What is inverse oflaplace (5+5s)/((s^2+25)^2) ?

The answer to inverse oflaplace (5+5s)/((s^2+25)^2) is 1/50 (sin(5t)-5tcos(5t))+(tsin(5t))/2

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inverse oflaplace (5+5s)/((s^2+25)^2)

inverse laplace

\frac{5 + 5 s}{( s ^{2} + 25 ) ^{2}}

\frac{1}{50} (sin (5 t) - 5 t cos (5 t)) + \frac{t sin ( 5 t )}{2}

Solution steps

L^{- 1} {\frac{5 + 5 s}{( s ^{2} + 25 ) ^{2}}}

Expand

= L^{- 1} {\frac{5}{( s ^{2} + 25 ) ^{2}} + \frac{5 s}{( s ^{2} + 25 ) ^{2}}}

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)}

= 5 L^{- 1} {\frac{1}{( s ^{2} + 25 ) ^{2}}} + 5 L^{- 1} {\frac{s}{( s ^{2} + 25 ) ^{2}}}

L^{- 1} {\frac{1}{( s ^{2} + 25 ) ^{2}}} : \frac{1}{250} (sin (5 t) - 5 t cos (5 t))

L^{- 1} {\frac{s}{( s ^{2} + 25 ) ^{2}}} : \frac{t sin ( 5 t )}{10}

= 5 \cdot \frac{1}{250} (sin (5 t) - 5 t cos (5 t)) + 5 \cdot \frac{t sin ( 5 t )}{10}

Refine

5 \frac{1}{250} (sin (5 t) - 5 t cos (5 t)) + 5 \frac{t sin ( 5 t )}{10} : \frac{1}{50} (sin (5 t) - 5 t cos (5 t)) + \frac{t sin ( 5 t )}{2}

= \frac{1}{50} (sin (5 t) - 5 t cos (5 t)) + \frac{t sin ( 5 t )}{2}

f (x) = 4 x - x^{2}, at x = 1

x \to - 2 - lim (\frac{x ^{2} - 4}{x + 2})

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

x \to 0 lim (\frac{π}{x})

f (x) = x x^{2} + 21

What is inverse oflaplace (5+5s)/((s^2+25)^2) ?
The answer to inverse oflaplace (5+5s)/((s^2+25)^2) is 1/50 (sin(5t)-5tcos(5t))+(tsin(5t))/2