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

The answer to inverse oflaplace s/((s^2+25)^2) is (tsin(5t))/(10)

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

inverse laplace

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

\frac{t sin ( 5 t )}{10}

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

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

= \frac{d}{d t} (L^{- 1} {\frac{1}{( s ^{2} + 25 ) ^{2}}}) + L^{- 1} {\frac{1}{( s ^{2} + 25 ) ^{2}}} (0)

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

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

Evaluate

\frac{1}{250} (sin (5 t) - 5 t cos (5 t))

at point

0 = 0

= \frac{t sin ( 5 t )}{10}

\int 3 x (5 - x) d x

x \to 0 lim (\frac{x}{2 tan ( x )})

\frac{d}{d x} (\frac{x ^{3}}{6})

\int_{0}^{1} \frac{1}{5 x - 3} d x

2 y^{^{''}} + y^{^{'}} - y = x + 1

What is inverse oflaplace s/((s^2+25)^2) ?
The answer to inverse oflaplace s/((s^2+25)^2) is (tsin(5t))/(10)