{1}{2} δ (t) + \frac{1}{2 10} sin (\frac{5}{2} t)

Solution steps

L^{- 1} {\frac{s ^{2} + 3}{2 s ^{2} + 5}}

Take the partial fraction of

\frac{s ^{2} + 3}{2 s ^{2} + 5} : \frac{1}{2} + \frac{1}{2 ( 2 s ^{2} + 5 )}

= L^{- 1} {\frac{1}{2} + \frac{1}{2 ( 2 s ^{2} + 5 )}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{2}} + L^{- 1} {\frac{1}{2 ( 2 s ^{2} + 5 )}}

L^{- 1} {\frac{1}{2}} : \frac{1}{2} δ (t)

L^{- 1} {\frac{1}{2 ( 2 s ^{2} + 5 )}} : \frac{1}{2 10} sin (\frac{5}{2} t)

= \frac{1}{2} δ (t) + \frac{1}{2 10} sin (\frac{5}{2} t)

\frac{d}{d x} ((\frac{3}{4})^{\frac{- 4}{3}})

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2} + 1})

\frac{\partial}{\partial x} (x^{0.5})

x \to 0 lim (\frac{( 2 + x ) ^{2}}{x})

What is inverse oflaplace (s^2+3)/(2s^2+5) ?
The answer to inverse oflaplace (s^2+3)/(2s^2+5) is 1/2 \delta(t)+1/(2sqrt(10))sin(sqrt(5/2)t)