What is inverse oflaplace 1/((s^2+1)(s^2+25)) ?

The answer to inverse oflaplace 1/((s^2+1)(s^2+25)) is 1/24 sin(t)-1/120 sin(5t)

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inverse oflaplace 1/((s^2+1)(s^2+25))

inverse laplace

\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 25 )}

\frac{1}{24} sin (t) - \frac{1}{120} sin (5 t)

Solution steps

L^{- 1} {\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 25 )}}

Take the partial fraction of

\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 25 )} : \frac{1}{24 ( s ^{2} + 1 )} - \frac{1}{24 ( s ^{2} + 25 )}

= L^{- 1} {\frac{1}{24 ( s ^{2} + 1 )} - \frac{1}{24 ( s ^{2} + 25 )}}

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}{24 ( s ^{2} + 1 )}} - L^{- 1} {\frac{1}{24 ( s ^{2} + 25 )}}

L^{- 1} {\frac{1}{24 ( s ^{2} + 1 )}} : \frac{1}{24} sin (t)

L^{- 1} {\frac{1}{24 ( s ^{2} + 25 )}} : \frac{1}{120} sin (5 t)

= \frac{1}{24} sin (t) - \frac{1}{120} sin (5 t)

y^{^{'}} = \frac{( 5 x ^{3} + y ^{3} )}{x y ^{2}}

f (x) = 6 x + 1

\frac{d}{d x} (2 3^{x})

r (x) = sin (x^{2}) + \frac{1 - x}{1 + x}

3 arctan (x - 1 + x^{2})

What is inverse oflaplace 1/((s^2+1)(s^2+25)) ?
The answer to inverse oflaplace 1/((s^2+1)(s^2+25)) is 1/24 sin(t)-1/120 sin(5t)